Here, 125° + m = 180° [Linear pair]
⇒ m = 180 – 125° = 55°
and 125°+n=180° [Linear pair]
⇒ n=180°-125° = 55°

∵ Exterior angle x° = Sum of opposite interior angles
∴x=55°+55°=110°

(b) Sum of angles of a pentagon = (n-2) x 180°
= (5-2) x 180°
= 3×180°=540°
By linear pairs of angles,
∠1 +90° = 180° …………(i)
∠2 +60° = 180° …………(ii)
∠3 +90° = 180° …………(iii)
∠4 +70° =180° …………(iv)
∠5 +x = 180° …………(v)

Adding eq. (i), (ii), (iii), (iv) and (v),

x + ( ∠1 + ∠2 + ∠3 + ∠4 +∠5) + 310° = 900
⇒ x+540°+310°=900° ⇒x+850°=900° ⇒x=900°-850°=50°

Class 8 Maths Chapter 3 Exercise 3.2 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-3/