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

Given:

From an external point \( P \), tangents \( P A=P B \) are drawn to a circle with centre \( O \).

\( \angle P A B=50^{\circ} \).

To do:

We have to find \( \angle A O B \).

Solution:

$PA = PB$    (Tangents drawn from an external point are equal)

$\angle PBA = \angle PAB = 50^o$     (Angles equal to opposite sides)

In $\triangle APB$,

$\angle PBA + \angle PAB + \angle APB =180^o$

$\angle APB = 180^o - 50^o - 50^o = 80^o$

In cyclic quadrilateral $OAPB$,

$\angle AOB + \angle APB = 180^o$   (sum of opposite angles of a cyclic quadrilateral is $180^o$)

$\angle AOB + 80^o = 180^o$

$\angle AOB = 180^o- 80^o = 100^o$

Therefore, $\angle AOB = 100^o$.

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Updated on: 10-Oct-2022

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