RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

Other Exercises

Mark the correct alternative in each of the following :
Question 1.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q such that OQ = 12 cm. Length PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) √119 cm
Solution:
(d) Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 1
OP ⊥ PQ
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (12)² = (5)2 + PQ²
=> 144 = 25 + PQ²
PQ² = 144 – 25 = 119
PQ = √119

Question 2.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:
(a) Let PQ be the tangent from Q to the circle with O as centre
PQ = 24 cm
OQ = 25 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 2
Let Radius OQ = r
OQ ⊥ PQ
Now in right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = r² + (24)²
=> 625 = r² + 576
=> r² = 625 – 576 = 49 = (7)²
r = 7
Radius of the circle = 7 cm

Question 3.
The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is
(a) √7 cm
(b) 7 cm
(c) 5 cm
(d) 25 cm
Solution:
(c) Let AB be the tangent from A to the circle of centre O, then
OB = 3 cm
BA = 4 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 3
OB ⊥ BA
In right ∆OBA,
OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²
OA = 5
Distance of A from the centre O = 5 cm

Question 4.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80° then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
(a) PA and PB are the tangents to the circle from P and ∠APB = 80°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 4
∠AOB = 180° – ∠APB = 180°- 80° = 100°
But OP is the bisector of ∠AOB
∠POA = ∠POB = \(\frac { 1 }{ 2 }\) ∠AOB
=> ∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°

Question 5.
If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:
(b) TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 5
But ∠POQ + ∠PTQ = 180°
=> 110° + ∠PTQ = 180°
=> ∠PTQ = 180° – 110° = 70°

Question 6.
PQ is a tangent to a circle with centre O at the point P. If ∆OPQ is an isosceles triangle, then ∠OQP is equal to
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(b) In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 6
OP is radius of the circle
OP ⊥ PQ
OP = PQ
∠POQ = ∠OQP
But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)
∠OQP = ∠POQ = 45°

Question 7.
Two equal circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(d) Two circles with centres O and O’ touch each other at C externally
A common tangent is drawn which touches the circles at A and B respectively.
Join OA, O’B and O’O which passes through C
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 7
AO = BO’ (radii of the equal circle)
AB || OO’
=> AOO’B is a rectangle
Draw another common tangent through C which intersects AB at D, then DA = DC = DB
ADCO and BDCO’ are squares
AC and BC are the diagonals of equal square
AC = BC
∠DAC = ∠DBC = 45°
∠ACB = 90°

Question 8.
ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in ∆ABC. The radius of the circle is
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
Solution:
(b) In a right ∆ABC, ∠B = 90°
BC = 6 cm, AB = 8 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 8
AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²
AC = 10 cm
An incircle is drawn with centre 0 which touches the sides of the triangle ABC at P, Q and R
OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle
OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA
OPBQ is a square
Let r be the radius of the incircle
PB = BQ = r
AR = AP = 8 – r,
CQ = CR = 6 – r
AC = AR + CR
=> 10 = 8 – r + 6 – r
10 = 14 – 2r
=> 2r = 14 – 10 = 4
=> r = 2
Radius of the incircle = 2 cm

Question 9.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Solution:
(c) PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 9
OQ is radius and PQ is tangent to the circle
OQ ⊥ QP or ∠OQP = 90°
But ∠QOP + ∠POR = 180° (Linear pair)
=> ∠QOP + 120° = 180°
∠QOP = 180° – 120° = 60°
Now in ∆POQ
∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)
=> 60° + 90° + ∠OPQ = 180°
=> 150° + ∠OPQ = 180°
=> ∠OPQ = 180° – 150° = 30°

Question 10.
If four sides of a quadrilateral ABCD are tangential to a circle, then
(a) AC + AD = BD + CD
(b) AB + CD = BC + AD
(c) AB + CD = AC + BC
(d) AC + AD = BC + DB
Solution:
(b) A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides
AB + CD = BC + AD

Question 11.
The length of the tangent drawn from a point 8 cm away from the centre of a circle of radius 6 cm is
(a) √7 cm
(b) 2√7cm
(c) 10 cm
(d) 5 cm
Solution:
(b) Radius of the circle = 6 cm
and distance of the external point from the centre = 8 cm
Length of tangent = √{(8)² – (6)²}
= √(64 – 36) = √28
= √(4 x 7) = 2√7 cm

Question 12.
AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm
Solution:
(c) AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 10
CD = 4 cm
DA and DC are tangents to the first circle from D
CD = AD = 4 cm
Similarly DC and DB are tangents to the second circle from D
CD = DB = 4 cm
AB = AD + DB = 4 + 4 = 8 cm

Question 13.
In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then,
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 11
(a) AD = AB + BC + CA
(b) 2AD = AB + BC + CA
(c) 3AD = AB + BC + CA
(d) 4AD = AB + BC + CA
Solution:
(b) AD, AE and BC are the tangents to the circle at D, E and F respectively
D and AE are tangents to the circle from A
AD = AE ……(i)
Similarly, CD = CF and BE = BF ….(ii)
Now AB + AC + BC = AE – BE + AD – CD + CF + BF
= AD – BE + AD – CD + BE + BE
= 2AD [From (i) and (ii)]
or 2 AD = AB + BC + CA

Question 14.
In the figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 12
(a) 8 cm
(b) 3 cm
(c) 2.5 cm
(d) 5 cm
Solution:
(d) In the figure, 0 is the centre of the circle
QR is tangent to the circle and QOS is a diameter SQ = 6 cm, QR = 4 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 13
OQ = \(\frac { 1 }{ 2 }\) QS = \(\frac { 1 }{ 2 }\) x 6 = 3 cm
OQ is radius
OQ ⊥ QR
Now in right ∆OQR
OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OR = 5 cm

Question 15.
In the figure, the perimeter of ∆ABC is
(a) 30 cm
(b) 60 cm
(c) 45 cm
(d) 15 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 14
Solution:
(a) ∆ABC is circumscribed of circle with centre O
AQ = 4 cm, CP = 5 cm and BR = 6 cm
AQ and AR the tangents to the circle AQ = AR = 4 cm
Similarly BP and BR are tangents,
BP = BR = 6 cm
and CP and CQ are the tangents
CQ = CP = 5 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 15
AB = AR + BR = 4 + 6 = 10 cm
BC = BP + CP = 6 + 5 = 11 cm
AC = AQ + CQ = 4 + 5 = 9 cm
Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

Question 16.
In the figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then, AP =
(a) 2√2 cm
(b) 2 cm
(c) 2√3 cm
(d) 3√2 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 16
Solution:
(c) In the figure, AP is the tangent to the circle with centre O such that
OP = 4 cm, ∠OPA = 30°
Join OA, let AP = x
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 17

Question 17.
AP and AQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =
(a) 12 cm
(b) 18 cm
(c) 24 cm
(d) 36 cm
Solution:
(c) OP is radius, PA is the tangent
OP ⊥ AP
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 18
Now in right ∆OAP,
OA² = OP² + AP²
(15)² = (9)² + AP²
225 = 81 + AP²
=> AP² = 225 – 81 = 144 = (12)²
AP = 12 cm
But AP = AQ = 12 cm (tangents from A to the circle)
AP + AQ = 12+ 12 = 24 cm

Question 18.
At one end of a diameter PQ of a circle of radius 5 cm, tangent XPY is drawn to the circle. The length of chord AB parallel to XY and at a distance of 8 cm from P is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm
Solution:
(d) In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm
From P, at a distance of 8 cm AB is a chord drawn parallel to XY
To find the length of AB
Join OA
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 19
XY is tangent and OP is the radius
OP ⊥ XY or PQ ⊥ XY
AB || XY
OQ is ⊥ AB which meets AB at R
Now in right ∆OAR,
OA² = OR² + AR²
(5)² = (3)² + AR²
25 = 9 + AR²
=> AR² = 25 – 9 = 16 = (4)²
AR = 4 cm
But R is mid-point of AB
AB = 2 AR = 2 x 4 = 8 cm

Question 19.
If PT is tahgent drawn froth a point P to a circle touching it at T and O is the centre of the circle, then ∠OPT + ∠POT =
(a) 30°
(b) 60°
(c) 90°
(d) 180°
Solution:
(c) In the figure, PT is the tangent to the circle with centre O.
OP and OT are joined
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 20
PT is tangent and OT is the radius
OT ⊥ PT
Now in right ∆OPT
∠OTP = 90°
∠OPT + ∠POT = 180° – 90° = 90°

