Question

In the given figure EF//DQ and AB//CD. If <FEB = 64°, <PDC = 23° then find <PDQ, <AED and <DEF.​

Answer 1

Explanation:

AED = PDC = 23° ( Alternate angle)

angle DEF = 180° -(64° + 23°)

= 180° - 87

= 93°

angle DEF = angle PDQ = 93° (alternate angles)