A) \[45{}^\circ \]
B) \[60{}^\circ \]
C) \[50{}^\circ \]
D) \[55{}^\circ \]
Correct Answer: C
Solution :
(c): In a figure, AOC is a diameter of the circle. Using the property that, diameter subtends an angle \[90{}^\circ \] to the circle. i.e., \[\angle PQR=90{}^\circ \] In \[\Delta \,PQR\], \[\angle P+\angle Q+\angle R=180{}^\circ ;\] \[\Rightarrow \]\[\text{ }\angle P+90{}^\circ +50{}^\circ =180{}^\circ \] \[\text{ }\angle P+180{}^\circ -140{}^\circ =40{}^\circ \] \[\because \] OP is perpendicular to PT; \[\therefore \]\[\angle OPT=90{}^\circ \] \[\Rightarrow \]\[\angle OPQ+\angle QPT=90{}^\circ ;\] \[\Rightarrow \]\[\angle QPT=90{}^\circ -40{}^\circ =50{}^\circ \]You need to login to perform this action.
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