Question 20.
In the adjacent figure, if AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =
(a) 5 cm
(b) 4 cm
(c) 6 cm
(d) 7 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 21
Solution:
(d) In the figure, ∆ABC is the circumscribed a circle
AB = 12 cm, BC = 8 cm and AC = 10 cm
Let AD = a, DB = b and EC = c, then
AF = a, BE = b and FC = c
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 22
But AB + BC + AC = 12 + 8 + 10 = 30
a + b + b + c + c + a = 30
=> 2 (a + b + c) = 30
a + b + c = 15
Subtracting BC or b + c from this a = 15 – 8 = 7
AD = 7 cm

Question 21.
In the figure, if AP = PB, then
(a) AC = AB
(b) AC = BC
(c) AQ = QC
(d) AB = BC
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 23
Solution:
(b) In the figure, AP = PB
But AP and AQ are the tangent from A to the circle
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 24
AP = AQ
Similarly PB = BR
But AP = PB (given)
AQ = BR ….(i)
But CQ and CR the tangents drawn from C to the circle
CQ = CR
Adding in (i)
AQ + CQ = BR + CR
AC = BC

Question 22.
In the figure, if AP = 10 cm, then BP =
(a) √91 cm
(b) √127 cm
(c) √119 cm
(d) √109 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 25
Solution:
(b) In the figure,
OA = 6 cm, OB = 3 cm and AP = 10 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 26
OA is radius and AP is the tangent
OA ⊥ AP
Now in right ∆OAP
OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136
Similarly BP is tangent and OB is radius
OP² = OB² + BP²
136 = (3)² + BP2
=> 136 = 9 + BP²
=> BP² = 136 – 9 = 127
BP = √127 cm

Question 23.
In the figure, if PR is tangent to the circle at P and Q is the centre of the circle, then ∠POQ =
(a) 110°
(b) 100°
(c) 120°
(d) 90°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 27
Solution:
(c) In the figure, PR is the tangent to the circle at P.
O is the centre of the circle ∠QPR = 60°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 28
OP is the radius and PR is the tangent OPR = 90°
=> ∠OPQ + ∠QPR = 90°
=> ∠OPQ + 60° = 90°
=> ∠OPQ = 90° – 60° = 30°
OP = OQ (radii of the circle)
∠OQP = 30°
In ∆OPQ,
∠OPQ + ∠OQP + ∠POQ = 180°
=> 30° + 30° + ∠PQR = 180°
=> 60° + ∠POQ = 180°
∠POQ = 180° – 60° = 120°

Question 24.
In the figure, if quadrilateral PQRS circumscribes a circle, then PD + QB =
(a) PQ
(b) QR
(c) PR
(d) PS
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 29
Solution:
(a) In the figure, quadrilateral PQRS is circumscribed a circle
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 30
PD = PA (tangents from P to the circle)
Similarly QA = QB
PD + QB = PA + QA = PQ

Question 25.
In the figure, two equal circles touch each other at T, if QP = 4.5 cm, then QR =
(a) 9 cm
(b) 18 cm
(c) 15 cm
(d) 13.5 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 31
Solution:
(a) In the figure, two equal circles touch, each other externally at T
QR is the common tangent
QP = 4.5 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 32
PQ = PT (tangents from P to the circle)
Similarly PT = PR
PQ = PT = PR
Now QR = PQ + PR = 4.5 + 4.5 = 9 cm

Question 26.
In the figure, APB is a tangent to a circle with centre O at point P. If ∠QPB = 50°, then the measure of ∠POQ is
(a) 100°
(b) 120°
(c) 140°
(d) 150°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 33
Solution:
(a) In the figure, APB is a tangent to the circle with centre O
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 34
∠QPB = 50°
OP is radius and APB is a tangent
OP ⊥ AB
=> ∠OPB = 90°
=> ∠OPQ + ∠QPB = 90°
∠OPQ + 50° = 90°
=> ∠OPQ = 90° – 50° = 40°
But OP = OQ
∠OPQ = OQP = 40°
∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

Question 27.
In the figure, if tangents PA and PB are drawn to a circle such that ∠APB = 30° and chord AC is drawn parallel to the tangent PB, then ∠ABC =
(a) 60°
(b) 90°
(c) 30°
(d) None of these
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 35
Solution:
(c) In the figure, PA and PB are the tangents to the circle with centre O
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 36
∠APB = 30°
Chord AC || BP,
AB is joined
PA = PB
∠PAB = ∠PBA
But ∠PAB + ∠PBA = 180° – 30° = 150°
=> ∠BPA + ∠PBA = 150°
=> 2 ∠PBA = 150°
=> ∠PBA = 75°
AC || BC
∠BAC = ∠PBA = 75°
But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)
∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

Question 28.
In the figure, PR =
(a) 20 cm
(b) 26 cm
(c) 24 cm
(d) 28 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 37
Solution:
(b) In the figure, two circles with centre O and O’ touch each other externally
PQ and RS are the tangents drawn to the circles
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 38
OQ and O’S are the radii of these circles and
OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm
Now in right ∆OQP
OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²
OP = 5 cm
Similarly in right ∆RSO’
(O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)²
O’R = 13 cm
Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm

Question 29.
Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. Then AO : AO’ =
(a) \(\frac { 3 }{ 2 }\)
(b) 2
(c) 3
(d) \(\frac { 1 }{ 4 }\)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 39
Solution:
(c) Two circles of equal radii touch each other externally at P. OO’ produced meets at A
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 40
From A, AT is the tangent to the circle (O, r)
O’Q ⊥ AT
Now AO : AO’ = 3r : r
= 3 : 1 = \(\frac { 3 }{ 1 }\)

Question 30.
Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord BC which touches the inner circle at P is equal to
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 41
Solution:
(c) In the figure, two concentric circles of radii 3 cm and 5 cm with centre O
Chord BC touches the inner circle at P
Draw a tangent AB to the inner circle
Join OQ and OA
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 42
OQ is radius and AQB is the tangent
OQ ⊥ AB and OQ bisects AB
AQ = QB
Similarly, BP = PC or P is mid-point of BC
But BQ and BP are tangents from B
QB = BP = AQ
In right ∆OAQ,
OA² = AQ² + OQ²
(5)² = AQ² + (3)²
=> AQ² = (5)² – (3)²
=> AQ² = 25 – 9 = 16 = (4)²
AQ = 4 cm
BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

Question 31.
In the figure, there are two concentric, circles with centre O. PR and PQS are tangents to the inner circle from point plying on the outer circle. If PR = 7.5 cm, then PS is equal to
(a) 10 cm
(b) 12 cm
(c) 15 cm
(d) 18 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 43
Solution:
(c) In the figure, two concentric circles with centre O
From a point P on the outer circle,
PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively
PR = 7.5 cm
Join OR and OQ
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 44
PT is chord and OR is radius
R is mid-point of PT
Similarly Q is mid-point of PS
But PR = PQ (tangents from P)
PT = 2 PR and PS = 2 PQ
PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

Question 32.
In the figure, if AB = 8 cm and PE = 3 cm, then AE =
(a) 11 cm
(b) 7 cm
(c) 5 cm
(d) 3 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 45
Solution:
(c) In the figure, AB and AC are the tangents to the circle from A
DE is another tangent drawn from P
AB = 8 cm, PE = 3 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 46
AB = AC (tangents drawn from A to the circle)
Similarly PE = EC and DP = DB
Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

Question 33.
In the figure, PQ and PR are tangents drawn from P to a circle with centre O. If ∠OPQ = 35°, then
(a) a = 30°, b = 60°
(b) a = 35°, b = 55°
(c) a = 40°, b = 50°
(d) a = 45°, b = 45°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 47
Solution:
(b) In the figure, PQ and PR are the tangents drawn from P to the circle with centre O
∠OPQ = 35°
PO is joined
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 48
PQ = PR (tangents from P to the circle)
∠OPQ = ∠OPR
=> 35° = a
=> a = 35°
OQ is radius and PQ is tangent
OQ ⊥ PQ
=> ∠OQP = 90°
In ∆OQP,
∠POQ + ∠QPO = 90°
=> b + 35° = 90°
=> b = 90° – 35° = 55°
a = 35°, b = 55°

Question 34.
In the figure, if TP and TQ are tangents drawn from an external point T to a circle with centre O such that ∠TQP = 60°, then
(a) 25°
(b) 30°
(c) 40°
(d) 60°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 49
Solution:
(b) In the figure, TP and TQ are the tangents drawn from T to the circle with centre O
OP, OQ and PQ are joined
∠TQP = 60°
TP = TQ (Tangents from T to the circle)
∠TQP = ∠TPQ = 60°
∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°
and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°
But OP = OQ (radii of the same circle)
∠OPQ = ∠OQP
But ∠OPQ + ∠OQP = 180° – 120° = 60°
But ∠OPQ = 30°

