Question

PQ is a tangent at a point C to a circle with center O. If AB is a diameter and \[\angle CAB={{30}^{\circ }}\], find \[\angle PCA\].

Answer 1

Hint: Join OC to form two triangles out of the given triangle AOB. Consider triangle AOC and find the measure of angle OCA using the property of isosceles triangle that angle opposite to the equal sides are equal. Consider OC perpendicular to PQ by using the property of tangent that radius is perpendicular to the tangent at the point of contact. Subtract the obtained angle OCA from \[{{90}^{\circ }}\] to get the answer.

Complete step by step answer:
In the above figure, let us join OC to form triangle AOC.
It is given that angle CAB = \[{{30}^{\circ }}\], therefore in triangle OAC, we have,
\[\angle OAC={{30}^{\circ }}\]

Now, considering triangle AOC,
OA = OC = radius of the circle.
Therefore, triangle OAC is an isosceles triangle with sides OA OC equal.
Using the property of isosceles triangle that “angles opposite to the equal sides are equal”, we get,
\[\angle OAC=\angle OCA={{30}^{\circ }}\] - (1)
Now, we know that, radius of a circle is perpendicular to the tangent. Therefore,
OC is perpendicular to PQ.
\[\Rightarrow \angle OCA={{90}^{\circ }}\]
We can clearly see that,
\[\angle OCP=\angle OCA+\angle PCA\]
\[\Rightarrow {{90}^{\circ }}=\angle PCA+{{30}^{\circ }}\] - [using (1)]
\[\begin{align}
  & \Rightarrow \angle PCA={{90}^{\circ }}-{{30}^{\circ }} \\
 & \Rightarrow \angle PCA={{60}^{\circ }} \\
\end{align}\]

Note: One may note that we can also find \[\angle PCA\] by first determining the measure of \[\angle QCB\] by the same above process. Once \[\angle QCB\] is determined, the measure of \[\angle ACB\] is \[{{90}^{\circ }}\] because it is an angle in the semi – circle. Take the sum of \[\angle QCB\] and \[\angle ACB\] and subtract it from \[{{180}^{\circ }}\] to get the value of \[\angle PCA\]. But this will be a length process as we have to determine an extra angle QCB which is not necessary.