Question

Solve questions no. 3 in fig. 6.41​

Answer 1

$\underline \bold \red{Answer}$

∠AED=∠BAC (alternate interior angles)

⇒∠AED=35°

⇒∠CED=35°

InΔDEC

∠CDE+∠DCE+∠CED=180° (by angle sum property)

53° +∠DCE+35° =180°

∠DCE+88°=180°

∠DCE=180°−88°

∠DCE=92°