# 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) $90^o$"

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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 $70^o$.

Updated on 10-Oct-2022 13:23:37