question_answer

In\[\Delta PQR,PS\] is the bisector of\[\angle P\] and\[PT\bot QR,\]then\[\angle TPS\]is equal to

A)  \[\angle Q+\angle R\]

B)  \[90{}^\circ +\frac{1}{2}\angle Q\]

C)  \[90{}^\circ -\frac{1}{2}\angle R\]        

D)  \[\frac{1}{2}(\angle Q-\angle R)\]

Correct Answer: D

Solution :

\[\Rightarrow \]\[\angle 1+\angle 2=\angle 3\]                     ?(i) \[\Rightarrow \]   \[\angle Q=90{}^\circ -\angle 1\] \[\angle R=90{}^\circ -\angle 2-\angle 3\] So,\[Q-\angle R=[90{}^\circ -\angle 1]-[90{}^\circ -\angle 2-\angle 3]\] \[\Rightarrow \]   \[\angle Q-\angle R=\angle 2+\angle 3-\angle 1\] \[=\angle 2+(\angle 1+\angle 2)-\angle 1\]    [from Eq. (i)] \[\Rightarrow \]   \[\angle Q-\angle R=2\angle 2\Rightarrow \frac{1}{2}(\angle Q-\angle R)=\angle TPS\]