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If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^o$, then $∠POA$ is equal to
(a) $50^o$
(b) $60^o$
(c) $70^o$
(d) $80^o$
Given:
Tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle of $80^o$.
To do:
We have to find $∠POA$.
Solution:
We know that,
The tangent to a circle is perpendicular to the radius through the point of contact.
The lengths of the tangents from an external point are equal.
This implies,
In $\triangle OAP$ and $\triangle OBP$,
$OA = OB$
$PA = PB$
$OP = OP$
Therefore, by SSS congruence,
$\triangle OAP \cong \triangle OBP$
This implies,
$\angle AOB + \angle APB = 180^o$
$\angle AOB + 80^o = 180^o$
$\angle AOB = 180^o - 80^o$
$\angle AOB = 100^o$
This implies,
$\angle POA=\frac{1}{2}\times100^o$
$=50^o$
Therefore, $∠POA$ is equal to $50^o$.
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