Sine Rule GCSE Questions, Worked Examples and Answers

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Start with a triangle

What you need to know about the sine rule

Imagine three straight paths joining three locations, A, B and C. The path from B to C lies across the triangle from A: it is opposite angle A. We call its length a. In the same way, b is the length opposite B, and c is the length opposite C. Capital letters label angles; lower-case letters label their facing side lengths.

A side and the angle facing it

Why the matching pairs matter

A sine value compares two lengths in a right-angled triangle. Drawing one height inside our triangle lets us connect two such comparisons.

Locate angle Alook across the trianglethe side that does not touch A is a

Locate angle Bmatch B with bdo not pair an angle with a neighbouring side

Compareside ÷ sine of facing anglethis ratio is equal for every pair

The dashed height belongs to both smaller right-angled triangles. Side a faces A; side b faces B. The drawing explains the labels, not the measurements in every question.

In a right-angled triangle, the hypotenuse is the side opposite the right angle. The sine of an angle is the length opposite that angle divided by the hypotenuse. If this is new, first review right-triangle trigonometry.

In the left smaller triangle, the height $h$ faces $A$ and $b$ is the hypotenuse. So $\sin A=h/b$ ; multiplying by $b$ gives $h=b\sin A$ . In the right smaller triangle, $h$ faces $B$ and $a$ is the hypotenuse, giving $h=a\sin B$ . Both expressions measure the same height:

b\sin A=a\sin B

Divide by $\sin A\sin B$ :

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

This height argument explains the rule for the acute triangle shown. The sine rule also holds for obtuse and right-angled triangles.

From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Find a missing side

What the problem asks: For example, a = 9 cm, A = 40° and B = 65°; find b.

How to solve it: The complete pair is a and A. Set b/sin B equal to a/sin A, then multiply both sides by sin B to leave b on its own.

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

Find a missing angle

What the problem asks: For example, a = 10 cm, A = 50° and b = 7 cm; find B.

How to solve it: This task has a complete known side–angle pair and a known side facing the wanted angle. Write sine over side: these ratios are equal because they are the reciprocals of the equal side-over-sine ratios. Multiply by b to isolate sin B, then use inverse sine in degree mode. Also test 180° minus the calculator angle: retain only candidates consistent with the given information and a triangle angle sum of 180°.

For this example:

\frac{\sin B}{7}=\frac{\sin50^\circ}{10}

\sin B=\frac{7\sin50^\circ}{10}

B=\sin^{-1}\left(\frac{7\sin50^\circ}{10}\right)\approx32.4^\circ

Inverse sine asks “which angle has this sine?”. The other candidate is about $147.6^\circ$ , but $147.6^\circ+50^\circ>180^\circ$ , so it cannot be an angle of this triangle.

Find a side after finding the third angle

What the problem asks: Two angles and the side opposite the unknown third angle are supplied.

How to solve it: Subtract the two known angles from 180° to complete a side–angle pair. Then use the same missing-side calculation.

Choose between sine and cosine rule

What the problem asks: A question gives two sides and the angle between them, but no complete opposite pair.

How to solve it: The sine rule cannot yet give a single equation with one unknown. Use the cosine rule to find the third side first.

A reliable routine

Find a side using a complete opposite pair

This method applies when one side and its opposite angle are known and the angle opposite the wanted side is known or can be found. It works because both side-to-sine ratios describe the same triangle.

Mark the known complete pair and the pair containing the missing side.
Write the two equal fractions with side lengths on top.
Substitute the known values, keeping each side with its facing angle.
Multiply by the sine underneath the unknown side.
Use degree mode and retain calculator precision until the final rounding.

Check: Bigger angles face longer sides. This is a useful check on a side answer; it does not replace the calculation.

Fully worked

Sine rule GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Find a missing side

Higher only3 marks

Question

In triangle ABC, $a=9$ cm, $A=40^\circ$ and $B=65^\circ$ . Find $b$ to 3 significant figures.

The $9$ cm side faces $40^\circ$ . The wanted side faces $65^\circ$ .

\frac b{\sin65^\circ}=\frac9{\sin40^\circ}

Multiply both sides by $\sin65^\circ$ .

b=\frac{9\sin65^\circ}{\sin40^\circ}

b=12.689681\ldots

\boxed{b=12.7\text{ cm}}

Check: $65^\circ>40^\circ$ , so $b$ should exceed $9$ cm.

