Lesson 15 ~35 min Unit 3 · Geometry +85 XP

Sine Rule

Unlock the power to solve any triangle. The sine rule connects every side to the sine of its opposite angle -- and it works for every triangle, not just right-angled ones.

Today's hook: A surveyor measures two angles of a triangular paddock as 50° and 60°, and one side as 100 m. Without a right angle anywhere, how can they find the other sides?

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Think First

warm-up

SOH CAH TOA only works in right-angled triangles. Imagine a triangle with angles 50°, 60° and 70° -- no right angle at all. How might you find a missing side if you know one side and all three angles? What relationship could connect sides to angles?

Record your answer in your workbook.

The Big Idea

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The sine rule states that in any triangle, the ratio of each side to the sine of its opposite angle is constant. This means larger sides face larger angles, and this relationship is precise and calculable.

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

The sine rule is essential when you know two angles and one side (AAS) and need to find another side, or when you know two sides and a non-included angle (SSA) and need to find another angle. It transforms angle information into side information and vice versa.

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

Label carefully

Capital letters for angles, lowercase for opposite sides.

AAS or SSA only

Need an angle opposite a known side.

Any triangle

Works for acute, obtuse and right triangles.

What You'll Master

objectives

Know

The sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
When the sine rule applies (AAS and SSA cases)

Understand

Why the sine rule works for any triangle, not just right-angled ones
The ambiguous case and why two triangles may exist

Can Do

Use the sine rule to find unknown sides and angles
Identify when the ambiguous case applies
Choose between sine rule and SOH CAH TOA

Words You Need

vocabulary

Sine rule$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$, relating sides to the sines of opposite angles.

Ambiguous caseWhen two different triangles can satisfy the given SSA information.

AASAngle-Angle-Side: two angles and one side known. Unique triangle.

SSASide-Side-Angle: two sides and a non-included angle known. May be ambiguous.

OppositeThe side directly across from a given angle.

Included angleThe angle between two known sides.

Spot the Trap

heads-up

Wrong: The sine rule can be used in any triangle, including right-angled triangles -- so use it everywhere.

Right: The sine rule can be used in right-angled triangles, but it is unnecessary. Use SOH CAH TOA for right-angled triangles and the sine rule for non-right-angled triangles.

Wrong: The sine rule is $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = \text{area}$.

Right: The sine rule is $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. It relates sides to the sines of their opposite angles, not to the area.

Deriving the Sine Rule

+5 XP

The sine rule can be derived from the area formula for a triangle. By calculating the area in three different ways (using each side as the base), we obtain three expressions that must be equal -- and the sine rule drops out.

Area = $\frac{1}{2}ab \sin C = \frac{1}{2}bc \sin A = \frac{1}{2}ca \sin B$. Divide all three by $\frac{1}{2}abc$: $$\frac{\sin C}{c} = \frac{\sin A}{a} = \frac{\sin B}{b}$$ Taking reciprocals gives the sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. The key insight: the height of a triangle can be expressed using any side and the sine of the adjacent angle.

Area = $\frac{1}{2}ab \sin C$ → $\frac{a}{\sin A} = \frac{b}{\sin B}$

Height = b sin A

The geometric heart of the proof.

Works for obtuse too

sin(180° - x) = sin x.

Constant = 2R

The ratio equals twice the circumradius.

Finding Sides

+5 XP

Use the sine rule to find an unknown side when you know two angles and one side (AAS). First find the third angle (angles sum to 180°), then set up the proportion and solve.

Step 1: Find the third angle: $C = 180° - A - B$.
Step 2: Write the sine rule with the known side-angle pair and the unknown side.
Step 3: Cross-multiply and solve.
Always use the side opposite the known angle as your reference pair. This gives you the constant ratio that connects to every other side-angle pair.

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

Find the third angle first

$C = 180° - A - B$. Always.

Keep calculator in degree mode

Check before every exam question.

Round at the end

Keep full precision during working.

Finding Angles and the Ambiguous Case

+5 XP

When you know two sides and a non-included angle (SSA), the sine rule can find the missing angle. But beware: there may be two possible answers. This is the famous ambiguous case.

From $\frac{a}{\sin A} = \frac{b}{\sin B}$, you get $\sin B = \frac{b \sin A}{a}$. Since $\sin x = \sin(180° - x)$, if the calculated angle is acute, $180°$ minus that angle is also a valid solution -- provided the resulting triangle angles still sum to 180°. Two triangles exist when: $a < b$ and $a > b \sin A$.

$\sin B = \frac{b \sin A}{a}$ → $B$ or $180° - B$

Check the sum

Both $B$ and $180° - B$ must leave room for angle $C$.

No ambiguity if a ≥ b

Then only one triangle exists.

No solution if a < b sin A

The side is too short to reach.

Common Pitfalls

heads-up

Using sine rule when SOH CAH TOA would do

Applying the full sine rule to a right-angled triangle. While it works, it is unnecessarily complicated and increases the chance of error.

