question_answer

           In the figure, AB is parallel to CD and BE is paralleled to FH. What is\[\angle FHE\]is equal to?            

A)  \[110{}^\circ \]

B)                     \[120{}^\circ \]           

C)  \[125{}^\circ \]           

D)  \[130{}^\circ \]  

Correct Answer: A

Solution :

\[\angle HEB=180{}^\circ -60{}^\circ -50{}^\circ =70{}^\circ \] Since HF || BE and HE is transversal line \[\therefore \]      \[\angle FHE+\angle HEB=180{}^\circ \](interior angle) \[\Rightarrow \]   \[\angle FHE+70{}^\circ =180{}^\circ \] \[\therefore \]                  \[\angle FHE=110{}^\circ \]