Question

In the given figure, AB || CD and AC || BD. If $\angle EAC={40}^{\circ },\angle FDG={55}^{\circ },\angle HAB=x^{\circ },$ then find the value of x.

• A. ${85}^{\circ }$
• B. ${75}^{\circ }$
• C. ${65}^{\circ }$
• D. ${55}^{\circ }$

Answer 1

The correct option is A ${85}^{\circ }$

$\angle DCK=\angle FDG={55}^{\circ }(corr.\angle s)$

$\therefore \angle AEC={180}^{\circ }-(\angle EAC+\angle ACE)\therefore \angle HAB=\angle AEC={85}^{\circ }(corr.\angle s)$
Hence, x = ${85}^{\circ }$