Question

From an external point P, tangents PA=PB are drawn to a circle with centre O. If $\angle PAB={50}^{\circ }$, then find $\angle AOB.$

Answer 1

In ∆PAB given that,

$\angle$PAB is 50°

and $\angle$ PAO=90°

PAO= $\angle$PAB + $\angle$OAB

90 = 50 + $\angle$OAB

then $\angle$ OAB=40°

Now, in $\Delta OAB$

OA = OB
Thus, $\angle$ OAB and $\angle$OBA will be equal

then in $\Delta OAB$

$\angle$OAB + $\angle$OBA + $\angle$AOB = ${180}^{\circ }$

40 + 40 + $\angle$AOB = ${180}^{\circ }$

$\angle$AOB =180 - 80

$\angle$AOB = 100°