Mathematics — CBSE Class 10 board question

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 △PAB: ∠PAB = 25°, ∠ABP = 180°−65° = 115°, ∴ ∠APB = 180°−115°−25° = 40°)

Let me redo cleanly:

• ∠AOB = 50° ⟹ ∠ACB = 25° (angle at circumference on major arc)
• ∠OAP = 90° (tangent); ∠OAB = ∠OBA = 65° (isosceles, ∠AOB=50°) ⟹ ∠PAB = 90°−65° = 25°
• ∠ABC = 65°, so ∠PBC = 180°−65° = 115°
• In △PAB: ∠APB = 180°−25°−115° = 40°

$$\boxed{\angle APB = 40°}$$

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Key steps examiners expect:

1. Find ∠AOB = 180° − 130° = 50° (since ∠AOC + ∠AOB = ∠BOC = 180°, as BC is diameter).
2. ∠ABC = ½ ∠AOC = 65° (central angle theorem); ∠PBC = 115° (straight line).
3. ∠OAB = 65° (isosceles △OAB); ∠PAB = 90° − 65° = 25° (since OA ⊥ PA).
4. Angle sum in △PAB gives ∠APB = 180° − 115° − 25° = 40°.