## 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**

- RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15A
- RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15B
- RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15C
- RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D

**Question 1.**

**Solution:**

In ∆ABC,

∠A = 75°, ∠B = 45°

side BC is produced to D

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°

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

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

∠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

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.

∠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,

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°

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

∠ACD = 130°, ∠A = y°, ∠B = x° and ∠ACB = z°.

x : y = 2 : 3

Now, in ∆ABC,

Exterior ∠ACD = ∠A + ∠B

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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