Question

In the figure, $\triangle APB$ is formed by three tangents to a circle with centre $O$. If $\angle APB=40^\circ$, then the measure of $\angle BOA$ is 

• A. $50^\circ$
• B. $55^\circ$
• C. $60^\circ$
• D. $70^\circ$

Hint:

For two tangents meeting at $P$ with contact points $A,B$, use $\angle APB=180^\circ-\angle AOB$. The acute angle between the radii is then $\tfrac12(180^\circ-\angle AOB)$.

Answer 1

The Correct Option is D

Step 1: In quadrilateral \(OAPB\), \(OA\perp\) tangent at \(A\) and \(OB\perp\) tangent at \(B\), so both angles at \(A\) and \(B\) are \(90^\circ\).

Step 2: Angle sum of the quadrilateral: \[ 360^\circ-90^\circ-90^\circ-40^\circ=140^\circ \]

Step 3: By symmetry about \(OP\), \(\angle BOA\) is half this value: \[ \angle BOA=\frac{140^\circ}{2}=\boxed{70^\circ} \]