Question

AB II DE BAC = 35° AND CDE=53°FIND DCE

​

Answer 1

Given, AB∥DE,

∠BAC=∠CED=35°

(Alternate angles of parallel lines AB and DE)

In △CDE,

∠CDE+∠DCE+∠CED=180°

53°+35°+∠DCE=180°

∠DCE=180°−88°

∠DCE=92°