Question

In the figure above (not to scale) two circles $C_1$ and $C_2$ intersect at S and Q PQN and RQM are tangents drawn to $C_1$ and $C_2$ respectively at Q MAB and ABN are the chords of the circles $C_1$ and $C_2$ If $\angle NQR={85}^{\circ }$ then find $\angle AQB$

• A. ${10}^{\circ }$
• B. ${15}^{\circ }$
• C. ${20}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is A ${10}^{\circ }$

Let us denote the centre of $C1$ as O.

Draw OQ as one of the radii and OM as the second radii.

Now as OQ is the radii, it is perpendicular to NQP.

So, $\angle OQM=5^{\circ }$ and $\angle MQP={90}^{\circ }-\angle OQM={85}^{\circ }$,

Now $\angle OMQ=\angle OQM$ as $OQ=OM$

being the radii of $C_1.$

Now , $\angle MOQ={180}^{\circ }-5^{\circ }-5^{\circ }={170}^{\circ }$, ( Sum of all angles of a triangle is ${180}^{\circ }$.)

so angle subtended at the center by the chord MQ is ${170}^{\circ }$,

and $\angle MBQ=\frac{1}{2}\angle MOQ={85}^{\circ }$..

Similarly, $\angle A=\angle B={85}^{\circ }$.

which implies $\angle AQB={10}^{\circ }$

therefore option A will be the answer.