What is not using equal radii?

Two radii make an isosceles triangle with equal base angles - an easy extra fact to forget.

SingaporeMathsSyllabus dot point

What special angle rules hold inside a circle, such as the angle in a semicircle and the radius-tangent angle?

Use the basic angle properties of the circle, including the angle in a semicircle and the angle between a tangent and a radius

A focused answer to the N(A)-Level Mathematics outcome on circle angles. The angle in a semicircle, the tangent-radius right angle, equal radii forming isosceles triangles, and finding unknown angles.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use the basic angle properties of a circle to find unknown angles: the angle in a semicircle, the right angle between a tangent and a radius, and the isosceles triangles formed by two radii. These properties, combined with the triangle angle sum, solve most N(A)-Level circle questions.

The answer

Parts of a circle

A few names you need: the centre is the middle point; a radius joins the centre to the circle; a diameter is a chord through the centre (twice the radius); a chord joins two points on the circle; and a tangent is a line that touches the circle at exactly one point.

The angle in a semicircle

If a triangle is drawn inside a circle so that one side is a diameter, then the angle at the point on the circle (opposite the diameter) is a right angle:

\text{angle in a semicircle} = 90^\circ

This is one of the most useful circle facts, because it creates a right-angled triangle you can then solve.

The tangent and radius

A tangent touches the circle at one point. At that point of contact, the tangent is perpendicular to the radius:

\text{angle between tangent and radius} = 90^\circ

So any radius drawn to the point where a tangent touches makes a right angle with the tangent.

Two radii form an isosceles triangle

Because all radii of a circle are equal, a triangle with two sides that are radii is isosceles. Its two base angles (opposite the equal radii) are therefore equal. This often supplies an extra equal angle in a circle problem.

Combining with the angle sum

Most circle questions use one property to find a right angle or an equal angle, then finish with the triangle angle sum of $180^\circ$ . Work one step at a time, naming each property.

Worked example

$AB$ is a diameter of a circle with centre $O$ , and $C$ is a point on the circle. The two radii $OA$ and $OC$ are drawn, and angle $OCA = 25^\circ$ . Find angle $COB$ .

Step 1: Use the equal radii

$OA$ and $OC$ are both radii, so triangle $OAC$ is isosceles with equal base angles. Thus $\angle OAC = \angle OCA = 25^\circ$ .

Step 2: Find angle AOC with the angle sum

In triangle $OAC$ :

\angle AOC = 180^\circ - 25^\circ - 25^\circ = 130^\circ.

Step 3: Use angles on a straight line

$AB$ is a diameter, so $A$ , $O$ and $B$ lie on a straight line. Angles $AOC$ and $COB$ are on this line:

\angle COB = 180^\circ - 130^\circ = 50^\circ.

Step 4: State the result

Angle $COB$ is $50^\circ$ , found using the equal radii, the triangle angle sum, and angles on a straight line.

Examples in context

Example 1. A wheel and its spoke. A straight road just touching a circular wheel is a tangent, and a spoke to the contact point is a radius, so the spoke meets the road at $90^\circ$ . This tangent-radius right angle appears whenever a line grazes a circle, such as a belt around a pulley.

Example 2. A triangle in a semicircle. A triangle drawn in a semicircle with the flat side as the diameter is automatically right-angled at the curved vertex. Designers use this to guarantee a right angle, and in exams it instantly gives you a $90^\circ$ angle to work with.

Try this

Cue. A triangle has a diameter as one side. The angle opposite the diameter is $90^\circ$ (angle in a semicircle).
Cue. A tangent meets a radius at the contact point. The angle between them is $90^\circ$ .
Cue. Two radii and a chord form a triangle with one base angle $40^\circ$ . The other base angle is also $40^\circ$ (isosceles, equal radii).

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marks

AB

is a diameter of a circle and

C

is a point on the circle. The angle

CAB

35^\circ

. Find the angle

ABC

Show worked answer →

The angle in a semicircle is $90^\circ$ , so the angle $ACB$ (at the point on the circle, subtended by the diameter) is $90^\circ$ .

In triangle $ABC$ , the angles sum to $180^\circ$ :

$\angle ABC = 180^\circ - 90^\circ - 35^\circ = 55^\circ$ .

What markers reward: recognising the angle in a semicircle as $90^\circ$ , then using the triangle angle sum. Stating the semicircle property is the key reasoning mark.

Original3 marksA tangent touches a circle at point

P

O

is the centre and

OP

is a radius. A line from

O

meets the tangent at

Q

, making angle

OQP = 40^\circ

. Find angle

QOP

Show worked answer →

The angle between a tangent and a radius at the point of contact is $90^\circ$ , so $\angle OPQ = 90^\circ$ .

In triangle $OPQ$ , the angles sum to $180^\circ$ :

$\angle QOP = 180^\circ - 90^\circ - 40^\circ = 50^\circ$ .

What markers reward: using the tangent-radius right angle of $90^\circ$ , then the triangle angle sum. The property that the radius meets the tangent at $90^\circ$ is the essential step.