Question

question no 4 and 5 ​

Answer 1

For question 4 refer to the attached image..

here's the solution of q 5

↓↓↓

If AB|| DE,

HENCE,<BAE= <AED

HENCE , <AED= 35°

NOW IN ∆ DCE, <D+<C+<E= 180°{$UM OF ANGLES OF A TRIANGLE IS 180° }

HENCE, 53°+<C+35°=180°

→88°+<C = 180°

→<C=180°-88°

<C=92°

therefore,<DCE=92°.

• HOPE IT HELPS U;-)