In the given figure, PA and PB are tangents to a circle centred at O. If \(\angle OAB = 15^\circ\), then \(\angle APB\) equals :
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A highly useful direct shortcut relation for this specific standard geometry configuration is:
\( \angle APB = 2 \cdot \angle OAB \).
Using this shortcut: \( \angle APB = 2 \times 15^\circ = 30^\circ \).
This direct formula can save a lot of time in competitive and board exams!
Step 1: Use the symmetry of the tangent-kite figure.
Since $PA$ and $PB$ are tangents from the same external point $P$, $PA=PB$, and $OA=OB$ as both are radii. This makes $OAPB$ a symmetric kite about line $OP$, so $OP$ bisects both $\angle AOB$ and $\angle APB$.
Step 2: Find angle AOB using the isosceles triangle OAB.
Since $OA=OB$, triangle $OAB$ is isosceles, so $\angle OBA=\angle OAB=15^\circ$. By the angle sum of the triangle:
\[ \angle AOB = 180^\circ-15^\circ-15^\circ = 150^\circ \]
Step 3: Halve the angle using the bisection property.
\[ \angle AOP = \frac{150^\circ}{2} = 75^\circ \]
Step 4: Use the right triangle OAP.
Since the radius is perpendicular to the tangent at the point of contact, $\angle OAP=90^\circ$. So:
\[ 90^\circ+75^\circ+\angle APO = 180^\circ \implies \angle APO = 15^\circ \]
Step 5: Double it back using the bisection property.
\[ \angle APB = 2\times\angle APO = 2\times15^\circ = 30^\circ \]
