## RS Aggarwal Class 9 Solutions Chapter 4 Lines and Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D.

**Other Exercises**

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4A

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4B

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4C

RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D

**Question 1.**

**Solution:**

In ∆ABC,

∠B = 76° and ∠C = 48°

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

=> ∠A + 76° + 48° = 180°

=> ∠ A + 124° = 180°

=> ∠A= 180° – 124° = 56°

**Question 2.**

**Solution:**

Angles of a triangle are in the ratio = 2:3:4

Let first angle = 2x

then second angle = 3x

and third angle = 4x

2x + 3x + 4x = 180°

(Sum of angles of a triangle)

=> 9x = 180°

=> x = \(\frac { { 180 }^{ o } }{ 9 } \) = 20°

First angle = 2x = 2 x 20° = 40°

Second angle = 3x = 3 x 20° = 60°

and third angle = 4x = 4 x 20° = 80° Ans.

**Question 3.**

**Solution:**

In ∆ABC,

3∠A = 4∠B = 6∠C = x (Suppose)

**Question 4.**

**Solution:**

In ∆ABC,

∠ A + ∠B = 108° …(i)

∠B + ∠C – 130° …(ii)

But ∠A + ∠B + ∠C = 180° …(iii)

(sum of angles of a triangle)

Subtracting (i) from (iii),

∠C = 180° – 108° = 72°

Subtracting (ii) from (iii),

∠A = 180°- 130° = 50°

But ∠ A + ∠B = 108° (from i)

50° + ∠B = 108°

=> ∠B = 108° – 50° = 58°

Hence ∠A = 50°, ∠B = 58° and ∠C = 72° Ans.

**Question 5.**

**Solution:**

In ∆ABC,

∠A+∠B = 125° …(i)

∠A + ∠C = 113° …(ii)

But ∠A + ∠B + ∠C = 180° …(iii)

(sum of angles of a triangles) Subtracting, (i), from (iii),

∠C = 180°- 125° = 55°

Subtracting (ii) from (iii),

∠B = 180°- 113° – 67°

∠A + ∠B = 125°

∠ A + 67° = 125°

=> ∠ A = 125° – 67°

∠A = 58°

Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.

**Question 6.**

**Solution:**

In ∆ PQR,

∠ P – ∠ Q = 42°

=> ∠P = 42°+∠Q …(i)

∠Q – ∠R = 21°

∠Q – 21°=∠R …(ii)

But ∠P + ∠Q + ∠R = 180°

(Sum of angles of a triangles)

42° + ∠Q + ∠Q + ∠Q – 21°= 180°

=> 21° + 3∠Q = 180°

=> 3∠Q = 180°- 21° = 159°

from ∠Q = \(\frac { { 159 }^{ o } }{ 3 } \) = 53°

(i)∠P = 42° + ∠Q = 42° + 53° = 95°

and from (ii) ∠R = ∠Q – 21°

= 53° – 25° = 32°

Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.

**Question 7.**

**Solution:**

Let ∠ A, ∠ B and ∠ C are the three angles of A ABC.

and ∠A + ∠B = 116° …(i)

**Question 8.**

**Solution:**

Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC

Let ∠ A = ∠ B = x

then ∠C = x + 48°

But ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle)

x + x + x + 18° = 180°

=> 3x + 18° = 180°

=> 3x = 180° – 18° = 162°

x = \(\frac { { 162 }^{ o } }{ 3 } \) = 54°

∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72°

Hence angles are 54°, 54 and 72° Ans.

**Question 9.**

**Solution:**

Let the smallest angle of a triangle = x°

their second angle = 2x°

and third angle = 3x°

But sum of angle of a triangle = 180°

x + 2x + 3x = 180°

=> 6x = 180°

=> x – \(\frac { { 180 }^{ o } }{ 6 } \) = 30°

Hence smallest angle = 30°

Second angle = 2 x 30° = 60°

and third angle = 3 x 30° = 90° Ans.

**Question 10.**

**Solution:**

In a right angled triangle.

one angle is = 90°

Sum of other two acute angles = 90°

But one acute angle = 53°

Second acute angle = 90° – 53° = 37°

Hence angle of the triangle with be 90°, 53°, 37° Ans.

**Question 11.**

**Solution:**

**Given :** In ∆ ABC,

∠ A = ∠B + ∠C

**To Prove :** ∆ABC is a right-angled

**Proof :** We know that in ∆ABC,

∠A + ∠B + ∠C = 180°

(angles of a triangle)

But ∠ A = ∠ B + C given

∠A + (∠B + ∠C) = 180°

=> ∠A + ∠A = 180°

=> 2∠A = 180°

=> ∠ A = \(\frac { { 180 }^{ o } }{ 2 } \) = 90°

∠ A = 90°

Hence ∆ ABC is a right-angled Hence proved.

