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In the figure, $PQ$ is a tangent at a point C to a circle with center O. if AB is a diameter and  $\angle CAB\ =\ 30^{o}$, find $\angle PCA$.
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Given: PQ is a tangent at a point C to a circle with center O. if AB is a diameter and $\vartriangle CAB = 30^{o}$, in the given figure.

To do: To find $\angle PCA=?$

Solution:

In the given figure,

In $\vartriangle\ ACO,$

$OA\ =\ OC\ \ \ \ \ \ \ ( Radii\ of\ the\ same\ circle)$

$\vartriangle\ ACO$ is an isosceles triangle.

$\angle CAB\ =\ 30^{o}\ \dotsc ( Given)$

$\angle CAO\ =\angle ACO\ =\ 30^{o}\ \ \ \ \ \ \\ ( angles\ opposite\ to\ equal\ sides\ of\ an\ isosceles\ triangle\ are\ equal)$

$\angle PCO\ =\ 90^{o}\ \ \  \ \ \ \ (  radius\ drawn\ at\ the\ point\ of\ contact\ is\ perpendicular\ to\ the\ tangent)$

Now $\angle PCA\ =\angle PCO–\angle CAO$

$\angle PCA=\ 90^{o} –30^{o}=\ 60^{o}$

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Updated on: 10-Oct-2022

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