The sum of the interior angles of any quadrilateral is 360^°. In the quadrilateral OAPB, we have four angles: OAP, APB, OBP, and AOB.

Substitute the known values into the equation: OAP = 90^°, APB = 70^°, and OBP = 90^°. Then, solve for AOB by subtracting the sum of the known angles from 360^°.

If $PA$ and $PB$ are tangents from an external point $P$ to a circle with centre $O$ such that $\angle APB = 70^\circ$ , find $\angle AOB$

Answer: The measure of $\angle AOB$ is $110^\circ$ .

Step-by-step solution

Step 1: Identify properties of tangents and radii

We know that the radius drawn to the tangent at the point of contact is perpendicular to the tangent. Therefore, the angle between the radius $OA$ and tangent $PA$ is $90^\circ$ , and similarly, the angle between radius $OB$ and tangent $PB$ is $90^\circ$ .

Step 2: Apply angle sum property of quadrilateral

The sum of the interior angles of any quadrilateral is $360^\circ$ . In the quadrilateral OAPB, we have four angles: $\angle OAP$ , $\angle APB$ , $\angle OBP$ , and $\angle AOB$ .

Step 3: Substitute known values and solve for $\angle$ AOB

Substitute the known values into the equation: $\angle OAP = 90^\circ$ , $\angle APB = 70^\circ$ , and $\angle OBP = 90^\circ$ . Then, solve for $\angle AOB$ by subtracting the sum of the known angles from $360^\circ$ .

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If PAPAPA and PBPBPB are tangents from an external point PPP to a circle with centre OOO such that ∠APB=70∘\angle APB = 70^\circ∠APB=70∘, find ∠AOB\angle AOB∠AOB