Question 35.
In the figure, the sides AB, BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is [CBSE 2012]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 50
(a) 11 cm
(b) 10 cm
(c) 14 cm
(d) 15 cm
Solution:
(b) In the figure,
PA = 4 cm, BP = 3 cm, AC = 11 cm
AP and AR are the tangents from A to the circle
AP = AR
=> AR = 4 cm
Similarly BP and BQ are tangents
BQ = BP = 3 cm
AC =11 cm
AR + CR = 11 cm
4 + CR =11 cm
CR = 11 – 4 = 7 cm
CQ and CR are tangents to the circle
CQ = CR = 7 cm
Now, BC = BQ + CQ = 3 + 7 = 10 cm

Question 36.
In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is [CBSE 2012]
(a) 18 cm
(b) 13.5 cm
(c) 12 cm
(d) 9 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 51
Solution:
(a) In ∆DEF
DF touches the circle at H
and circle touches ED and EF Produced at K and M respectively
EK = 9 cm
EK and EM are the tangents to the circle
EM = EK = 9 cm
Similarly DH and DK are the tangent
DH = DK and FH and FM are tangents
FH = FM
Now, perimeter of ∆DEF
= ED + DF + EF
= ED + DH + FH + EF
= ED + DK + EM + EF
= EK + EM
= 9 + 9
= 18 cm

Question 37.
In the figure DE and DF are tangents from an external point D to a circle with centre A. If DE = 5 cm and DE ⊥ DF, then the radius of the circle is [CBSE 2013]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 52
(a) 3 cm
(b) 5 cm
(c) 4 cm
(d) 6 cm
Solution:
(b) If figure, DE and DF are tangents to the circle drawn from D.
A is the centre of the circle.
DE = 5 cm and DE ⊥ DF
Join AE, AF
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 53
DE is the tangent and AE is radius
AE ⊥ DE
Similarly, AF ⊥ DF
But ∠D = 90° (given)
AFDE is a square
AE = DE (side of square)
But DE = 5 cm
AE = 5 cm
Radius of circle is 5 cm

Question 38.
In the figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is [CBSE 2013]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 54
(a) 11
(b) 18
(c) 6
(d) 15
Solution:
(a) In the figure, a circle touches the sides of a quadrilateral ABCD
∠B = 90°, OP = OQ = r
AB = 29 cm, AD = 23 cm, DS = 5 cm
∠B = 90°
BA is tangent and OQ is radius
∠OQB = 90°
Similarly OP is radius and BC is tangents
∠OPB = 90°
But ∠B = 90° (given)
PBQO is a square
DS = 5 cm
But DS and DR are tangents to the circles
DR = 5 cm
But AD = 23 cm
AR = 23 – 5= 18 cm
AR = AQ (tangents to the circle from A)
AQ = 18 cm
But AB = 29 cm
BQ = 29 – 18 = 11 cm
OPBQ is a square
OQ = BQ = 11 cm
Radius of the circle = 11 cm

Question 39.
In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
(a) 4
(b) 3
(c) 2
(d) 1
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 55

Question 40.
Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of ∠APB is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
(d) We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)
∠TAP = ∠TPA = x (say)
and ∠TBP = ∠TPB = y (say)
Also, in triangle APB,
x + x + x + y + y = 180°
=> 2x + 2y = 180°
=> x + y = 90°
=> ∠APB = 90°

Question 41.
In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals
(a) 67°
(b) 134°
(c) 44°
(d) 46°
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 56
Solution:
(b) ∠OQP = 90°
[Tangent is ⊥ to the radius through the point of contact]
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 42.
In the figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is [CBSE2014]
(a) 3.8
(b) 7.6
(c) 5.7
(d) 1.9
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 57
Solution:
(b) In the figure, QR is common tangent to the two circles touching each other externally at T
Tangent at T meets QR at P
PT = 3.8 cm
PT and PQ are tangents from P
PT = PQ = 3.8 cm
Similarly PT and PR are tangents
PT = PR = 3.8 cm
QR = 3.8 + 3.8 = 7.6 cm

Question 43.
In the figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = x cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then x =
(a) 10
(b) 9
(c) 8
(d) 7 (CBSE 2014)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 58
Solution:
(b) In the given figure,
ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively
AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm
CR and CQ are tangents to the circle from C
CR = CQ = 3 cm
BQ = BC – CQ = 7 – 3 = 4 cm
BQ = and BP are tangents from B
BP = BQ = 4 cm
AS and AP are tangents from A
AP = AS = 5 cm
AB = AP + BP = 5 + 4 = 9 cm
x = 9 cm

Question 44.
If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is (NCERT Exemplar)
(a) 90°
(b) 50°
(c) 70°
(d) 40°
Solution:
(b) O is the centre of the circle.
Given, ∠POQ = 130°
PT and QT are tangents drawn from external point T to the circle.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 59
∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]
In quadrilateral OPTQ,
∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°
=> ∠PTQ + 90° + 90° + 130° = 360°
=> ∠PTQ = 360° – 310° = 50°
Thus, the angle between the tangents is 50°.

Question 45.
If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to [NCERT Exemplar]
(a) \(\frac { 3\surd 3 }{ 2 }\) cm
(b) 6 cm
(c) 3 cm
(d) 3√3 cm
Solution:
(d) Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.
Join OA and OP.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 60
Also, OP is a bisector of line ∠APC
∠APO = ∠CPO = 30°
Also, OA ⊥ AP
Tangent at any point of a circle is perpendicular to the radius through the point of contact.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 61
Hence, the length of each tangent is 3√3 cm

Question 46.
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is [NCERT Exemplar]
(a) 3 cm
(b) 6 cm
(c) 9 cm
(d) 1 cm
Solution:
(b) Let O be the centre of two concentric circles C1 and C2, whose radii are r1 = 4 cm and r2 = 5 cm.
Now, we draw a chord AC of circle C2, which touches the circle C1 at B.
Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 62
Now, in right angled ∆OBC, by using Pythagoras theorem,
OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]
=> 5² = BC² + 4²
=> BC² = 25 – 16 = 9
=> BC = 3 cm
Length of chord AC = 2 BC = 2 x 3 = 6 cm

Question 47.
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is [NCERT Exemplar]
(a) 4 cm
(b) 5 cm
(b) 6 cm
(d) 8 cm
Solution:
(d) First, draw a circle of radius 5 cm having centre O.
A tangent XY is drawn at point A.
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 63
A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.
Now, ∠OAY = 90°
[Tangent and any point of a circle is perpendicular to the radius through the point of contact]
∠OAY + ∠OED = 180°
[sum of cointerior is 180°]
=> ∆OED = 180°
Also, AE = 8 cm, Join OC
Now, in right angled ∆OBC
OC² = OE² + EC²
=> EC² = OC² – OE² [by Pythagoras theorem]
EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]
EC² = 25 – 9 = 16
=> EC = 4 cm
Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm
[Since, perpendicular from centre to the chord bisects the chord]

Question 48.
From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangent PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is [NCERT Exemplar]
(a) 60 cm²
(b) 65 cm²
(c) 30 cm²
(d) 32.5 cm²
Solution:
(a) Firstly, draw a circle of radius 5 cm having centre O.
P is a point at a distance of 13 cm from O.
A pair of tangents PQ and PR are drawn.
Thus, quadrilateral PQOR is formed.
OQ ⊥ QP [since, AP is a tangent line]
In right angled ∆PQO,
OP² = OQ² + QP²
=> 13² = 5² + QP²
=> QP² = 169 – 25 = 144 = 12²
=> QP = 12 cm
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 64
Now, area of ∆OQP = \(\frac { 1 }{ 2 }\) x QP x QO = \(\frac { 1 }{ 2 }\) x 12 x 5 = 30 cm²
Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

Question 49.
If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to
(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution:
(a) Given, PA and PB are tangent lines.
PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 65
In ∆PAB,
∠P + ∠A + ∠B = 180°
[since, sum of angles of a triangle = 180°
50°+ θ + θ = 180°
2θ = 180° – 50° = 130°
θ = 65°
Also, OA ⊥ PA
[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]
∠PAO = 90°
=> ∠PAB + ∠BAO = 90°
=> 65° + ∠BAO = 90°
=> ∠BAO = 90° – 65° = 25°

Question 50.
The pair of tangents AP and AQ drawn from an external point to a circle with centre O are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is [NCERT Exemplar]
(a) 10 cm
(b) 7.5 cm
(c) 5 cm
(d) 2.5 cm
Solution:
(c)
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 66

Question 51.
In the figure, if ∠AOB = 125°, then ∠COD is equal to [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 67
(a) 45°
(b) 35°
(c) 55°
(d) 62\(\frac { 1 }{ 2 }\)°
Solution:
(c) We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°
=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

