How would you use the following figure to justify the statement that the angle in a semicircle is 90°?

Step-by-step solution

Idea: The equal tick marks say OP = OA = OQ (all radii). Each of the two triangles OAP and OAQ is isosceles, so its base angles are equal. Then use the angle sum of the big triangle.

1. Call the ends of the diameter P (left) and Q (right). The tick marks show OP = OA = OQ (radii).½ mark
2. In ΔOAP, OA = OP, so ∠OAP = ∠OPA = a. In ΔOAQ, OA = OQ, so ∠OAQ = ∠OQA = b.1 mark
3. So ∠PAQ = ∠OAP + ∠OAQ = a + b. Angle sum of ΔPAQ: a + b + (a + b) = 180°, so 2(a + b) = 180°.1 mark
4. a + b = 90°, so ∠PAQ = 90°: the angle in a semicircle is a right angle.½ mark

Answer to write in the exam

Common mistakes that cost marks

• Assuming a = b. They are equal only when A is at the top of the semicircle.
• Adding only a + b + 90° = 180° (assuming the answer). The angle at A must be written as a + b.
• Not explaining the tick marks: they show the three segments are radii, which is what makes the triangles isosceles.

How this can come in the exam

Try one yourself

P and Q are the ends of a diameter with centre O, and A is on the circle with ∠OAP = 40°. Find ∠PAQ and ∠AQP.

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