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01 Nov 2021, 08:15
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C
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In the figure shown, O is the centre of the circle and PQ, PR and ST are tangents. What is the angle QOP?
(A) 40 degrees
(B) 45 degrees
(C) 65 degrees
(D) 75 degrees
(E) 90 degrees
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1.png [ 40.96 KiB | Viewed 2888 times ]
01 Nov 2021, 09:20
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Bunuel
Half the given information is not required.
Draw OP. OP will bisect \(\angle QPR\), so \(\angle QPO=\frac{50}{2}=25.\)
Take \(\triangle OPQ\).
\(\angle OQP = 90\), as tangent QP will make a 90 degree angle with Center O.
\(\angle QPO=25\)
Thus \(\angle QOP=90-25=65\degree\)
C
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01 Nov 2021, 09:24
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Here PQO = PRO = 90 degrees as they are tangents
Now since PQ = PR and OQ = OR and the imaginary line OP is common we can consider triangle POQ and POR to be congruent
As a result POQ = POR = (180 - 50)/2 = 65 degrees
Option C
Now since PQ = PR and OQ = OR and the imaginary line OP is common we can consider triangle POQ and POR to be congruent
As a result POQ = POR = (180 - 50)/2 = 65 degrees
Option C
