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 :

PS is bisector \[\angle QPS=\angle SPR\] \[\angle PST=\angle SPR+\angle R\]                  ..... (i) \[\angle QPT=90{}^\circ -\angle Q\]                     ..... (ii) Now, In \[\Delta \,PTS\] \[\angle TPS+90{}^\circ +\angle PST=180{}^\circ \] \[\angle TPS+90{}^\circ +\angle SPR+\angle R=180{}^\circ \] (From (i) \[\angle TPS+90{}^\circ +\angle QPS+\angle R=180{}^\circ \] \[\angle TPS+90{}^\circ +\angle QPT+\angle TPS+\angle R=180{}^\circ \] (from eq. (ii)) \[2\text{ }\angle TPS+90{}^\circ +90{}^\circ -\angle Q+\angle R=180{}^\circ \] \[\angle TPS=\frac{\angle \mathbf{Q}-\angle \mathbf{R}}{\mathbf{2}}\]