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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