(i) 50°+ x =120° [Exterior angle property of a ∆ ]
⇒ x =120°- 50° = 70°
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 70°+ y = 180°
⇒ 120° + y = 180°
⇒ y =180° – 120° = 60°

(ii) y = 80° ……….(i) [Vertically opposite angle]
Now, 50° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 80° + y = 180° [From equation (i)]
⇒ 130° + y =180°
⇒ y =180° – 130° = 50°

(iii) 50°+ 60°= x [Exterior angle property of a ∆ ]
⇒ x =110°
Now 50° + 60°+ y =180° [Angle sum property of a ∆ ]
⇒ 110° + y =180°
⇒ y = 180° – 110°
⇒ y = 70°

(iv) x = 60° ……….(i) [Vertically opposite angle]
Now, 30° + x + y =180° [Angle sum property of a ∆ ]
⇒ 50° + 60° + y =180° [From equation (i)]
⇒ 90° + y =180°
⇒ y =180° – 90° = 90°

(v) y = 90° ……….(i) [Vertically opposite angle]
Now, y + x + x =180° [Angle sum property of a ∆ ]
⇒ 90°+ 2x =180° [From equation (i)]
⇒ 2x = 180°- 90°
⇒ 2x = 90°
⇒ x = 90°/2 = 45°

(vi) x = y ……….(i) [Vertically opposite angle]
Now, x + x + y =180° [Angle sum property of a  ]
⇒ 2x + x =180° [From equation (i)]
⇒ 3x = 180°
⇒ 180° / 3 = 60°

Class 7 Maths Chapter 6 Exercise 6.3

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