question_answer

In the given figure below. PQ is a diameter of the circle whose centre is at O. If \[\angle \mathbf{ROS}=\mathbf{4}{{\mathbf{6}}^{{}^\circ }}\]and OR is a bisector of \[\angle \mathbf{PQR}\], then \[\angle \mathbf{RTS}\] is equal to,

A)  \[{{46}^{{}^\circ }}\]                        

B)  \[{{64}^{{}^\circ }}\]

C)  \[{{69}^{{}^\circ }}\]                       

D)  \[{{67}^{{}^\circ }}\]

Correct Answer: D

Solution :

(d):" Since, OR is a bisector of \[\angle PRQ\]. \[\therefore \]\[\angle PRO=\angle ORQ={{45}^{{}^\circ }}\] Also,  OP = OR \[\therefore \] \[\angle OPR={{45}^{{}^\circ }}\] In \[\Delta ORS\], \[OR=OS\Rightarrow \angle ORS=\angle OSR=\frac{{{180}^{{}^\circ }}-~{{46}^{{}^\circ }}}{2}={{67}^{{}^\circ }}\] \[\therefore \]\[\angle MRS={{67}^{{}^\circ }}-{{45}^{{}^\circ }}={{22}^{{}^\circ }}\] \[\Rightarrow \angle PRS={{90}^{{}^\circ }}+{{22}^{{}^\circ }}={{112}^{{}^\circ }}\] By properties of cyclic quadrilateral, \[\angle PRS+\angle PQS=180{}^\circ \] \[\Rightarrow \angle PQS={{180}^{{}^\circ }}-{{112}^{{}^\circ }}-{{68}^{{}^\circ }}\] In \[\Delta PTQ\],     \[\angle QPT+\angle PQT+\angle PTQ={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle PTQ-{{180}^{{}^\circ }}-{{45}^{{}^\circ }}-{{68}^{{}^\circ }}={{67}^{{}^\circ }}\]