Question

Find $\angle ABF$ from the given figure.

Answer 1

$\angle BCD+\angle BCH={180}^{\circ }$ [ Linear pair]

$\angle BCD$ = ${180}^{\circ }$ - ${35}^{\circ }$ = ${145}^{\circ }$

In quadrilateral BEDC,

$\angle BED$ + $\angle EDC$ + $\angle DCB$ + $\angle CBE$ = ${360}^{\circ }$

$\angle CBE$ = ${360}^{\circ }$ - (${105}^{\circ }$ + ${50}^{\circ }$ + ${145}^{\circ }$) = ${60}^{\circ }$

So, $\angle ABF$ = $\angle CBE$ = ${60}^{\circ }$(Vertically opposite angles)