^@ PA ^@ and ^@ PB ^@ are tangents to a circle with center ^@ O | Math Question and Answer

Answer:

Step by Step Explanation:

Given:
$PA$ and $PB$ are the tangents to the circle with center $O$ from an external point $P$ .

Here, we have to check if quadrilateral $AOBP$ is cyclic or not.

We know that in a cyclic quadrilateral the sum of opposite angles is $180^\circ$ .
Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Thus, $\begin{aligned} & PA \perp OA \implies \angle OAP = 90^\circ \\ & \text{ and } \\ & PB \perp OB \implies \angle OBP = 90^\circ \end{aligned}$ So, $\angle OAP + \angle OBP = 90^\circ + 90^\circ = 180^\circ \ldots \text{(i)}$
Now, the sum of all the angles of a quadrilateral is $360^\circ$ . Thus, $\begin{aligned} & \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \\ & \angle AOB + \angle APB + ( \angle OAP + \angle OBP )= 360^\circ \\ \implies & \angle AOB + \angle APB = 180^\circ && \text{ [Using } eq \text{(i)] } \end{aligned}$
Both pairs of opposite angles have the sum $180^\circ$ . Thus, we can say that quadrilateral $AOBP$ is $cyclic.$

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