(i) In ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180° [By angle sum property of a triangle]
⇒ x +50° + 60° =180°
⇒ x +110°+180°
⇒ x =180°-110° = 70°

(ii) In ∆ PQR,
∠RPQ + ∠PQR + ∠RPQ = 180° [By angle sum property of a triangle]
⇒ 90° + 30° + x =180°
⇒ x = 120°= 180°
⇒ x =180°- 120° = 60°

(iii) In ∆ XYZ,
∠ZXY + ∠XYZ + ∠YZX = 180° [By angle sum property of a triangle]
⇒ 30° +110° + x = 180°
⇒ x + 140°= 180°
⇒ x =180°- 140°= 40°

(iv) In the given isosceles triangle,
x + x + 50° = 180° [By angle sum property of a triangle]
⇒ 2x + 50° = 180°
⇒ 2x =180° – 50°
⇒ 2x =130°
⇒x=130°/2 = 65°

(v) In the given equilateral triangle,
x + x + x = 180° [By angle sum property of a triangle]
⇒ 3x = 180°
⇒ x = 180° / 3 = 60°

(vi) In the given right angled triangle,
X + 2x + 90° = 180° [By angle sum property of a triangle]
⇒ 3x + 90°=180°
⇒ 3x =180°-90°
⇒3x=90°
⇒ x = 90°/3 = 30°

Class 7 Maths Chapter 6 Exercise 6.2

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-6/