Question

In figure if PQR is tangent to circle at Q whose centre is O. AB is a chord parallel to PR and $\angle BQR={70}^{\circ }$ then $\angle AQB$ is equal to

• A. ${20}^{\circ }$
• B. ${40}^{\circ }$
• C. ${35}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

Answer: (B)

${40}^{\circ }$

PQR is a tangent. Thus, $\angle OQR={90}^{\circ }$(Angle between tangent and radius)
Thus, $\angle OQB=90-\angle BQR$
$\angle OQB={20}^{\circ }$
Let the line from O meet AB at M. Now In $△MQA$ and $△MQB$
$\angle QMA=\angle QMB$ (Each 90, as $AB\parallel PR$)
$QM=QM$ (common)
$MA=MB$ (Perpendicular from centre bisects the chord)
Thus, $△QMA\cong △QMB$ (SAS rule)
Hence, $\angle MQA=\angle MQB={20}^{\circ }$
Hence, $\angle AQB={40}^{\circ }$