**Question 12.**

**Solution:**

**Given.** In ∆ ABC, ∠A = 90°

AL ⊥ BC.

**To Prove :** ∠BAL = ∠ACB

**Proof :** In ∆ ABC, AL ⊥ BC

In right angled ∆ALC,

∠ ACB + ∠ CAL = 90° …(i)

( ∴∠L = 90°)

But ∠ A = 90° ‘

=> ∠ BAL + ∠ CAL = 90° …(ii)

From (i) and (ii),

∠BAL + ∠ CAL= ∠ ACB+ ∠CAL

=> ∠ BAL = ∠ ACB Hence proved.

**Question 13.**

**Solution:**

**Given.** In ∆ABC,

Each angle is less than the sum of the other two angles

∠A< ∠B + ∠C

∠B < ∠C + ∠A

and ∠C< ∠A + ∠C

**Proof :** ∠ A < ∠B + ∠C

Adding ∠ A both sides,

∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180°

(∴ ∠A+∠B+∠C=180°)

∠A < \(\frac { { 180 }^{ o } }{ 2 } \) => ∠A< 90

Similarly, we can prove that,

∠B < 90° and ∠C < 90°

∴ each angle is less than 90°

Hence, triangle is an acute angled triangle. Hence proved.

**Question 14.**

**Solution:**

**Given.** In ∆ABC,

∠B > ∠A + ∠C

**Question 15.**

**Solution:**

In ∆ABC

∠ ABC = 43° and Ext. ∠ ACD = 128°

**Question 16.**

**Solution:**

∠ ABC + ∠ ABD = 180°

(Linear pair)

**Question 17.**

**Solution:**

(i)In the figure, ∠BAE =110° and ∠ACD = 120°.

**Question 18.**

**Solution:**

In the figure,

∠A = 55°, ∠B = 45°, ∠C = 30° Join AD and produce it to E

**Question 19.**

**Solution:**

In the figure,

∠EAC = 108°,

AD divides ∠ BAC in the ratio 1 : 3

and AD = DB

∠EAC + ∠ BAC = 180°

(Linear pair)

**Question 20.**

**Solution:**

Sides BC, CA and AB

are produced in order forming exterior

angles ∠ ACD,

**Question 21.**

**Solution:**

**Given :** Two ∆ s DFB and ACF intersect each other as shown in the figure.

**To Prove :** ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°

**Proof :** In ∆ DFB,

∠D + ∠F + ∠B = 180°

(sum of angles of a triangle)

Similarly,in ∆ ACE

∠A + ∠C + ∠E = 180° …(ii)

Adding (i) and (ii), we get :

∠D + ∠F + ∠B+ ∠A+ ∠C + ∠E = 180° + 180°

=> ∠A+∠B+∠C+∠D+∠E + ∠ F = 360°

Hence proved.

**Question 22.**

**Solution:**

In the figure,

ABC is a triangle

and OB and OC are the angle

bisectors of ∠ B and ∠ C meeting each other at O.

∠ A = 70°

In ∆ ABC,

∠A + ∠B + ∠C = 180°

(sum of angles of a triangle)

**Question 23.**

**Solution:**

In ∆ABC, ∠ A = 40°

Sides AB and AC are produced forming exterior angles ∠ CBD and ∠ BCE

**Question 24.**

**Solution:**

In the figure, ∆ABC is triangle and ∠A : ∠B : ∠C = 3 : 2 : 1

AC ⊥ CD.

∠ A + ∠B + ∠C = 180°

(sum of angles of a triangle)

But ∠A : ∠B : ∠C = 3 : 2 : 1

**Question 25.**

**Solution:**

In ∆ ABC

AN is the bisector of ∠ A

∠NAB =\(\frac { 1 }{ 2 } \) ∠A.

Now in right angled ∆ AMB,

∠B + ∠MAB = 90° (∠M = 90°)

**Question 26.**

**Solution:**

(i) False: As a triangle has only one right angle

(ii) True: If two angles will be obtuse, then the third angle will not exist.

(iii) False: As an acute-angled triangle all the three angles are acute.

(iv) False: As if each angle will be less than 60°, then their sum will be less than 60° x 3 = 180°, which is not true.

(v) True: As the sum of three angles will be 60° x 3 = 180°, which is true.

(vi) True: A triangle can be possible if the sum of its angles is 180°

But the given triangle having angles 10° + 80° + 100° = 190° is not possible.

Hope given RS Aggarwal Solutions Class 9 Chapter 4 Lines and Triangles Ex 4D are helpful to complete your math homework.

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