Fix: check for a right angle first. If there is one, use SOH CAH TOA. Only use the sine rule for non-right-angled triangles.

Mixing up sides and angles

Using side $a$ with angle $B$ in the sine rule, when $a$ is opposite angle $A$. This produces completely wrong answers.

Fix: label your triangle first. Always pair lowercase side with uppercase opposite angle. Write the pairings before setting up the equation.

Forgetting the ambiguous case

Finding one angle from SSA information and stopping there. The second solution ($180° - \text{angle}$) may also be valid.

Fix: whenever you find an angle using inverse sine, always check if $180° - \text{angle}$ gives a valid triangle. Check that all three angles sum to 180°.

Watch Me Solve It · 3 examples

Watch Me Solve It · Finding a side

+15 XP per step

PROBLEM

In $\triangle ABC$, $\angle A = 50°$, $\angle B = 60°$, and side $a = 10$ cm. Find side $b$ correct to one decimal place.

1

Find angle C

$\angle C = 180° - 50° - 60° = 70°$

Angles in a triangle sum to 180°.
2

Write the sine rule

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

We know $a$ and $A$, and want $b$ given $B$.
3

Substitute and solve

$\frac{10}{\sin 50°} = \frac{b}{\sin 60°}$

$b = \frac{10 \times \sin 60°}{\sin 50°}$

$b = \frac{10 \times 0.8660}{0.7660} \approx 11.3$ cm
4

Check reasonableness

$B = 60° > A = 50°$, so $b$ should be greater than $a = 10$

$11.3 > 10$ ✓ Larger angle faces larger side.

Answer$b \approx 11.3$ cm

Watch Me Solve It · Finding an angle

+15 XP per step

PROBLEM

In $\triangle ABC$, side $a = 8$ cm, side $b = 12$ cm, and $\angle A = 30°$. Find $\angle B$ correct to one decimal place.

1

Write the sine rule for angles

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

We know two sides and one opposite angle.
2

Rearrange to find sin B

$\sin B = \frac{b \sin A}{a} = \frac{12 \times \sin 30°}{8}$

$\sin B = \frac{12 \times 0.5}{8} = \frac{6}{8} = 0.75$
3

Find angle B

$B = \sin^{-1}(0.75) \approx 48.6°$

Using inverse sine on your calculator.
4

Check for the ambiguous case

$180° - 48.6° = 131.4°$

If $B = 131.4°$, then $C = 180° - 30° - 131.4° = 18.6° > 0$ ✓

Since $a < b$ and $a > b \sin A$ ($8 > 12 \times 0.5 = 6$), two triangles exist. $B \approx 48.6°$ or $131.4°$.

Answer$B \approx 48.6°$ or $131.4°$

Watch Me Solve It · Complete triangle

+15 XP per step

PROBLEM

In $\triangle ABC$, $\angle A = 40°$, $\angle B = 70°$, and side $c = 12$ cm. Find all remaining sides and angles.

1

Find angle C

$\angle C = 180° - 40° - 70° = 70°$

Angles in a triangle sum to 180°.
2

Find side a

$\frac{a}{\sin 40°} = \frac{12}{\sin 70°}$

$a = \frac{12 \times \sin 40°}{\sin 70°} = \frac{12 \times 0.6428}{0.9397} \approx 8.2$ cm
3

Find side b

$\frac{b}{\sin 70°} = \frac{12}{\sin 70°}$

$b = 12$ cm

Since $\angle B = \angle C = 70°$, the triangle is isosceles with $b = c$.
4

Verify

Angles: $40° + 70° + 70° = 180°$ ✓

Sides: $a < b = c$ since $A < B = C$ ✓

Largest angle faces largest side. The triangle is consistent.

Answer$C = 70°$, $a \approx 8.2$ cm, $b = 12$ cm

Copy Into Your Books

Sine Rule

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Side / sin(opposite angle) = constant
Works for any triangle

Finding Sides (AAS)

Find third angle first
Set up proportion with known pair
Cross-multiply and solve

Finding Angles (SSA)

Rearrange to find sin(angle)
Use inverse sine
Check for ambiguous case

Ambiguous Case

Two solutions if $a < b$ and $a > b \sin A$
Second angle = $180° - \text{first angle}$
Check angles sum to 180°

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Sine rule drill

4 problems

Four quick problems on the sine rule. Work each, then reveal the answer.

1 In $\triangle ABC$, $\angle A = 40°$, $\angle B = 70°$, and $a = 12$ cm. Find side $b$.

$\angle C = 70°$. $\frac{12}{\sin 40°} = \frac{b}{\sin 70°}$ → $b = \frac{12 \times \sin 70°}{\sin 40°} \approx 17.5$ cm.$b \approx 17.5$ cm
2 In $\triangle ABC$, $a = 10$ cm, $b = 15$ cm, and $\angle A = 35°$. Find $\angle B$.