Question 52.
In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles MCQS 68
(a) 20°
(b) 40°
(c) 35°
(d) 45°
Solution:
(b) Given, AB || PR
∠ABQ = ∠BQR = 70° [alternate angles]
Also QD is perpendicular to AB and QD bisects AB.
In ∆QDA and ∆QDB
∠QDA = ∠QDB [each 90°]
AD = BD
QD = QD [common side]
∆ADQ = ∆BDQ [by SAS similarity criterion]
Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]
Also, ∠ABQ = ∠BQR [alternate interior angle]
∠ABQ = 70° [∠BQR = 70°]
Hence, ∠QAB = 70° [from Eq. (i)]
Now, in ∆ABQ,
∠A + ∠B + ∠Q = 180°
=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Other Exercises

Mark the correct alternative in each of the following :

Question 1.
If 6 θ is an acute angle such that cos θ =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 1
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 2
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 3

Question 2.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 4
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 5

Question 3.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 6
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 7
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 8

Question 4.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 9
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 10
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 11

Question 5.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 13

Question 6.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 14
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 15

Question 7.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 16
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 17

Question 8.
If 6 is an acute angle such that tan2 6 = 8 \(\frac { 8 }{ 7 }\), then the value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 18
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 19
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 20

Question 9.
If 3 cos θ = 5 sin 6, then the value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 21
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 22
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 23
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 24
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 25

Question 10.
If tan2 45° – cos2 30° = x sin 45° cos 45°, then x =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 26
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 27

Question 11.
The value of cos217° – sin2 73° is
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 28
Solution:
cos2 17° – sin2 73° = cos2 (90° – 73°) – sin2 73°
= sin2 73° – sin2 73° = 0 (c)

Question 12.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 29
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 30

Question 13.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 31
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 32

Question 14.
If A and B are complementary angles then
(a) A – sin B
(b) cos A = cos B
(c) A = tan B
(d) sec A = cosec B
Solution:
∵ A and B are complementary angles
∴ A + B = 90°
⇒ A – 90° – B
sec A = sec (90° – B) = cosec B (d)

Question 15.
If x sin (90° – θ) cot (90° – 6) – cos (90° – θ), then x =
(a) 0
(b) 1

(c) -1
(d) 2,

Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 33

Question 16.
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 34
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 35

Question 17.
If angles A, B, C of a AABC form an increasing AP, then sin B =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 36
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 37

Question 18.
If 6 is an acute angle such that sec2 θ =
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 38
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 39

Question 19.
The value of tan 1° tan 2° tan 3s tan 89° is
(a) 1
(b) -1

(c) 0
(d) None of these

Solution:
tan 1° tan 2° tan 3° tan 44° tan 45° tan 46° tan 89°
= tan 1° tan 2° tan 3° tan 44° tan 45° tan
(90° – 44°) tan (90° – 43°) tan (90° – 1°)’
= tan 1° tan 2° tan 3° tan 44° tan 45° cot
44° cot 43°….. cot 1°
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 40

Question 20.
The value of cos 1″ cos 2° cos 3° cos 180° is
(a) 1
(b) 0
(c) -1
(d) None of these
Solution:
None of these, because we deal here only an angle O<θ ≤ 90° (d)

Question 21.
The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these
Solution:
tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° – 15°) tan (90° – 10°) .
= tan 10° tan 15° cot 15° cot 10°
= tan 10° cot 10° tan 15° cot 15°
{∵ tan θ cot θ = 1}
= 1×1 = 1 (C)

Question 22.
The value of
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 41
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 42
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 43

Question 23.
If 6 and 20 – 45° arc acute angles such that sin 0 = cos (20 – 45°), then tan 0 is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 44
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 45

Question 24.
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ-√3 tan 3θ is equal to
(a) 1
(b) 0
(c) -1
(d) 1 + √3
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 46
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 47

Question 25.
If A + B = 90°, then
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 48

Question 26.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 49
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 50
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 51

Question 27.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 52
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 53

Question 28.
sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°

(c) 45°
(d) 60°

Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 54

Question 29.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 55
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 56

Question 30.
If A, B and C are interior angles of a
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 57
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 58

Question 31.
If cos θ = \(\frac { 2 }{ 3 }\), then 2 sec2 θ + 2 tan2 θ-7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4
Solution:

Question 32.
tan 5° x tan 30° x 4 tan 85° is equal to
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 59
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 60

Question 33.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 61
(a)-2
(b) 2
(c) 1
(d) 0
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 62

Question 34.
In the figure, the value of cos Φ is
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 63
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 64

Question 35.
In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, find cot θ.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 65
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS 66

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3

Other Exercises

Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
Steps of construction :
(i)
Draw a circle with O centre and 6 cm radius.
(ii) Take a point P, 10 cm away from the centre O.
(iii) Join PO and bisect it at M.
(iv) With centre M and diameter PO, draw a circle intersecting the given circle at T and S.
(v) Join PT and PS.
Then PT and PS are the required tangents.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 1

Question 2.
Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Solution:
Steps of construction :
(i)
Draw a circle with centre O and radius 3 cm.
(ii) Draw a diameter and produce it to both sides.
(iii) Take two points P and Q on this diameter with a distance of 7 cm each from the centre O.
(iv) Bisect PO at M and QO at N
(v) With centres M and N, draw circle on PO and QO as diameter which intersect the given circle at S, T and S’, T’ respectively.
(vi) Join PS, PT, QS’ and QT’.
Then PS, PT, QS’ and QT’ are the required tangents to the given circle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 2

Question 3.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. [CBSE 2013]
Solution:
Steps of construction :
(i)
Draw a line segment AB = 8 cm.
(ii) With centre A and radius 4 cm and with centre B and radius 3 cm, circles are drawn.
(iii) Bisect AB at M.
(iv) With centre M and diameter AB, draw a circle which intersects the two circles at S’, T’ and S, T respectively.
(v) Join AS, AT, BS’and BT’.
Then AS, AT, BS’ and BT’ are the required tangent.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 3

Question 4.
Draw two tangents to a circle of raidus 3.5 cm from a point P at a distance of 6.2 cm from its centre.
Solution:
Steps of construction :
(i) Draw a circle with centre O and radius 3.5 cm
(ii) Take a point P which is 6.2 cm from O.
(iii) Bisect PO at M and draw a circle with centre M and diameter OP which intersects the given circle at T and S respectively.
(iv) Join PT and PS.
PT and PS are the required tangents to circle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 4

Question 5.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°.            [CBSE 2013]
Solution:
Steps of construction :
Angle at the centre 180° – 45° = 135°
(i) Draw a circle with centre O and radius 4.5 cm.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 5
(ii) At O, draw an angle ∠TOS = 135°
(iii) At T and S draw perpendicular which meet each other at P.
PT and PS are the tangents which inclined each other 45°.

Question 6.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.
Solution:
Steps of Construction :
Draw a line segment BC = 8 cm
From B draw an angle of 90°
Draw an arc \(\breve { BA }\)  = 6cm cutting the angle at A.
Join AC.
ΔABC is the required A.
Draw ⊥ bisector of BC cutting BC at M.
Take M as centre and BM as radius, draw a circle.
Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.
AB and AE are the required tangents.
Justification :
∠ABC = 90°                                            (Given)
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also AE is a tangent to the circle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 6

Question 7.
Draw two concentric circles of radii 3 cm and 5 cm. Construct a tangent to the smaller circle from a point on the larger circle. Also, measure its length.                      [CBSE 2016]
Solution:
Given, two concentric circles of radii 3 cm and 5 cm with centre O. We have to draw pair of tangents from point P on outer circle to the other.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 7
Steps of construction :       
(i) Draw two concentric circles with centre O and radii 3 cm and 5 cm.
(ii) Taking any point P on outer circle. Join OP.
(iii) Bisect OP, let M’ be the mid-point of OP.
Taking M’ as centre and OM’ as radius draw a circle dotted which cuts the inner circle as M and P’.
(iv) Join PM and PP’. Thus, PM and PP’ are the required tangents.
(v) On measuring PM and PP’, we find that PM = PP’ = 4 cm.
Actual calculation:
In right angle ΔOMP, ∠PMO = 90°
∴ PM2 = OP2 – OM2
[by Pythagoras theorem i.e. (hypotenuse)2 = (base)2 + (perpendicular)2]
⇒ PM2 = (5)2 – (3)2 = 25 – 9 = 16
⇒ PM = 4 cm
Hence, the length of both tangents is 4 cm.

