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**Properties of Triangles – Maharashtra Board Class 6 Solutions for Mathematics (English Medium)**

MathematicsGeneral ScienceMaharashtra Board Solutions

### Exercise-76

**Solution 1:**

By the property of exterior angle of a triangle,

m∠XZP = m∠XYZ + m∠YXZ

∴ 90° = m∠XYZ + 55°

∴ m∠XYZ = 90° – 55°

∴ m∠XYZ = 35°

**Solution 2:**

By the property of exterior angle of a triangle,

m∠BCD = m∠BAC + m∠ABC

∴ 130° = m∠BAC + 60°

∴ m∠BAC = 130° – 60°

∴ m∠BAC = 70°

**Solution 3:**

The sum of the measures of the three angles of a triangle is 180°.

Thus, in ∆PST,

m∠P + m∠S + m∠PTS = 180°

∴ 60° + 80° + m∠PTS = 180°

∴ 140° + m∠PTS = 180°

∴ m∠PTS = 180° – 140°

∴ m∠PTS = 40°

Now, By the property of exterior angle of a triangle,

m∠PTN = m∠P + m∠S

∴ m∠PTN = 60° + 80°

∴ m∠PTN = 140°

**Solution 4:**

The sum of the measures of the three angles of a triangle is 180°.

Thus, in ∆RST,

m∠R + m∠S + m∠T = 180°

∴ 70° + 30° + m∠T = 180°

∴ 100° + m∠T = 180°

∴ m∠T = 180° – 100°

∴ m∠T = 80°

**Solution 5:**

By the property of exterior angle of a triangle,

m∠TRM = m∠N + m∠T

∴ m∠TRM = 30° + 80°

∴ m∠TRM = 110°

**Solution 6:**

By the property of exterior angle of a triangle,

m∠ACD = m∠A + m∠B

∴ m∠ACD = m∠A + m∠A (∵ m∠A = m∠B)

∴ m∠ACD = 2 × m∠A

∴ 140° = 2 × m∠A

∴ m∠A = 70° (dividing both sides by 2)

∴ m∠A = m∠B = 70°

### Exercise-77

**Solution 1:**

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

- l(AP) + l(BP) >
__l(AB)__ - l(AP) + l(CP)
__>__l(AC) __l(BP)__+ l(CP) > l(BC)- l(AB) +
__l(AC)__> l(BC)