Question

Please answer these question with method.

Answer 1

angle FQB=angle DPQ ( corresponding angles)
angle QPC+angle DPQ= 180°(linear pairs)
(5x)°+(4x)°=180°
( 9x)=180°
x=180/9
x= 20°
angle DPQ=4x= 4×20=80......hope it helps u............mark as brainliest

Answer 2

Given: AB║CD and EF is the transversal.

To find: The measure of ∠DPQ.

Answer:

∠FQB = 4x° = ∠AQP (Vertically Opposite Angles are equal).

∠AQP + ∠CPQ = 180° (Co - interior Angles are Supplementary).

4x° + 5x° = 180°

9x° = 180°

x° = 180°/9

x = 20°

Therefore ∠AQP = 4*20 = 80° and ∠CPQ = 5*20 = 100°.

∠CPQ + ∠DPQ = 180° (Linear Pair).

100° + ∠DPQ = 180°

∠DPQ = 180° - 100°

∠DPQ = 80°

Hope it helps :)