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

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

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

Correct Answer: D

Solution :

            \[\angle 1+\angle 2=\angle 3\]             \[\angle Q={{90}^{o}}-\angle 1\]             \[\angle R={{90}^{o}}-\angle 2-\angle 3\] So,       \[\angle Q-\angle R=({{90}^{o}}-\angle 1)\]        \[-({{90}^{o}}-\angle 2-\angle 3)\] \[\Rightarrow \]   \[\angle Q-\angle R=({{90}^{o}}-\angle 1)\]                  From (i)             \[=\angle 2+(\angle 1+\angle 2)-\angle 1\] \[\Rightarrow \]   \[\angle Q-\angle R=2\angle 2\] \[\Rightarrow \]   \[\frac{1}{2}(\angle Q-\angle R)=\angle TPS\]