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

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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 $\tfrac12(180^\circ-\angle AOB)$.

Updated On: Jul 16, 2026

$50^\circ$
$55^\circ$
$60^\circ$
$70^\circ$

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The Correct Option is D

Approach Solution - 1

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 \[ \angle BOA \;=\; \frac{1}{2}\big(180^\circ-\angle AOB\big) = \frac{1}{2}\big(180^\circ-140^\circ\big) = 70^\circ . \]

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Approach Solution -2

An alternate way to reach the same result is to use the angle sum of the quadrilateral formed by the centre and the two points of tangency, instead of the "$180^\circ$ minus the angle" shortcut, and then check each option.

Option A ($50^\circ$): Since $OA$ and $OB$ are radii drawn to the points where the tangents touch the circle, each of them is perpendicular to the tangent it meets, so in quadrilateral $OAPB$, the angles at $A$ and $B$ are each $90^\circ$. With $\angle APB=40^\circ$ given, the angle sum of the quadrilateral gives the far-side angle at $O$ as $360^\circ-90^\circ-90^\circ-40^\circ=140^\circ$. Taking the required angle $\angle BOA$ as half of this (fixed by the symmetry of the figure) gives $70^\circ$, not $50^\circ$, so this option is incorrect.
Option B ($55^\circ$): This also does not match the value $70^\circ$ obtained below, so it is incorrect.
Option C ($60^\circ$): This too does not match $70^\circ$, so it is incorrect.
Option D ($70^\circ$): From the quadrilateral $OAPB$ with right angles at $A$ and $B$, \[ 360^\circ-90^\circ-90^\circ-40^\circ=140^\circ. \] By the symmetry of the figure about line $OP$, the required angle $\angle BOA=\dfrac{140^\circ}{2}=70^\circ$, matching this option.

The quadrilateral angle-sum approach confirms that $\angle BOA=70^\circ$.

Hence, the correct answer is option D: $70^\circ$.