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 \[\Delta PTQ\], \[\angle QPT+\angle PQT+\angle PTQ={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle PTQ-{{180}^{{}^\circ }}-{{45}^{{}^\circ }}-{{68}^{{}^\circ }}={{67}^{{}^\circ }}\]You need to login to perform this action.
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