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, \[\angle HCX=68\frac{1{}^\circ }{2}-43{}^\circ =25\frac{1{}^\circ }{2}\]You need to login to perform this action.
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