question_answer

If X is a point on the line AB, Y and Z are points outside such that \[\angle AXY=45{}^\circ \] and \[\angle YXZ=150{}^\circ \], then \[\angle AXZ\] is equal to  

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

B) \[135{}^\circ \]

C) \[150{}^\circ \]

D) \[165{}^\circ \]  

Correct Answer: D

Solution :

                \[\angle YXB={{180}^{o}}-\angle AXY\]                                 \[={{180}^{o}}-{{45}^{o}}={{135}^{o}}\]                 \[\therefore \]   \[\angle AXZ=\angle YXZ-\angle YXB\]                                 \[{{150}^{o}}-{{135}^{o}}={{15}^{o}}\] Hence    \[\angle AXZ={{180}^{o}}-\angle BXZ\]                 \[={{180}^{o}}-{{15}^{o}}={{165}^{o}}\]