Question

If the tangents PA and PB from an external point P to circle with centre O are inclined to each other at an angle of ${40}^0$, then $\angle$POA =

• A. 70°
• B. 80°
• C. 50°
• D. 60°

Answer 1

The correct option is A

70°

$\Delta$POA $\cong$\ (\Delta\)POB (R.H.S)

So, $\angle$OPB = ${20}^0$ (C.P.C.T)

$\angle$POA = ${70}^0$ (sum of angles of $\Delta$ is ${180}^0$)