Question

In the given figure if PA and PB are tangents to the circle with center O such that \[\angle APB={{50}^{0}}\], then $\angle OAB$ is equal to?
(a) ${{25}^{0}}$
(b) ${{30}^{0}}$
(c) ${{40}^{0}}$
(d) ${{50}^{0}}$

Answer 1

Hint: Here, we will use the property that the sum of all the angles of a quadrilateral is equal to ${{360}^{0}}$ to find $\angle AOB$. Using the value of $\angle AOB$ we will find $\angle OAB$.

Complete step-by-step solution -

A tangent is a line which touches a circle at exactly one point. An important is the result is that the radius from the center of a circle to the point of tangency is perpendicular to the tangent line. Also, we know that the angles opposite the equal sides of a triangle are equal to each other.
Here, the tangent is PA and the radius is OA. Therefore, we can write that $OA\bot PA$. Similarly, PB is also a tangent and OB is the radius. So, $OB\bot PB$.
It means that $\angle OAP=\angle OBP={{90}^{0}}...........\left( 1 \right)$
Since, the sum of all the angles of a quadrilateral is equal to${{360}^{0}}$ and OAPB is a quadrilateral. Therefore, we get:
$\angle OAP+\angle APB+\angle OBP+\angle AOB={{360}^{0}}...........\left( 2 \right)$
 On putting the values of $\angle OAP$ and $\angle OBP$ from equation (1) in equation (2), we get:
${{90}^{0}}+\angle APB+{{90}^{0}}+\angle AOB={{360}^{0}}$
Since, the value of $\angle APB$ is give as ${{50}^{0}}$. So, we get:
$\begin{align}
  & {{90}^{0}}+{{50}^{0}}+{{90}^{0}}+\angle AOB={{360}^{0}} \\
 & \Rightarrow {{230}^{0}}+\angle AOB={{360}^{0}} \\
 & \Rightarrow \angle AOB={{360}^{0}}-{{230}^{0}}={{130}^{0}}...........\left( 3 \right) \\
\end{align}$
Since, OA and OB both are the radius of the same circle, they must be equal , so OA =OB.
In the triangle OAB:
Since the angles opposite to equal sides of a triangle are equal, we get:
Angle opposite to side OA = angle opposite to side OB.
So, $\angle OAB=\angle OBA$
We know that the sum of all the angles of a triangle is ${{180}^{0}}$.
Let us consider that angle OAB measures ‘x’, then angle OBA also measures x.
So, in this triangle OAB, we have:
$\begin{align}
  & \angle OAB+\angle OBA+\angle AOB={{180}^{0}} \\
 & \Rightarrow x+x+{{130}^{0}}={{180}^{0}} \\
 & \Rightarrow 2x={{180}^{0}}-{{130}^{0}} \\
 & \Rightarrow x=\dfrac{{{50}^{0}}}{2}={{25}^{0}} \\
\end{align}$
So, \[\angle OAB=\angle OBA={{25}^{0}}\].
Hence, option (a) is the correct answer.

Note: Students should note here that the sum of all the angles of a quadrilateral is equal to 360 degrees. It should also be kept in mind that the radius of a circle is always is perpendicular to the tangent of the circle.