Question

In the given figure, O is the centre of the circle, P and Q are the points on the circle. AB is the tangent drawn for the circle at P. If $\angle POQis{\theta }^{\circ }$, then what is the value of cosec $(\theta -10{)}^{\circ }?$

• A. 1
• B. 100
• C. 0
• D. -1

Answer 1

The correct option is A

1

We know that, radius of the circle will always be perpendicular to the tangent at the point of contact.
$\ast \angle OPB={90}^{\circ }\ast \angle OPQ+\angle QPB={90}^{\circ }(Given\angle QPB={50}^{\circ })\ast \angle OPQ={40}^{\circ }$
Since OP = OQ (Radili of the circle, their opposite angless also are equal)
$\ast \angle OPQ=\angle OQP={40}^{\circ }\ast \angle POQ={180}^{\circ }-(\angle OPQ+\angle OQP)\ast \angle POQ={180}^{\circ }-({40}^{\circ }+{40}^{\circ })\ast \angle POQ={180}^{\circ }-{80}^{\circ }={100}^{\circ }\ast \theta ={100}^{\circ }cosec(\theta -10{)}^{\circ }=cosec(90{)}^{\circ }=1$