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In figure, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $∠POQ = 110^o$, then $∠PTQ$ is equal to
(a) $60^o$
(b) $70^o$
(c) $80^o$
(d) $90^o$"
Given:
$TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $∠POQ = 110^o$.
To do:
We have to find $∠PTQ$.
Solution:  
We know that,
The tangent to a circle is perpendicular to the radius through the point of contact.
This implies,
$\angle OPT = 90^o$
$\angle OQT = 90^o$
This implies,
$POQT$ is a quadrilateral.
Therefore,
$\angle PTQ + \angle POQ = 180^o$
$\angle PTQ + 110^o = 180^o$
$\angle PTQ = 180^o - 110^o$
$\angle PTQ = 70^o$
Therefore, $\angle PTQ$ is equal to $70^o$.
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