Question

In the given figure, AB$||$CD and $\angle$F$={30}^o$. Then $\angle$FCD is ___________.

• A. ${90}^o$
• B. ${120}^o$
• C. ${60}^o$
• D. ${45}^o$

Answer 1

Answer: (B)

${120}^o$

$\angle AEF=90-\angle F={60}^{\circ }$ (as $△AFE$ is a rightangle triangle)

$\angle FCD=\angle BEF$ (From the figure, as $AB||CD$)

$\angle AEF+\angle BEF={180}^{\circ }\Rightarrow \angle BEF=180-60={120}^{\circ }$

$\therefore \angle FCD={120}^{\circ }$