Question

In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If $\angle$ POQ = ${70}^o$, then $\angle$ TPQ is equal to

(a) ${35}^o$ (b) ${45}^o$ (c) ${55}^o$ (d) ${70}^o$

Answer 1

Given a circle with centre O. PQ is a chord and PT is tangent.∠POQ = 70°

To find: $\angle TPQ$

Solution:

In \Delta POQ, OP = OQ = radii

Since, angles opposite to equal sides are equal,

$\angle OPQ=\angle OQP$

By angle sum property,

$\angle OPQ+\angle POQ+\angle OQP={180}^o$

$2\angle OPQ+{70}^o={180}^o$

$2\angle OPQ={110}^o$

$\angle OPQ={55}^o$

Now, since the radius is perpendicular to the tangent at the point of contact,

$\angle TPO={90}^o$

$\angle TPQ=\angle TPO-\angle OPQ\angle TPQ={90}^o-{55}^o={35}^o$

$\angle TPQ={35}^o$