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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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