question_answer

                         ABC is an isosceles triangle in which AB=AC, CH=CB and HK ||BC. If the exterior\[\angle CZX=137{}^\circ ,\]then what is the measure of \[\angle HCK?\]

A)  \[68\frac{1{}^\circ }{2}\]                    

B)  \[43{}^\circ \]

C)  \[25\frac{1{}^\circ }{2}\]

D)  \[137{}^\circ \]

Correct Answer: C

Solution :

\[\angle CAX=137{}^\circ \] \[\therefore \]      \[\angle ABC=\frac{1}{2}(137{}^\circ )=68\frac{1{}^\circ }{2}\] \[\therefore \] Again, BC = CH and \[\angle ABC=68\frac{1{}^\circ }{2}\] Therefore,                      \[\angle CHB=68\frac{1{}^\circ }{2}\] Therefore,                      \[\angle HCB=43{}^\circ \] Hence, \[\angle HCX=68\frac{1{}^\circ }{2}-43{}^\circ =25\frac{1{}^\circ }{2}\]