$\sin B = \frac{15 \times \sin 35°}{10} = \frac{15 \times 0.5736}{10} \approx 0.860$. $B \approx 59.4°$ or $120.6°$.$B \approx 59.4°$ or $120.6°$
3 In $\triangle ABC$, $\angle A = 50°$, $\angle B = 60°$, and $a = 8$ cm. Find $\angle C$ and side $c$.

$\angle C = 180° - 50° - 60° = 70°$. $c = \frac{8 \times \sin 70°}{\sin 50°} \approx 9.8$ cm.$C = 70°$, $c \approx 9.8$ cm
4 In $\triangle ABC$, $a = 5$ cm, $b = 8$ cm, and $\angle A = 30°$. How many possible triangles exist? Explain.

$b \sin A = 8 \times 0.5 = 4$. Since $a = 5$ and $4 < 5 < 8$, two triangles exist. If $a < 4$, no triangle. If $a \ge 8$, one triangle.Two triangles

Complete in your workbook.

Multiple Choice · 5 questions

MC1

The sine rule formula

+10 XP

The sine rule states:

MC2

When to use the sine rule

+10 XP

The sine rule is used when you know:

MC3

Sine rule calculation

+10 XP

In $\triangle ABC$, $A = 30°$, $B = 60°$, $a = 5$. Find $b$:

MC4

The ambiguous case

+10 XP

The ambiguous case occurs with:

MC5

Sine rule applicability

+10 XP

The sine rule applies to:

Short Answer · 3 questions

Complete triangle

+15 XP

SHORT ANSWER

In $\triangle ABC$, $\angle A = 40°$, $\angle B = 70°$, and side $c = 12$ cm.
(a) Find $\angle C$.
(b) Use the sine rule to find side $a$.
(c) Use the sine rule to find side $b$.

Write your working in your book.

Ambiguous case

+15 XP

SHORT ANSWER

In $\triangle ABC$, side $a = 8$ cm, side $b = 12$ cm, and $\angle A = 30°$.
(a) Use the sine rule to find $\angle B$.
(b) Explain why there might be two possible values for $\angle B$.
(c) Sketch both possible triangles.

Write your working in your book.

Sine rule vs SOH CAH TOA

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SHORT ANSWER

Explain when you would use the sine rule instead of SOH CAH TOA. Give an example of a triangle where the sine rule is needed.

Write your working in your book.

Stretch Challenge · Surveyor's problem

+20 XP

STRETCH

A surveyor stands at point $A$ and measures the angle to a tower at point $B$ as $35°$ east of north. Walking 200 m east to point $C$, the surveyor measures the angle to the tower as $22°$ west of north.
(a) Find $\angle ABC$ and $\angle ACB$.
(b) Find $\angle BAC$.
(c) Use the sine rule to find the distance from $A$ to the tower ($AB$).
(d) Find the distance from $C$ to the tower ($CB$).
(e) Explain why there is no ambiguous case in this problem.

Record in your book -- full marks require clear working.

Reveal solution

(a) $\angle ABC = 180° - 35° = 145°$ (bearing geometry: the tower is 35° east of north from A, so the interior angle at B is 180° - 35° = 145°... wait, let us be more careful. From A, the tower is 35° east of north. From C, the tower is 22° west of north. The line AC runs east. The angle between north and AB is 35°. The angle between north and CB is 22°. Since AC is east-west, the angles on the same side of north sum: $\angle BAC = 90° - 35° = 55°$ and $\angle BCA = 90° - 22° = 68°$.

(b) $\angle BAC = 180° - 55° - 68° = 57°$... Actually, using the bearing geometry more carefully: the angle at A inside the triangle is $90° - 35° = 55°$ and at C is $90° - 22° = 68°$. So $\angle ABC = 180° - 55° - 68° = 57°$.

(c) Using sine rule: $\frac{AB}{\sin 68°} = \frac{200}{\sin 57°}$. $AB = \frac{200 \times \sin 68°}{\sin 57°} \approx \frac{200 \times 0.9272}{0.8387} \approx 221$ m.

(d) $\frac{CB}{\sin 55°} = \frac{200}{\sin 57°}$. $CB = \frac{200 \times \sin 55°}{\sin 57°} \approx \frac{200 \times 0.8192}{0.8387} \approx 195$ m.

(e) There is no ambiguous case because we know two angles (ASA effectively), which uniquely determines a triangle. The ambiguous case only arises with SSA (two sides and a non-included angle).

Quick Review

Sine rule

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

Finding sides

Use AAS, find third angle first

Finding angles

Use SSA, check ambiguous case

Ambiguous

$B$ or $180° - B$

Label

Capital = angle, lowercase = side

Right triangle

Use SOH CAH TOA instead

Interactive: Sine Rule Solver

Practice using the sine rule to find missing sides and angles. Test your calculations with instant feedback.

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