Example 2

Find an acute angle

Higher only3 marks

Question

A triangle has $a=10$ cm, $A=50^\circ$ and $b=7$ cm. Find angle $B$ to 1 decimal place.

Use sine over side so the unknown sine is in the numerator.

\frac{\sin B}{7}=\frac{\sin50^\circ}{10}

\sin B=\frac{7\sin50^\circ}{10}

B=\sin^{-1}\left(\frac{7\sin50^\circ}{10}\right)

\boxed{B=32.4^\circ}

Inverse sine means “which angle has this sine?”, not divide by sine. The alternative $147.6^\circ$ would make the angle sum exceed $180^\circ$ , so it is impossible here.

Example 3

Find the third angle first

Higher only4 marks

Question

In triangle PQR, $P=38^\circ$ , $Q=67^\circ$ and $PQ=12$ cm. Find $QR$ to 3 significant figures.

Side PQ faces R. Complete this pair first.

R=180^\circ-38^\circ-67^\circ=75^\circ

Side QR faces P, so

\frac{QR}{\sin38^\circ}=\frac{12}{\sin75^\circ}

QR=\frac{12\sin38^\circ}{\sin75^\circ}

\boxed{QR=7.65\text{ cm}\quad(3\text{ s.f.})}

Tip: label the opposite corner before choosing an angle.

Example 4

Use an obtuse angle

Higher only3 marks

Question

A triangle has $a=15$ cm, $A=100^\circ$ and $b=10$ cm. Find $B$ to 1 decimal place.

The sine rule still applies when a triangle contains an obtuse angle.

\sin B=\frac{10\sin100^\circ}{15}

B=\sin^{-1}\left(\frac{10\sin100^\circ}{15}\right)

\boxed{B=41.0^\circ}

The other possible sine angle is too large: with $100^\circ$ already used, $B$ must be below $80^\circ$ .

Example 5

Check two possible triangles

Higher only4 marks

Question

A triangle has $a=9$ cm, $A=30^\circ$ and $b=12$ cm. Find both possible values of $B$ to 1 decimal place.

\sin B=\frac{12\sin30^\circ}{9}=\frac23

The calculator returns the acute angle.

B_1=\sin^{-1}(2/3)=41.8103\ldots^\circ

The supplement of an angle is $180^\circ$ minus that angle. These two angles have the same sine, so there is a second candidate to check.

B_2=180^\circ-B_1=138.1896\ldots^\circ

Both leave a positive third angle when combined with $30^\circ$ .

\boxed{B=41.8^\circ\text{ or }138.2^\circ}

Tip: a two-sides-and-an-opposite-angle question can describe two triangles.

Example 6

Measure across a lake

Higher only4 marks

Question

Surveyors stand at A and B on one bank of a lake, $18$ m apart. They sight a marker C. Angles CAB and ABC are $45^\circ$ and $55^\circ$ . Find BC to 3 significant figures.

The known baseline AB faces angle C.

C=180^\circ-45^\circ-55^\circ=80^\circ

\frac{BC}{\sin45^\circ}=\frac{18}{\sin80^\circ}

BC=\frac{18\sin45^\circ}{\sin80^\circ}

\boxed{BC=12.9\text{ m}}

The required distance faces the $45^\circ$ angle, not the angle at B.

10 original questions · total 28 marks

Sine rule GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 35 minutes · Use degree mode. Mark opposite pairs and show the substituted equation. Round only the final answer. · answers start collapsed

1

Match a side and an angle

Higher only1 mark

In triangle XYZ, which angle is opposite side XZ?

Show worked answer

Side XZ joins X and Z, so it faces the remaining corner: angle Y.

2

Find the longer side

Higher only3 marks

$a=10$ cm, $A=50^\circ$ , $B=70^\circ$ . Find $b$ to 3 significant figures.

Show worked answer

\frac b{\sin70^\circ}=\frac{10}{\sin50^\circ}

b=\frac{10\sin70^\circ}{\sin50^\circ}

\boxed{b=12.3\text{ cm}}

It is longer than $a$ , as expected from the larger opposite angle.

3

Find the shorter side

Higher only3 marks

$a=16$ cm, $A=70^\circ$ , $B=55^\circ$ . Find $b$ to 3 significant figures.

Show worked answer

\frac b{\sin55^\circ}=\frac{16}{\sin70^\circ}

b=\frac{16\sin55^\circ}{\sin70^\circ}

\boxed{b=13.9\text{ cm}}

4

Find an angle

Higher only3 marks

$a=12$ cm, $A=65^\circ$ , $b=8$ cm. Find $B$ to 1 decimal place.

