In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to t…

Question

In the given figure, PA is a tangent to the circle drawn from the external point P and PBC is the secant to the circle with BC as diameter. If $\angle AOC = 130°$, then find the measure of $\angle APB$, where O is the centre of the circle.

Accepted Answer

## Model Answer **Step 1:** Since BC is a diameter, ∠BOC = 180°. ∠AOC = 130° (given) ∴ ∠AOB = 180° − 130° = 50° **Step 2:** The inscribed angle (angle in alternate segment) or using the property: ∠ABC = ½ × ∠AOC = ½ × 130° = 65° (angle at centre = twice angle at circumference) **Step 3:** Since PA is a tangent, OA ⊥ PA, so ∠OAP = 90°. In △OAB: ∠OAB = 90° − ∠OAP ... Using the tangent–secant angle: ∠APB = ∠ABC − ∠PAB ... Using the standard result: ∠APB = ½ |∠BOC − ∠AOB| ... **Correct approach:** ∠AOB = 50°, so arc AB subtends ∠ACB = 25° at circumference. ∠ABP: Since BC is diameter, ∠BAC = 90° (angle in semicircle). In △PAB: ∠PAB = 90° (tangent ⊥ radius OA, and ∠OAB relates...) **Direct method:** ∠APB = ½(∠AOC − ∠AOB) ... ∠OAP = 90°, ∠AOB = 50°, ∠AOC = 130° ∠ABC (angle in segment) = ½ × ∠AOC = 65° ∠BAP = ∠OAP − ∠OAB; ∠OAB = ∠OBA = (180°−50°)/2 = 65° ∴ ∠BAP = 90° − 65° = 25° In △APB: ∠APB = 180° − ∠ABP − ∠BAP = 180° − 65° − 90°... ∠APB = 180° − 65° − (180°−65°−∠APB)... **∠APB = 25°** (In…

Mathematics — Standard (041) — CBSE Class 10 board question