Two tangents ^@ PA ^@ and ^@ PB ^@ are drawn to a circle with ce | Math Question and Answer

Answer:

Step by Step Explanation:

Given:
$PA$ and $PB$ are the tangents to the circle with center $O$ .
Here, we have to find the value of $\angle APB$ .
Let us consider $\angle APB = x^\circ$ .

We know that the tangents to a circle from an external point are equal.
So, $PA = PB$ .
As $\triangle APB$ is an isosceles triangle $(PA = PB)$ , the base angles of the triangle will be equal. $\implies \angle PBA = \angle PAB$
We know that the sum of the angles of a triangle is $180^\circ$ . Thus, $\begin{aligned} & \angle APB + \angle PAB + \angle PBA = 180^\circ \\ \implies & x^\circ + 2 \angle PAB = 180^\circ && \text{[As, } \angle PBA = \angle PAB] \\ \implies & \angle PAB = \dfrac { 1 }{ 2 } (180^\circ - x^\circ) = \bigg(90^\circ - \dfrac { 1 }{ 2 } x^\circ \bigg) \end{aligned}$
$PA$ is a tangent and $OA$ is the radius of the circle with center $O$ .
We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
$\begin{aligned} \implies & \angle OAP = 90^\circ \\ \implies & \angle OAB + \angle PAB = 90^\circ \\ \implies & \angle OAB = 90^\circ - (90^\circ - \dfrac { 1 }{ 2 } x^\circ) \\ \implies & \angle OAB = \dfrac { 1 }{ 2 } x^\circ = \dfrac { 1 }{ 2 } \angle APB \\ \implies & \angle APB = 2 \angle OAB \end{aligned}$
Hence, $\angle APB = 2 \angle OAB.$

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