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.3 are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2

Other Exercises

Question 1.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it.
Solution:
Steps of construction :
(i)
Draw a line segment BC = 5 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 6 cm, draw arcs intersecting each other at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 3 equal parts making BB1 = B1B2= B2B3.
(v) Join B3C.
(vi) Draw B’C’ parallel to B3C and C’A’ parallel to CA then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 1

Question 2.
Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)th of the corresponding sides of ΔABC. It is given that AB = 5 cm, BC = 7 cm and ∠ABC = 50°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 7 cm.
(ii) Draw a ray BX making an angle of 50° and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 7 equal parts making BB, =B1B2=B2B3=B3B4=B4Bs=B5B6=B6B7
(v) Join B7 and C
(vi) Draw B5C’ parallel to B7C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 2

Question 3.
Construct a triangle similar to a given ∠ABC such that each of its sides is \(\frac { 2 }{ 3 }\)rd of the corresponding sides of ΔABC. It is given that BC = 6 cm, ∠B = 50° and ∠C = 60°.
Solution:
Steps of construction :
(i) Draw a line segment BC = 6 cm.
(ii) Draw a ray BX making an angle of 50° and CY making 60° with BC which intersect each other at A. Then ABC is the triangle.
(iii) From B, draw another ray BZ making an acute angle below BC and intersect 3 equal parts making BB1 =B1B2 = B2B2
(iv) Join B3C.
(v) From B2, draw B2C’ parallel to B3C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 3

Question 4.
Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to \(\frac { 3 }{ 4 }\)th of the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i)
Draw a line segment BC = 6 cm.
(ii) With centre B and radius 4 cm and with centre C and radius 5 cm, draw arcs’intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle,
(iv) Draw a ray BX making an acute angle with BC and cut off 4 equal parts making BB1=  B1B= B2B3 = B3B4.
(v) Join B4 and C.
(vi) From B3C draw C3C’ parallel to B4C and from C’, draw C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 4

Question 5.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac { 7 }{ 5 }\) of the corresponding sides of the first triangle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 5 cm.
(ii) With centre B and radius 6 cm and with centre C and radius 7 cm, draw arcs intersecting eachother at A.
(iii) Join AB and AC. Then ABC is the triangle.
(iv) Draw a ray BX making an acute angle with BC and cut off 7 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
(v) Join B5 and C.
(vi) From B7, draw B7C’ parallel to B5C and C’A’ parallel CA. Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 5

Question 6.
Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°. Draw a triangle similar to ΔABC with its sides equal to (\(\frac { 5 }{ 4 }\))th ot the corresponding sides of ΔABC.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 4.5 cm.
(ii) At A, draw a ray AX perpendicular to AB and cut off AC = AB = 4.5 cm.
(iii) Join BC. Then ABC is the triangle.
(iv) Draw a ray AY making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 =A3A4 = A4A5
(v) Join A4 and B.
(vi) From 45, draw 45B’ parallel to  A4B and  B’C’ parallel to BC.
Then ΔAB’C’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 6

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are \(\frac { 5 }{ 3 }\) times the corresponding sides of the given triangle.  (C.B.S.E. 2008)
Solution:
Steps of construction :
(i)
Draw a line segment BC = 5 cm.
(ii) At B, draw perpendicular BX and cut off BA = 4 cm.
(iii )join Ac , then ABC is the triangle
(iv) Draw a ray BY making an acute angle with BC, and cut off 5 equal parts making BB1 = B1B2 = B2B3 = B3B4 = B4B5
(v) Join B3 and C.
(vi) From B5, draw B5C’ parallel to B3C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 7

Question 8.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(\frac { 3 }{ 2 }\) times the corresponding sides of the isosceles triangle.
Solution:
Steps of construction :
(i) Draw a line segment BC = 8 cm and draw its perpendicular bisector DX and cut off DA = 4 cm.
(ii) Join AB and AC. Then ABC is the triangle.
(iii) Draw a ray DY making an acute angle with OA and cut off 3 equal parts making DD1 = D1D2 =D2D3 = D3D4
(iv) Join D2
(v) Draw D3A’ parallel to D2A and A’B’ parallel to AB meeting BC at C’ and B’ respectively.
Then ΔB’A’C’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 8

Question 9.
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a trianglewhose sides are \((\frac { 3 }{ 4 } )\)th of the corresponding sides of the ΔABC.
Solution:
Steps of construction :
(i)
Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle of 60° with BC and cut off BA = 5 cm.
(iii) Join AC. Then ABC is the triangle.
(iv) Draw a ray BY making an acute angle with BC and cut off 4 equal parts making BB1= B1B2  B2B3=B3B4.
(v) Join B4 and C.
(vi) From B3, draw B3C’ parallel to B4C and C’A’ parallel to CA.
Then ΔA’BC’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 9

Question 10.
Construct a triangle similar to ΔABC in which AB = 4.6 cm, BC = 5.1 cm,∠A = 60° with scale factor 4 : 5.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 4.6 cm.
(ii) At A, draw a ray AX making an angle of 60°.
(iii) With centre B and radius 5.1 cm draw an arc intersecting AX at C.
(iv) Join BC. Then ABC is the triangle.
(v) From A, draw a ray AX making an acute angle with AB and cut off 5 equal parts making AA1 = A1A2 = A2A3 = A3A4=A4A5.
(vi) Join A4 and B.
(vii) From A5, drawA5B’ parallel to A4B and B’C’ parallel to BC.
Then ΔC’AB’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 10

Question 11.
Construct a triangle similar to a given ΔXYZ with its sides equal to \((\frac { 3 }{ 2 })\) th of the corresponding sides of ΔXYZ. Write the steps of construction.                      [CBSE 1995C]
Solution:
Steps of construction :
(i)
Draw a triangle XYZ with some suitable data.
(ii) Draw a ray YL making an acute angle with XZ and cut off 5 equal parts making YY1= Y1Y2 = Y2Y3 = Y3Y4.
(iii) Join Y4 and Z.
(iv) From Y3, draw Y3Z’ parallel to Y4Z and Z’X’ parallel to ZX.
Then ΔX’YZ’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 11

Question 12.
Draw a right triangle in which sides (other than the hypotenuse) are of lengths 8 cm and 6 cm. Then construct another triangle whose sides are \(\frac { 3 }{ 4 }\) times the corresponding sides of the first triangle.
Solution:
(i) Draw right ΔABC right angle at B and BC = 8 cm and BA = 6 cm.
(ii) Draw a line BY making an a cut angle with BC and cut off 4 equal parts.
(iii) Join 4C and draw 3C’ || 4C and C’A’ parallel to CA.
The BC’A’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 12

Question 13.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle. [CBSE 2014]
Solution:
Steps of construction:
(i) Draw a line segment BC = 5.5 cm.
(ii) With centre B and radius 5 cm and with centre C and radius 6.5 cm, draw arcs which intersect each other at A
(iii) Join BA and CA.
ΔABC is the given triangle.
(iv) At B, draw a ray BX making an acute angle and cut off 5 equal parts from BX.
(v) Join C5 and draw 3D || 5C which meets BC at D.
From D, draw DE || CA which meets AB at E.
∴ ΔEBD is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 13

Question 14.
Construct a triangle PQR with side QR = 7 cm, PQ = 6 cm and ∠PQR = 60°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ΔPQR. [CBSE 2014]
Solution:
Steps of construction:
(i)
Draw a line segment QR = 7 cm.
(ii) At Q draw a ray QX making an angle of 60° and cut of PQ = 6 cm. Join PR.
(iii) Draw a ray QY making an acute angle and cut off 5 equal parts.
(iv) Join 5, R and through 3, draw 3, S parallel to 5, R which meet QR at S.
(v) Through S, draw ST || RP meeting PQ at T.
∴ ΔQST is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 14

Question 15.
Draw a ΔABC in which base BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of ΔABC.    [CBSE 2017]
Solution:
Steps of construction:

  1. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
  2.  Draw a ray BX, which makes an acute angle ∠CBX below the line BC.
  3. Locate four points B1, B2, B3and Bon BX such that BB1 = B1B2=B2B3 = B3B4.
  4. Join B4C and draw a line through B3 parallel to B4C intersecting BC to C’.
  5. Draw a line through C’ parallel to the line CA to intersect BA at A’.

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 15

Question 16.
Draw a right triangle in which the sides (other than the hypotenuse) arc of lengths 4 cm and 3 cm. Now, construct another triangle whose sides are \(\frac { 5 }{ 3 }\) times the corresponding sides of the given triangle. [CBSE 2017]
Solution:
Steps of construction:

  1. Draw a right triangle ABC in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. ∠B = 90°.
  2. Draw a line BX, which makes an acute angle ∠CBX below the line BC.
  3. Locate 5 points B1, B2, B3, B4 and B5 on BX such that BB1 = B1B2=B2B3=B3B4=B4B5.
  4. Join B3 to C and draw a line through B5 parallel to B3C, intersecting the extended line segment BC at C’.
  5. Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’.

