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
Show 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 $\tfrac12(180^\circ-\angle AOB)$.
Let the tangents meeting at $P$ touch the circle at $B$ and $A$. For two tangents meeting at an external point, the angle between them is \[ \angle APB \;=\; 180^\circ-\angle AOB . \] Thus $\angle AOB=180^\circ-40^\circ=140^\circ$. Radii $OA$ and $OB$ are perpendicular to the respective tangents, hence each bisects the angle between the tangent through it and the line joining $O$ to $P$. Consequently, the acute angle between $OA$ and $OB$ is \[ \angle BOA \;=\; \frac{1}{2}\big(180^\circ-\angle AOB\big) = \frac{1}{2}\big(180^\circ-140^\circ\big) = 70^\circ . \]
