What is sign slip in the cosine rule?

The formula subtracts 2bccos A; keep the minus sign, and note that an obtuse angle gives a negative cosine.

What are calculator in radians?

Trigonometry here is in degrees, so set the calculator to degree mode.

Find the area of a triangle with sides 6\ cm and 9\ cm and an included angle of 30^. [2 marks]

In a triangle, b = 4, c = 6 and the included angle A = 60^. Find a, to 2 decimal places. [2 marks]

SEAB Original-style practice: In triangle ABC, AB = 7\ cm, AC = 9\ cm and the angle BAC = 50^. Find the length of BC, to 2 decimal places.

Two sides and the included angle are known, so use the cosine rule with BC opposite the 50^ angle. BC^2 = 7^2 + 9^2 - 2(7)(9)cos 50^ = 49 + 81 - 126 × 0.6428 = 130 - 80.993 = 49.007. BC = sqrt(49.007) = 7.0005, which is 7.00\ cm to 2 decimal places. Markers reward selecting the cosine rule for two sides and the included angle, correct substitution, and the length of BC.

SEAB Original-style practice: In triangle PQR, angle P = 40^, angle Q = 75^ and side PQ = 10\ cm (the side r opposite R). Find the length of side PR, to 2 decimal places.

First find angle R: R = 180^ - 40^ - 75^ = 65^. Side PR is opposite angle Q, and PQ = 10\ cm is opposite angle R. Use the sine rule: PRsin Q = PQsin R. PR = 10 × sin 75^sin 65^ = 10 × 0.96590.9063 = 10.657, which is 10.66\ cm to 2 decimal places. Markers reward finding the third angle, setting up the sine rule with the correct opposite pairs, and the length of PR.

SingaporeMathsSyllabus dot point

How do we find sides and angles in triangles that are not right-angled?

Apply the sine rule and the cosine rule to find sides and angles in any triangle, and find the area using the sine formula

A focused answer to the O-Level E-Maths outcome on the sine and cosine rules. When to use each rule, finding sides and angles in non-right-angled triangles, and the area of a triangle using half ab sine C.

Generated by Claude Opus 4.89 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 sine rule and the cosine rule to find sides and angles in triangles that are not right-angled, and to find a triangle's area from two sides and the included angle. Choosing the correct rule from the information given is the key decision.

The answer

The sine rule

The sine rule relates each side to the sine of its opposite angle:

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Use it when you have a side and its opposite angle paired up, plus one more piece, that is two angles and a side, or two sides and a non-included angle.

The cosine rule

The cosine rule generalises Pythagoras to any triangle:

a^2 = b^2 + c^2 - 2bc\cos A

Use it when you have two sides and the included angle (to find the third side), or all three sides (to find an angle, by rearranging for $\cos A$ ).

Choosing between the rules

Look at what is given. If a side is paired with its opposite angle, the sine rule is usually quickest. If the only angle is between two known sides, or you know all three sides, use the cosine rule. After the cosine rule gives one part, the sine rule often finishes the triangle.

The area of a triangle

When two sides and the included angle are known, the area is:

\text{area} = \frac{1}{2}ab\sin C

where $C$ is the angle between sides $a$ and $b$ . This works for any triangle, not just right-angled ones.

Worked example

In triangle $ABC$ , the three sides are $a = 8\ \text{cm}$ , $b = 5\ \text{cm}$ and $c = 7\ \text{cm}$ . Find the angle $A$ opposite the longest side, to 1 decimal place.

Step 1: Choose the cosine rule

All three sides are known and an angle is wanted, so rearrange the cosine rule for $\cos A$ .

Step 2: Rearrange the cosine rule

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{5^2 + 7^2 - 8^2}{2(5)(7)}

Step 3: Evaluate the fraction

\cos A = \frac{25 + 49 - 64}{70} = \frac{10}{70} = 0.142857\ldots

Step 4: Apply the inverse cosine

A = \cos^{-1}(0.142857) = 81.79\ldots^\circ \approx 81.8^\circ

The angle $A$ is about $81.8^\circ$ .

Examples in context

Example 1. Surveying a plot. A triangular plot of land with all three sides measured can have its angles found by the cosine rule, and its area by $\frac{1}{2}ab\sin C$ once an angle is known. Surveyors compute land areas this way without right angles.

Example 2. Navigation legs. A ship sailing two legs at a known angle between them forms a triangle whose third side (the direct distance home) comes from the cosine rule. This is a classic bearings-and-distance application.

Try this

Q1. State which rule to use to find a side, given two angles and one side. [1 mark]

Cue. The sine rule, since a side is paired with its opposite angle.

Q2. Find the area of a triangle with sides $6\ \text{cm}$ and $9\ \text{cm}$ and an included angle of $30^\circ$ . [2 marks]

Cue. $\frac{1}{2} \times 6 \times 9 \times \sin 30^\circ = \frac{1}{2} \times 54 \times 0.5 = 13.5\ \text{cm}^2$ .

Q3. In a triangle, $b = 4$ , $c = 6$ and the included angle $A = 60^\circ$ . Find $a$ , to 2 decimal places. [2 marks]

Cue. $a^2 = 16 + 36 - 2(4)(6)\cos 60^\circ = 52 - 48 \times 0.5 = 28$ , so $a = \sqrt{28} = 5.29$ .

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.

Original4 marksIn triangle

ABC

AB = 7\ \text{cm}

AC = 9\ \text{cm}

and the angle

BAC = 50^\circ

. Find the length of

BC

, to 2 decimal places.

Show worked answer →

Two sides and the included angle are known, so use the cosine rule with $BC$ opposite the $50^\circ$ angle.

$BC^2 = 7^2 + 9^2 - 2(7)(9)\cos 50^\circ = 49 + 81 - 126 \times 0.6428 = 130 - 80.993 = 49.007$ .

$BC = \sqrt{49.007} = 7.0005\ldots$ , which is $7.00\ \text{cm}$ to 2 decimal places.

Markers reward selecting the cosine rule for two sides and the included angle, correct substitution, and the length of $BC$ .

Original4 marksIn triangle

PQR

, angle

P = 40^\circ

, angle

Q = 75^\circ

and side

PQ = 10\ \text{cm}

(the side

r

opposite

R

). Find the length of side

PR

, to 2 decimal places.

Show worked answer →

First find angle $R$ : $R = 180^\circ - 40^\circ - 75^\circ = 65^\circ$ .

Side $PR$ is opposite angle $Q$ , and $PQ = 10\ \text{cm}$ is opposite angle $R$ . Use the sine rule: $\dfrac{PR}{\sin Q} = \dfrac{PQ}{\sin R}$ .

$PR = \dfrac{10 \times \sin 75^\circ}{\sin 65^\circ} = \dfrac{10 \times 0.9659}{0.9063} = 10.657\ldots$ , which is $10.66\ \text{cm}$ to 2 decimal places.

Markers reward finding the third angle, setting up the sine rule with the correct opposite pairs, and the length of $PR$ .