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 16
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 17

Question 17.
Construct a ΔABC in which AB = 5 cm, ∠B = 60°, altitude CD = 3 cm. Construct a ΔAQR similar to ΔABC such that side of ΔAQR is 1.5 times that of the corresponding sides of ΔACB.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 5 cm.
(ii) At A, draw a perpendicular and cut off AE = 3 cm.
(iii) From E, draw EF || AB.
(iv) From B, draw a ray making an angle of 60 meeting EF at C.
(v) Join CA. Then ABC is the triangle.
(vi) From A, draw a ray AX making an acute angle with AB and cut off 3 equal parts making A A1= A1A2 = A2A3.
(vii) Join A2 and B.
(viii) From A , draw A^B’ parallel to A2B and B’C’ parallel toBC.
Then ΔC’AB’ is the required triangle.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 18

Hope given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :
Question 1.
In the figure, PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length PC ?
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 1
Solution:
In the figure, PA and PB are the tangents to the circle drawn from P
CD is the third tangent to the circle drawn at Q
PB = 10 cm, CQ = 2 cm
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 2
PA and PB are tangents to the circle
PA = PB = 10 cm
Similarly CQ and CA are tangents to the circle
CQ = CA = 2 cm
PC = PA – CA = 10 – 2 = 8 cm

Question 2.
What is the distance between two parallel tangents of a circle of radius 4 cm ?
Solution:
TT’ and SS’ are two tangents of a circle with centre O and radius 4 cm and TT’ || SS’
OP and OQ are joined
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 3
Now OP is the radius and TPT’ is the tangent
OP ⊥ TPT’
Similar OQ ⊥ SS’
But TT’ || SS’
POQ is the diameter
Which is 4 x 2 = 8 cm
Distance between the two parallel tangents is 8 cm

Question 3.
The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle ?
Solution:
PA is a tangent to the circle from P at a distance of 5 cm from the centre O
PA = 4 cm
OA is joined and let OA = r
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 4
Now in right ∆OAP,
OP² = OA² + PA²
=> (5)² = r² + (4)²
=> 25 = r + 16
=> r² = 25 – 16 = 9 = (3)²
r = 3
Radius of the circle = 3 cm

Question 4.
Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in the following figure. If they are inclined to each other at an angle of 100°, then what is the value of ∠POQ ?
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 5
Solution:
TP and TQ are the tangents from T to the circle with centre O and ∠PTQ = 100°
OT, OP and OQ are joined
OP and OQ are radius
OP ⊥ PT and OQ ⊥ QT
Now in quadrilateral OPTQ,
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360° (Sum of angles of a quadrilateral)
=> ∠POQ + 90° + 100° + 90° = 360°
=> ∠POQ + 280° = 360°
=> ∠POQ = 360° – 280° = 80°
Hence ∠POQ = 80°

Question 5.
What is the distance between two parallel tangents to a circle of radius 5 cm?
Solution:
In a circle, the radius is 5 cm and centre is O
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 6
TT’ and SS’ are two tangents at P and Q to the circle
Such that TT’ || SS’
Join OP and OQ
OP is radius and TPT’ is the tangent
OP ⊥ TT’
Similarly OQ ⊥ SS’
POQ is the diameter of the circle
Now length of PQ = OP + OQ = 5 + 5 = 10 cm
Hence distance between the two parallel tangents = 10 cm

Question 6.
In Q. No. 1, if PB = 10 cm, what is the perimeter of ∆PCD ?
Solution:
In the figure, PB = 10 cm, CQ = 2 cm
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 7
PA and PB are tangents to the give from P
PA = PB = 10 cm
Similarly, CA and CQ are the tangents
CA = CQ = 2 cm
and DB and DQ are the tangents
DB = DQ
Now, perimeter of ∆PCD
PC + PD + CQ + DQ
= PC + CQ + PD + DQ
= PC + CA + PD + DB {CQ = CA and DQ = DB}
= PA + PB = 10 + 10 = 20 cm

Question 7.
In the figure, CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then find the length of BR. (C.B.S.E. 2009)
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 8
Solution:
Given : In the figure, CP and CQ are tangents to a circle with centre O
ARB is a third tangent to the circle at R
CP = 11 cm, BC = 7 cm
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 9
To find : The length of BR
BQ and BR are tangents to the circle drawn from B
BQ = BR ….(i)
Similarly CQ = CP
=> BC + BQ = CP = 11 (CP = 11 cm and BC = 7 cm)
=> 7 + BQ = 11
=> BQ = 11 – 7
BQ = 4 cm
But BQ = BR
BR = 4 cm

Question 8.
In the figure, ∆ABC is circumscribing a circle. Find the length of BC. (C.B.S.E. 2009)
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 10
Solution:
∆ABC is circumscribing a circle which touches it at P, Q and R
AC = 11 cm, AR = 4 cm, BR = 3 cm
Now we have to find BC
AR and AQ are tangents to the circle from A
AQ = AR = 4 cm
Then CQ = AC – AQ = 11 – 4 = 7 cm
Similarly,
CP and CQ are tangents from C
CP = CQ = 7 cm
and BP and BR are tangents from B
BP = BR = 3 cm
Now BC = BP + CP = 3 + 7 = 10 cm

Question 9.
In the figure, CP and CQ are tangents from an external point C to a circle with centre O. AB is another tangent which touches the circle at R. If CP = 11 cm and BR = 4 cm, find the length of BC. [CBSE 2010]
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 11
Solution:
CP and CQ are the tangents to the circle from C.
AB is another tangent to the same circle which touches at R and meets the first two tangents at A and B. O is the centre of the circle.
OC is joined
CP = 11 cm, BR = 4 cm
CP and CQ are tangents to the circle
CP = CQ = 11 cm
Similarly from B, CR and BQ are the tangents
BQ = BR = 4 cm
Now BC = CQ – BQ = 11 – 4 = 7 cm

Question 10.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles with centre O, have radii 5 cm and 3 cm
AB is a chord which touches the smaller circle at P
OP is joined which is radius of smaller circle
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 12
P is mid-point of AB
OP = 3 cm and OA = 5 cm
Now in right ∆OAP
OA² = OP² + AP²
(5)² = (3)² + AP²
=> 25 = 9 + AP²
=> AP² = 25 – 9 = 16 = (4)²
AP = 4 cm
AB = 2 AP = 2 x 4 cm = 8 cm

Question 11.
In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. [CBSE 2015]
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 13
Solution:
In the given figure,
PA and PB are tangents to the circle from P
PA = PB
∠APB = 50°, OA is joined
To find ∠OAB
In ∆PAB
PA = PB
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 14

Question 12.
In the figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then, find ∠PRQ. [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 15
Solution:
∠OPQ = ∠OQP = 30°, i.e., ∠POQ = 120°
Also, ∠PRQ = \(\frac { 1 }{ 2 }\) reflex ∠POQ

Question 13.
In the figure, PQL and PRM are tangents to the circle with centre O at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, find ∠QSR. [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 16
Solution:
Here ∠OSQ = ∠OQS = 90° – 50° = 40°
and ∠RSO = ∠SRO = 90° – 60° = 30°.
Therefore, ∠QSR = 40° + 30° = 70°

Question 14.
In the figure, BOA is a diameter of a circle and the tangent at a point P meets BA produced at T. If ∠PBO = 30°, then find ∠PTA. [NCERT Exemplar]
RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS 17
Solution:
As ∠BPA = 90°,
∠PAB = ∠OPA = 60°
Also OP ⊥ PT.
Therefore, ∠APT = 30°
and ∠PTA = 60° – 30° = 30°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS

Other Exercises

Answer each of the following questions either in one word or one sentence or as per requirement of the questions :

Question 1.
Write the maximum and minimum values of sin θ.
Solution:
Maximum value of sin θ = 1
and minimum value of sin θ = 0

Question 2.
Write the maximum and minimum values of cos 0.
Solution:
Maximum value cos θ=1 and minimum value of cos θ = θ

Question 3.
What is the maximum value of \(\frac { 1 }{ sec\theta }\) ?
Solution:
Maximum value of \(\frac { 1 }{ sec\theta }\) or cos θ = 1

Question 4.
What is the maximum value of \(\frac { 1 }{ cosec\theta }\)
Solution:
Maximum value of \(\frac { 1 }{ cosec\theta }\) or sin θ = 1

Question 5.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 1
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 2

Question 6.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 3
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 4

Question 7.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 5
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 6

Question 8.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 7
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 8
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 9

Question 9.
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 10
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 11

Question 10.
If tan A = \(\frac { 3 }{ 4 }\) and A + B = 90°, then what is the value of cot B ?
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 12

Question 11.
If A + B = 90°, cos B = \(\frac { 3 }{ 5 }\), what is the value of sin A.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 13

Question 12.
Write the acute angle θ satisfying √3 sin θ = cos θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 14

Question 13.
Write the Value of cos 1° cos 2° cos 3° ……. cos 179° cos 180°.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 15

Question 14.
Write the Value of tan 10′ tan 15° tan 75° tan 80°.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 16
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 17

Question 15.
If A + B = 90° and tan A = \(\frac { 3 }{ 4 }\) what is cot B?
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 18

Question 16.
If tan A = \(\frac { 5 }{ 12 }\), find the value of (sin A + cos A) sec A. (C.B.S.E. 2008)
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 19
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS 20

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS are helpful to complete your math homework.