Show worked answer

\frac{\sin B}{8}=\frac{\sin65^\circ}{12}

B=\sin^{-1}\left(\frac{8\sin65^\circ}{12}\right)

\boxed{B=37.2^\circ}

The supplementary angle cannot fit alongside $65^\circ$ .

5

Keep an exact length

Higher only3 marks

$a=7$ cm, $A=30^\circ$ , $B=60^\circ$ . Find $b$ exactly.

Show worked answer

b=\frac{7\sin60^\circ}{\sin30^\circ}

b=\frac{7(\sqrt3/2)}{1/2}

\boxed{b=7\sqrt3\text{ cm}}

Keep the square root because a rounded decimal is not exact.

6

Complete a pair

Higher only4 marks

$A=42^\circ$ , $C=70^\circ$ and $a=11$ cm. Find $b$ to 3 significant figures.

Show worked answer

B=180^\circ-42^\circ-70^\circ=68^\circ

\frac b{\sin68^\circ}=\frac{11}{\sin42^\circ}

b=\frac{11\sin68^\circ}{\sin42^\circ}

\boxed{b=15.2\text{ cm}}

7

Recover an obtuse angle

Higher only3 marks

Angle $B$ is obtuse and $\sin B=0.8$ . Find $B$ to 1 decimal place.

Show worked answer

The inverse-sine button returns the acute angle.

\sin^{-1}(0.8)=53.1301\ldots^\circ

B=180^\circ-53.1301\ldots^\circ

\boxed{B=126.9^\circ}

8

Spot impossible measurements

Higher only3 marks

Can a triangle have $a=4$ cm, $A=30^\circ$ and $b=10$ cm? Justify your answer.

Show worked answer

\sin B=\frac{10\sin30^\circ}{4}

\sin B=1.25

A real angle cannot have sine greater than $1$ , so no such triangle exists.

9

Choose a valid first rule

Higher only2 marks

A triangle has sides 6 cm and 8 cm with an included angle of $50^\circ$ . Which rule finds the third side directly, and why?

Show worked answer

Cosine rule. We know two sides and the angle between them. There is no complete known opposite side–angle pair for a direct sine-rule calculation.

10

Use a right-angled triangle

Higher only3 marks

$A=35^\circ$ , $a=8$ cm and $B=90^\circ$ . Use the sine rule to find $b$ to 3 significant figures.

Show worked answer

\frac b{\sin90^\circ}=\frac8{\sin35^\circ}

Since $\sin90^\circ=1$ ,

b=\frac8{\sin35^\circ}

\boxed{b=13.9\text{ cm}}

This is the hypotenuse; SOHCAHTOA gives the same result.

Examiner-style feedback

Common sine rule mistakes

Pairing neighbouring sides and angles

The opposite side does not touch the angle’s vertex. Write its endpoint letters if the picture is confusing.

Dividing by the angle instead of its sine

sin 40° is a calculator value, not 40. Keep sin attached to the angle.

Rounding before inverse sine

Use the full stored sine value to calculate the angle, then round.

Accepting the acute answer automatically

If an obtuse angle is required, test its supplement. Always check the triangle’s angle sum.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

Match each angle with the side facing it.
Use a complete known pair.
Find a third angle first when necessary.
Check angle possibilities, units and rounding.

Quick answers

Sine rule FAQ

Is the sine rule Higher tier?

Yes. Applying the sine rule to find unknown sides and angles is Higher-tier GCSE Mathematics content.

Does the sine rule work in every triangle?

Yes, but you need enough information to use it. A direct missing-side calculation needs a known opposite pair and the angle facing the missing side.

Why can inverse sine give two answers?

An acute angle and its supplement have the same sine. The remaining information decides whether one or both triangles are possible.

Build connected skills

What to revise next

Prerequisite

Trigonometry

Understand sine in right-angled triangles.

Revise trigonometry

Alternative rule

Cosine rule

Work with three sides or an included angle.

Revise cosine rule

Exact answers

Exact trig values

Use special-angle ratios without rounding.

Revise exact trig values

Content standards

Curriculum and rights review

Reviewed 7 September 2026 against GCSE G22. Original questions, worked answers and diagram; no exam-board questions reproduced. Suggested practice marks are Pass an Exam estimates.

Official specification references

AQA GCSE geometry: G20–G22