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)) \[2\text{ }\angle TPS+90{}^\circ +90{}^\circ -\angle Q+\angle R=180{}^\circ \] \[\angle TPS=\frac{\angle \mathbf{Q}-\angle \mathbf{R}}{\mathbf{2}}\]You need to login to perform this action.
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