RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B.

Other Exercises

Question 1.
Solution:
In ∆ABC,
∠A = 75°, ∠B = 45°
side BC is produced to D
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 1
forming exterior ∠ ACD
Exterior ∠ACD = ∠A + ∠B (Exterior angle is equal to sum of its interior opposite angles)
= 75° + 45° = 120°

Question 2.
Solution:
In ∆ABC, BC is produced to D forming an exterior angle ACD
∠ B = 68°, ∠ A = x°, ∠ ACB = y° and ∠ACD = 130°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 2
In triangle,
Exterior angles is equal to sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 130° = x + 68°
⇒ x = 130° – 68° = 62°
But ∠ACB + ∠ACD = 180° (Linear pair)
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
Hence x = 62° and y = 50°

Question 3.
Solution:
In ∆ABC, side BC is produced to D forming exterior angle ACD.
∠ACD = 65°, ∠A = 32°
∠B = x, ∠ACB = y
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 3
In a triangle, the exterior angles is equal to the sum of its interior opposite angles
∠ACD = ∠A + ∠B
⇒ 65° = 32° + x
⇒ x = 65° – 32° = 33°
But ∠ ACD + ∠ ACB = 180° (Linear pair)
⇒ 65° + y = 180°
⇒ y = 180°- 65° = 115°
x = 33° and y = 115°

Question 4.
Solution:
In ∆ABC, side BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 4
∠ACD = 110°, and ∠A : ∠B = 2 : 3
In a triangle, exterior angles is equal to the sum of its interior opposite angles
⇒ ∠ACD = ∠A + ∠B
⇒ ∠A + ∠B = 110°
But ∠A : ∠B = 2 : 3
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 5
But ∠A + ∠B + ∠C = 180° (sum of angles of a triangle)
⇒ 44° + 66° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
Hence ∠ A = 44°, ∠ B = 66° and ∠ C = 70°

Question 5.
Solution:
In ∆ABC, side BC is produced to forming exterior angle ACD.
∠ACD = 100° and ∠A = ∠B
Exterior angle of a triangle is equal to the sum of its interior opposite angles.
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 6
∠ACD = ∠A + ∠B But ∠A = ∠B
∠A + ∠A = ∠ACD = 100°
⇒ 2 ∠A = 100°
⇒ ∠A = 50°
∠B = ∠A = 50°
But ∠A + ∠B + ∠ ACB = 180° (sum of angles of a triangle)
⇒ 50° + 50° + ∠ ACB = 180°
⇒ 100° + ∠ ACB = 180°
⇒ ∠ ACB = 180° – 100° = 80°
Hence ∠ A = 50°, ∠ B = 50° and ∠ C = 80°

Question 6.
Solution:
In ∆ABC, side BC is produced to D From D, draw a line meeting AC at E so that ∠D = 40°
∠A = 25°, ∠B = 45°
In ∆ABC,
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 7
Exterior ∠ACD = ∠A + ∠B = 25° + 45° = 70°
Again, in ∆CDE,
Exterior ∠ AED = ∠ ECD + ∠ D = ∠ACD + ∠D = 70° + 40° = 110°
Hence ∠ACD = 70° and ∠AED = 110°

Question 7.
Solution:
In ∆ABC, sides BC is produced to D and BA to E
∠CAD = 50°, ∠B = 40° and ∠ACB = 100°
∠ ACB + ∠ ACD = 180° (Linear pair)
⇒ 100° + ∠ ACD = 180°
⇒ ∠ ACD = 180° – 100° = 80°
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 8
In ∆ACD,
∠ CAD + ∠ ACD + ∠ ADC = 180° (sum of angles of a triangle)
⇒ 50° + 80° + ∠ ADC = 180°
⇒ 130° + ∠ ADC = 180°
⇒ ∠ ADC = 180° – 130° = 50°
Now, in ∆ABD, BA is produced to E
Exterior ∠DAE = ∠ACD + ∠ADC = 80° + 50° = 130°
Hence ∠ ACD = 80°, ∠ ADC = 50° and ∠DAE = 130°

Question 8.
Solution:
In ∆ABC, BC is produced to D forming exterior ∠ ACD
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 9
∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.
x : y = 2 : 3
Now, in ∆ABC,
Exterior ∠ACD = ∠A + ∠B
RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15B 10
But ∠A + ∠B + ∠ACB = 180° (sum of angles of a triangle)
⇒ 78° + 52° + ∠ACB = 180°
⇒ 130° + ∠ACB = 180°
⇒ ∠ACB = 180° – 130°
⇒ ∠ACB = 50°
⇒ ∠ = 50°
Hence x = 52°, y = 78° and z = 50°

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B are helpful to complete your math homework.

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