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# In figure, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $âˆ POQ = 110^o$, then $âˆ PTQ$ 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: â€Š

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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