Question

Triangle PAB is formed by three tangents to circle O and $\angle APB={40}^{\circ }$ then angle AOB

• A. ${45}^{\circ }$
• B. ${50}^{\circ }$
• C. ${60}^{\circ }$
• D. ${70}^{\circ }$

Answer 1

Answer: (D)

${70}^{\circ }$

In the given figure,
$\angle APB={40}^{\circ }$
Now, in quadrilateral ORPT,
$\angle ORP=\angle OTP={90}^{\circ }$ (Angle between tangent and the radius)
Hence, $\angle TPR+\angle ROT=180$ (Angle sum property)
$40+\angle ROT=180$
$\angle ROT={140}^{\circ }$
Now, join OQ.
In $△OQB$ and $△OBR$
$\angle OQB=\angle ORB$ (Each 90)
$OB=OB$ (Common)
$OQ=OR$ (Radius of circle)
thus, $△OQB\cong △ORB$(SAS rule)
Hence, $\angle QOB=\angle ROB=x$
Similarly, $\angle QOA=\angle TOA=y$
We know, $\angle TOA=140$
$\angle QOB+\angle ROB+\angle TOA+\angle QOA=140$
$2(x+y)=140$
$x+y={70}^{\circ }$
$\angle AOB={70}^{\circ }$