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RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1

RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1

Other Exercises

Question 1.
Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3. Also justify your construction.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 12 cm.
(ii) Draw a ray AX at A making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) Cut off 2 equal parts from AX and 3 equal parts from BY.
(v) Join 2 and 3 which intersects AB at P.
P is the required point which divides AB in the ratio of 2 : 3 internally.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1 1

Question 2.
Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 9 cm.
(ii) Draw a ray AX making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) Cut off 4 equal parts from AX and 3 parts from BY.
(v) Join 4 and 3 which intersects AB at P.
P is the required point which divides AB in the ratio of 4 : 3 internally.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1 2

Question 3.
Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also justify your construction.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 14 cm.
(ii) Draw a ray AX making an acute angle with AB.
(iii) From B, draw another ray BY parallel to AX.
(iv) From AX, cut off 2 equal parts and from B, cut off 5 equal parts.
(v) Join 2 and 5 which intersects AB at P.
P is the required point which divides AB in the ratio of 2 : 5 internally.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1 3

Question 4.
Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5.
Solution:
Steps of construction :
(i)
Draw a line segment AB = 8 cm.
(ii) Draw a ray AX making an acute angle with ∠BAX = 60° withAB.
RD Sharma Class 10 Solutions Chapter 9 Constructions Ex 9.1 4
(iii) Draw a ray BY parallel to AX by making an acute angle ∠ABY = ∠BAX.
(iv) Mark four points A1, A2, A3, A4 on AX and five points B1, B2, B3, B4, Bs on BY in such a way that AA1 = A1A2 = A2A3 = A3A4 .
(v) Join A4B5.
(vi) Let this line intersect AB at a point P.
Thus, P is the point dividing the line segment AB internally in the ratio of 4 : 5.

 

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RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS

Other Exercises

Mark the correct alternative in each of the folloiwng :
Question 1.
If sec θ + tan θ = x, then sec θ =
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 2
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 3

Question 2.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 4
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 5

Question 3.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 5.1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 7

Question 4.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 8
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 9

Question 5.
sec4 A – sec2 A is equal to

(a) tan2 A – tan4 A
(b) tan4 A – tan2 A
(c) tan4 A + tan2 A          
(d) tan2 A + tan4 A
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 10
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 11

Question 6.
cos4 A – sin4 A is equal to
(a) 2 cos2 A + 1        
(b) 2 cos2 A – 1
(c) 2 sin2 A – 1           
 (d) 2 sin2 A + 1
Solution:
cos4 A – sin4 A = (cos2 A + sin2 A) (cos2 A – sin2 A)
= 1 (cos2 A – sin2 A) = cos2 A – (1 – cos2 A)
= cos2 A – 1 + cos2 A
= 2 cos2 A – 1            (b)

Question 7.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 12
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 13

Question 8.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 14
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 15
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 16

Question 9.
The value of (1 + cot θ – coscc θ) (1 + tan θ + sec θ) is
(a) 1                          
(b) 2
(c) 4                          
(d) 0
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 17

Question 10.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 18
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 19
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 20

Question 11.
(cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) is equal 
(a) 0                          
(b) 1
(c) -1
(d) None of these
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 21

Question 12.
If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =
(a) a2b2                             
(b) ab
(c) a4b4                      
(d) a2 + b2
Solution:
x = a cos θ, y = b sin θ                       …(i)
bx = ab cos θ, ay = ab sin θ          ….(ii)
Adding (i) and (ii) we get,
b2x2+ a2y2 = a2b2 cos θ + a2b2 sin θ
= a2b2 (cos θ + sin θ)
= a2b2 x 1
= a2b2                         (a)

Question 13.
If x = a sec θ and y-b tan θ, then b2x2 – a2y2
(a) ab
(b) a2 – b2
(c) a2 + b2
(d) a2b2
Solution:
x = a sec θ and y = b tan θ
b2x2 – a2y2 = b2 (a sec θ)2 – a2 (b tan θ)2
= a2b2 sec2 θ – a2b2 tan θ
= a2b2 (sec θ – tan θ)
= a2b2 x 1
= a2b2                       (d)

Question 14.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 22
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 23

Question 15.
2 (sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) is equal to
(a) 0                
(b) 1
(c) -1               
(d) None of these
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 24.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 25

Question 16.
If a cos θ + b sin θ = 4 and a sin θ – b cos θ = 3, then a2 + b2 =
(a) 7                 
(b) 12
(c) 25                         
(d) None of these
Solution:
a cos θ + b sin θ = 4
a sin θ – b cos θ = 3
Squaring and adding
a2 cos2  θ + b2  sin2 θ  + 2ab sin θ cos θ=16
a2 sinθ + b2  cos2 θ  – 2ab sin θ  cos θ = 9
a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ) = 25 (∵ sin2 θ + cos2 θ=1)
⇒ a2 x 1 + b2 x 1 = 25
⇒  a2 + b2 = 25                                           (c)

Question 17.
If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p2 – q2 =
(a)   a2 – b2                  
(b) b2 – a2
(c)  a2 + b2                  

(d)  b – a
Solution:
a cot θ + b cosec θ = p
b cot θ + a cosec θ = q
Squaring and subtracting,
p2 – q2  = (a cot θ + b cosec θ)2 – (b cot θ + a cosec θ)2
= a2 cot2 θ + b2 cosec2 θ + 2ab cot θ cosec θ – (b2 cot2 θ + a2 cosec2 θ + 2ab cot θ cosec θ)
= a2 cot2 θ +  b2  cosec2 θ + lab cot θ cosec θ – b2 cot2  θ – a2 cosec2 θ – lab cot θ cosec θ
= a2 (cot2 θ – cosec2 θ) + b2 (cosec2 θ – cot2 θ)
= -a2 (cosec2 θ – cot2 θ) + b2 (cosec2 θ – cot2 θ)
= -a2 x 1 + b2 x 1 = b2 – a2                                    (b)

Question 18.
The value of sin2 29° + sin2 61° is
(a) 1
(b) 0
(c) 2sin2 29“                 
(d)  2cos2 61°
Solution:
sin2 29° + sin2 61° = sin2 29° + sin2 (99° – 29°)
= sin2 29 + cos2 29°              (a)
(sin2 θ + cos2 θ=1)

Question 19.
If x = r sin θ cos φ, y = r sin θ sin φ and z – r cos θ, then
(a) x2 +y2 + z2 = r2      
(b)   x2 +y2 – z2 = r2
(c) x2– y2+ z2 = r2      

(d)   z2 + y2 – x2 = r2
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 26

Question 20.
If sin θ + sin θ=1, then cos θ + cos θ
(a) -1                         

(b) 1
(c) 0                           
(d) None of these
Solution:
sin θ + sin2 θ=1
⇒ sin θ = 1- sin2 θ
⇒ sin θ = cos2 θ
cos2 θ + cos4 θ = sin θ + sin2 θ     {∵ cos2 θ = sin θ}
⇒  cos2 θ + cos4 θ=1                                 (b)
{∵ sin θ + sin2 θ = 1 (given)}

Question 21.
If a cos θ + b sin θ = m and a sin θ – b cos θ = it, then a2 + b2 =
(a) m2 – it2        
(b) m2n2
(c) n2 – m2        

(d) m2 + n2
Solution:
a cos θ + b sin θ = m
a sin θ – b cos θ = n
Squaring and adding
a2 cos2 θ + b2 sin2 θ   + lab sin  θ cos θ = m2
a2 sin2 θ + b2 cos2 θ   – 2ab sin  θ cos θ = n2
a2 (cos2 θ + sin2 θ) + b2 (sin2 θ + cos2 θ) = m2 + n2     {sin2 θ + cos2 θ=1}
⇒   a2 + 1 + b2 x 1 = m2 – n2
⇒  a2 + b2 = m2 + n2
Hence a2 + b2 = m2 + n2          (d)

Question 22.
If cos A + cos2 A = 1, then sin2 A + sin4 A = 
(a) -1                         
(b) 0                            
(c) 1                           
(d) None of these
Solution:
cos A + cos2 A = 1
⇒ cos A = 1 – cos2 A
⇒  cos A = sin2 A
Now, sin2 A + sin4 A = cos A + cos2 A = 1 (∵ cos A + cos2 A = 1) (given)
∴ sin2 A + sin4 A = 1                              (c)

