Question

Triangle PAB is formed by three tangents to circle O and $measuredangleAPB={40}^0$; then angle AOB equals:

• A. ${45}^0$
• B. ${50}^0$
• C. ${55}^0$
• D. ${60}^0$
• E. ${70}^0$

Answer 1

The correct option is E ${70}^0$

construction join $OP$

Now, $PR=PT$ so $PO$ is the angle bisector of $\angle RPA$

so, $\angle OPR$=${20}^0$.

alos $\angle ORP$=$\angle OTP$=${90}^0$(radius and tangent are perpendicular to each other)

In triangle ORP,

${90}^0$+${20}^0$+$\angle POR$=${180}^0$ (angle sum property)

$\angle POR$=${70}^0$

also $\angle ROP$=$\angle POT$(equal tangent subtends equal angles at the center)

similarly $\angle ROB$=$\angle BOS$ and $\angle TOA$=$\angle AOS$

$\angle ROB$+$\angle BOS$+$\angle AOS$+$\angle TOA$=${70}^0$+${70}^0$=${140}^0$

$2\angle BOS$+$2\angle AOS$=${140}^0$

so, angle$\angle AOB=$70^{0}$

so option E is correct.