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In fig. 1, O is the centre of a circle, PQ is a chord and PT is the tangent on P. if $\angle POQ=70^{o}$, then $\angle TPQ$ is equal to:
$( A) \ 55^{o}$
$( B) \ 70^{o}$
$( C) \ 45^{o}$
$( D) \ 35^{o}$"
Given: fig.1 a circle with centre O, a chord PQ, a tangent PT on P and ∠POQ=70°
To do: To find out $\angle TPQ=?$
Solution:
here in the given figure
O is the centre of the circle
OP is the radius of the circle and also OQ is the radius of the circle.
$\therefore \ OP=OQ$
In $\vartriangle OPQ$, OP and OQ are equal.
$\therefore \ \angle OPQ=\angle OQP$
$\Rightarrow \angle POQ+\angle OPQ+\angle OQP=180^{o}$
$\Rightarrow 70^{o} +\angle OPQ+\angle OQP=180^{o}$ $( \because \angle POQ=70^{o}\ as\ given\ in\ the\ question)$
$\Rightarrow 2\angle OPQ=180^{o} -70^{o} =110$ $( \because \angle OPQ=\angle OQP)$
$\Rightarrow \angle OPQ=\frac{110^{o} }{2} =55^{o}$
It is also given the question that TP is a tangent on P.
$\therefore\ \angle OPT=90^{o}$
And $\angle OPQ+\angle TPQ=\angle OPT=90^{o}$
$\Rightarrow 55^{o} +\angle TPQ=90^{o}$
$\Rightarrow \angle TPQ=90^{o} -55^{o} =35^{o}$
$\therefore$ Option $( D)$ is correct.
Updated on: 10-Oct-2022
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