Question 23.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 27
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 28

Question 24.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 29
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 30
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 31

Question 25.
9sec2 A – 9tan2 A is equal to
(a) 1
(b) 9
(c) 8                           
(d) 0
Solution:
9sec2 A – 9tan2 A = 9 (sec2 A – tan2 A)
= 9 x 1       (∵ sec2 A – tan2 A = 1)
= 9                        (b)

Question 26.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(a) 0
(b) 1
(c) 1                           
(d) -1
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 32

Question 27.
(sec A + tan A) (1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A                        
(d) cos A
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 33

Question 28.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 34
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 35

Question 29.
If sin θ cos θ = 0, then the value of sin4 θ + cos4 θ is
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 36
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 37
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 38

Question 30.
The value of sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cos θ                  
(b) 0
(c) 2 sin θ
(d) 1
Solution:
sin (45° + θ) – cos (45° – θ)
= sin (45° + θ) – sin (90° – 45° + θ)
= sin (45° + θ) – sift (45° + θ)
= 0                                                     (b)

Question 31.
If ΔABC is right angled at C, then the value of cos (A + B) is
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 39
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 40

Question 32.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities MCQS 41
Solution:
cos (9θ) = sin θ
⇒ sin (90° – 9θ) = sin θ
⇒ 90° – 90 = θ
⇒ 9θ = 90°
⇒  θ= 10
tan 6θ = tan 6
= tan 60° = \(\sqrt { 3 } \)        (b)

Question 33.
If cos (α + β) =0 , then sin (α – β) can be reduced to 
(a) cos β
(b) cos 2β
(c) sin α
(d) sin 2α
Solution:
cos (α + β) = 0
⇒ α + β = 90°                       [∵ cos 90° = 0]
⇒  θ = 90° – β                                          …(i)
sin (α – β) = sin (90° – ββ) [using (i)]
= sin (90° – 2β)
= cos 2β [∵ sin (90° – θ) = cos θ]         (b)

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RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3

Other Exercises

Question 1.
Evaluate the following :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 1
Solution:

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 2

Question 2.
Evaluate the following :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 3
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 4
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 5
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 6

Question 3.
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°
(ii) tan 65° + cot 49“
(iii) sec 76° + cosec 52°
(iv) cos 78° + sec 78°
(v) cosec 54° + sin 72°
(vi) cot 85″ + cos 75°
(vii) sin 67° + cos 75°
Solution:
(i) sin 59° + cos 56°
= sin (90° – 31°) + cos (90° – 34°)
= cos 31° +sin 34°
(ii) tan 65° + cot 49°
= tan (90° – 25°) + cot (90° – 41°)
= cot 25° + tan 41°
(iii) sec 76° + cosec 52°
= sec (90° – 14°) + cosec (90 0 – 38°)
= cosec 14° + sec 38°
(iv) cos 78° + sec 78°
= cos (90° – 12°) + sec (90°- 12°)
= sin 12° + cosec 12°
(v) cosec 54° + sin 72°
= cosec (90° – 36°) + sin (90°-18°)
= sec 36° + cos 18°
(vi) cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
(vii) sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

Question 4.
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Solution:
cos 75° + cot 75° = cos (90° – 15°) + cot (90°-15°)
= sin 15° + tan 15°

Question 5.
If sin 3A = cos (A – 26°), where 3A is an acute angle, And the value of A.
Solution:
sin 3A = cos (A – 26°)
⇒ cos (90° – 3A) = cos (A – 26°)
Comparing,
90° – 3A = A – 26°
⇒ 90° + 26° = A + 3A ⇒ 4A = 116°
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 7

Question 6.
If A, B, C are the interior angles of a triangle ABC, prove
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 8
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 9

Question 7.
Prove that :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 10
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 11
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 12
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 13

Question 8.
Prove the following :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 14
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 15
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 16
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 17
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 18

Question 9.
Evaluate :
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 19
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 20
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 21
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 22
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 23
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 24
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 25
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 26
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 27
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 28

Question 10.
If sin θ= cos (θ – 45°), where θ and (θ – 45°) are acute angles, find the degree measure of θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 29

Question 11.
If A, B, C are the interior angles of a AABC, show that :
(i) \(sin\frac { B+C }{ 2 } cos\frac { A }{ 2 }\)
(ii) \(cos\frac { B+C }{ 2 } sin\frac { A }{ 2 }\)
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 30

Question 12.
If 2θ + 45° and 30° – θ are acute angles, find the degree measures of θ satisfying sin (20 + 45°) = cos (30° – θ).
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 31

Question 13.
If θ is a positive acute angle such that sec θ = cosec 60°, And the value of 2 cos2 θ-1.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 32

Question 14.
If cos 2 θ – sin 4 θ, where 2 θ and 4 θ are acute angles, find the value of θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 33

Question 15.
If sin 3 θ = cos (θ – 6°), where 3 θ and θ – 6° are acute angles, find the value of θ.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 34

Question 16.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 35
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 36

Question 17.
If sec 2A = cosec (A – 42°), where 2A is an acute angle, find the value of A. (C.B.S.E. 2008)
Solution:
RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3 37

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RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1

RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1

Other Exercises

Question 1.
Fill in the blanks :
(i) The common point of a tangent and the circle is called ……….
(ii) A circle may have ………. parallel tangents.
(iii) A tangent to a circle intersects it in ……….. point(s).
(iv) A line intersecting a circle in two points is called a …………
(v) The angle between tangent at a point on a circle and the radius through the point is ………..
Solution:
(i) The common point of a tangent and the circle is called the point of contact.
(ii) A circle may have two parallel tangents.
(iii) A tangents to a circle intersects it in one point.
(iv) A line intersecting a circle in two points is called a secant.
(v) The angle between tangent at a point, on a circle and the radius through the point is 90°.

Question 2.
How many tangents can a circle have ?
Solution:
A circle can have infinitely many tangents.

Question 3.
O is the centre of a circle of radius 8 cm. The tangent at a point A on the circle cuts a line through O at B such that AB = 15 cm. Find OB.
Solution:
Radius OA = 8 cm, ST is the tangent to the circle at A and AB = 15 cm
RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1 1
OA ⊥ tangent TS
In right ∆OAB,
OB² = OA² + AB² (Pythagoras Theorem)
= (8)² + (15)² = 64 + 225 = 289 = (17)²
OB = 17 cm

Question 4.
If the tangent at a point P to a circle with centre O cuts a line through O at Q such that PQ = 24 cm and OQ = 25 cm. Find the radius of the circle.
Solution:
OP is the radius and TS is the tangent to the circle at P
RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1 2
OQ is a line
OP ⊥ tangent TS
In right ∆OPQ,
OQ² = OP² + PQ² (Pythagoras Theorem)
=> (25)² = OP² + (24)²
=> 625 = OP² + 576
=> OP² = 625 – 576 = 49
=> OP² = (7)²
OP = 7 cm
Hence radius of the circle is 7 cm

 

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RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2

RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2

Other Exercises

Question 1.
If cos θ = \(\frac { 4 }{ 5 }\) , find all other trigonometric ratios of angle θ
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 1
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 2
Question 2.
If sin θ = \(\frac { 1 }{ \sqrt { 2 } }\) , find all other trigono­metric ratios of angle θ
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 3
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 4

Question 3.
If tan θ = \(\frac { 1 }{ \sqrt { 2 } }\), find the value of \(\frac { { cosec }^{ 2 }-{ sec }^{ 2 } }{ { cosec }^{ 2 }+cot^{ 2 } }\) 
Solution:
<RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 5
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 6

Question 4.
If tan θ = \(\frac { 3 }{ 4 }\), find the value of \(\frac { 1-cosheta }{ 1+cosheta }\) 
 (C.B.S.E. 1995)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 7
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 8

Question 5.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 9
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 10
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 11

Question 6.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 12
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 13
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 14

Question 7.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 15
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 16

Question 8.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 17
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 18
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 19

Question 9.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 20
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 21
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 22

Question 10.
If \(\sqrt { 3 } \) tan θ = 3sin θ , find the value of sin2 θ  – cos2 θ .          (C.B.S.E. 2001)
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 23

Question 11.
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 24
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 25
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 26

Question 12.
If sin θ + cos θ = \(\sqrt { 2 } \)cos (90° – θ), find cot θ
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 27

Question 13.
If 2sin2 θ – cos2 θ = 2, then find the value of θ. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 28

Question 14.
If \(\sqrt { 3 } \) tan θ-1=0, find the value of sin2 – cos2 θ. [NCERT Exemplar]
Solution:
RD Sharma Class 10 Solutions Chapter 11 Trigonometric Identities Ex 11.2 29

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