Arcs and angles · 2 marks

The angle subtended by a diameter at any point on the circle is 90°.

Answer: If AB is a diameter and D is any other point on the circle, the arc AB not containing D subtends a straight angle ∠ACB = 180° at the centre C, so ∠ADB = ½ × 180° = 90°.

Step-by-step solution

Given: AB is a diameter of a circle with centre C; D is a point on the circle (D ≠ A, B)
To find: Show that ∠ADB = 90°

Idea: This follows at once from the result that the angle at the centre is double the angle at the circle. For a diameter, the angle at the centre is a straight angle, 180°.

Take the arc from A to B that does not contain D. Moving from CA to CB along this arc sweeps a straight angle: ∠ACB = 180° (A, C, B are on one line).1 mark
D is on the circle outside this arc, so ∠ADB = ½∠ACB = ½ × 180° = 90°.1 mark
This is often stated as: the angle in a semicircle is a right angle.

The arc cut off by a diameter subtends 180° at the centre, so it subtends ½ × 180° = 90° at any point on the circle: the angle in a semicircle is a right angle.

Answer to write in the exam

AB is a diameter ⇒ ∠ACB = 180° (straight angle at the centre C)

D lies on the circle outside the arc AB

∠ADB = ½∠ACB (angle at the centre is double the angle at the circle)

∴ ∠ADB = ½ × 180° = 90°

Common mistakes that cost marks

Taking the arc that contains D. The angle at D is half the angle subtended by the arc not containing D.
Saying the angle at the centre is 360° for a diameter. Each semicircle subtends a straight angle, 180°.
Thinking the right angle is at the centre. The right angle is at the point D on the circle.

How this can come in the exam

MCQ (1 mark)

AB is a diameter of a circle and C is a point on the circle with ∠CAB = 35°. Then ∠CBA is

35°
55°
90°
145°

Show answer

(B) 55°
∠ACB = 90° (angle in a semicircle), so ∠CBA = 180° − 90° − 35° = 55°.

Try one yourself

PQ is a diameter of a circle of radius 6.5 cm and R is a point on the circle with PR = 5 cm. Find QR.