In the given figure, PA and PB are tangents to a circle centred at O. If \(\angle AOB = 130^\circ\), then \(\angle APB\) is equal to :
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The angle between two tangents drawn from an external point to a circle and the angle subtended by the line segment joining the points of contact at the center are always supplementary.
This means: \(\angle APB + \angle AOB = 180^\circ\).
Therefore, you can directly calculate: \(\angle APB = 180^\circ - 130^\circ = 50^\circ\).
Using this relation saves time during exams!
Step 1: Split the angle at the centre using the join OP.
Join $O$ to $P$. Since $PA=PB$ (tangents from an external point) and $OA=OB$ (radii), $OP$ divides the kite $OAPB$ into two congruent right triangles, $OAP$ and $OBP$, and bisects both $\angle AOB$ and $\angle APB$. Let $\angle APO=\angle BPO=x$, so $\angle APB=2x$.
Step 2: Use the right angle at each point of tangency.
Since a tangent is perpendicular to the radius at the point of contact, $\angle OAP=\angle OBP=90^\circ$. In right triangle $OAP$:
\[ \angle AOP = 180^\circ-90^\circ-x = 90^\circ-x \]
By the same reasoning, $\angle BOP=90^\circ-x$ as well.
Step 3: Add the two halves of angle AOB.
\[ \angle AOB = (90^\circ-x)+(90^\circ-x) = 180^\circ-2x \]
Given $\angle AOB=130^\circ$:
\[ 130^\circ = 180^\circ-2x \implies 2x=50^\circ \implies x=25^\circ \]
