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hscscience Maths Std · Y12
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MST-12-S2-04 ~55 min ⚡ +95 XP available

The Sine Rule

Right-angled trigonometry needs a right angle. Most real triangles do not have one. The sine rule is the tool that works on any triangle, as long as you can pair one side with the angle opposite it.

Today's hook, A surveyor needs the distance across a river. She cannot walk across it. Standing on one bank she measures a baseline of $500\text{ m}$ and two angles. No right angle anywhere. Can she still find the distance?
0/5QUESTS
1

Orient to the sine rule

Meet the side and opposite-angle pairing, set the goal and settle the key terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any question in this focus area.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any question
01
Recall, your gut answer first
+5 XP warm-up

A surveyor wants the distance from her position to a tree on the far bank of a river. She paces out a baseline of $500\text{ m}$ along her own bank and measures the angle to the tree from each end of that baseline. The triangle she has made contains no right angle at all.

Without calculating write down whether you think she has enough information, and say what makes this different from every triangle you solved in the last five lessons.

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02
The sine rule, sides and their opposite angles
+5 XP to read

Everything you have used so far, SOH CAH TOA and Pythagoras, needs a right angle. The sine rule needs none. It works in any triangle, and it links each side to the angle sitting opposite that side.

Labelling convention. Capital letters are the angles, lower-case letters are the sides, and side $a$ is always the side opposite angle $A$. Getting this pairing right is most of the work.

You need a matched pair. The rule only starts if you know one side AND the angle opposite it. That pair is the anchor; everything else is solved against it.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Finding a SIDE
Put the unknown side on top: $b = \dfrac{a\sin B}{\sin A}$. The unknown ends up as a multiplication.
Finding an ANGLE
Flip the rule so the sines are on top: $\sin B = \dfrac{b\sin A}{a}$, then take the inverse sine.
Angles add to 180
If the question gives two angles, the third is free: $C = 180° - A - B$. Often that is the pair you need.
03
What you'll master
Know

Key facts

  • The sine rule: $a/\sin A = b/\sin B = c/\sin C$
  • Side $a$ is opposite angle $A$, always
  • The rule needs one matched side-and-opposite-angle pair to start
  • The three angles of any triangle add to $180°$
Understand

Concepts

  • Why the sine rule works when there is no right angle
  • Why $\sin\theta = \sin(180° - \theta)$, and what that costs you
  • Why a calculator cannot tell you whether an angle is acute or obtuse
Can do

Skills

  • Find an unknown side, including when the third angle is needed first
  • Find an unknown acute angle, to a degree or to the nearest minute
  • Handle the case where the question states the angle is obtuse
  • Check that your triangle actually closes
04
Key terms
Sine ruleA relationship holding in every triangle, that each side divided by the sine of its opposite angle gives the same value. Like this: in a triangle with $a = 12$ and $A = 40°$, every other side over the sine of its own opposite angle also equals $12 \div \sin 40°$.
Opposite sideThe side that does not touch the angle. Like this: angle $A$ is formed by sides $b$ and $c$ meeting, so the side left over, $a$, is the one opposite $A$.
Matched pairA side and the angle opposite it, both known. Like this: knowing $a = 9$ and $A = 35°$ is a matched pair, but knowing $a = 9$ and $B = 35°$ is not, and the sine rule cannot start from it.
Obtuse angleAn angle between $90°$ and $180°$. Like this: $119°$ is obtuse, and because $\sin 119° = \sin 61°$, a calculator asked for the inverse sine will hand back $61°$ every time.
2

Choose the arrangement

Put the unknown on top for a side, underneath for an angle, and watch for the obtuse case.

05
Choosing the arrangement, and the obtuse problem
core concept

Write the rule the way that puts your unknown on top. Two arrangements cover every question in this lesson:

unknown SIDE: $\quad b = \dfrac{a \sin B}{\sin A}$
unknown ANGLE: $\quad \sin B = \dfrac{b \sin A}{a} \quad\Rightarrow\quad B = \sin^{-1}\!\left(\dfrac{b\sin A}{a}\right)$

The second arrangement carries a hazard the first does not. Sine is positive for acute angles and for obtuse ones, and in fact $\sin\theta = \sin(180° - \theta)$. So $\sin 61°$ and $\sin 119°$ are the same number, and your calculator, handed that number, will always return the acute one.

What the syllabus asks of you. You are not asked to resolve the ambiguous case unaided. You are asked to find an acute angle when that is what the triangle has, and to find the obtuse angle when the question tells you the angle is obtuse. If it does, take the calculator's answer and subtract it from $180°$.

The sine rule is $a/\sin A = b/\sin B = c/\sin C$, where each lower-case side is opposite its capital-letter angle. To start you need one side and the angle opposite it. For an unknown angle, $\sin\theta = \sin(180° - \theta)$, so the calculator returns the acute value and you subtract from $180°$ if the question says the angle is obtuse.

Pause, copy the sine rule, the labelling convention (side $a$ opposite angle $A$), and one line explaining why an obtuse answer needs $180°$ minus the calculator value, into your book.

Quick check: In a triangle $A = 30°$, $B = 45°$ and $a = 10$. Find $b$, to 2 decimal places.

3

Work three sine rule examples

Follow an unknown side, a third angle first, and an angle with an obtuse alternative.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · UNKNOWN SIDE

In triangle $ABC$, $A = 40°$, $B = 75°$ and $a = 12$. Find $b$, correct to 2 decimal places.

1
Matched pair: $a = 12$ with $A = 40°$. Unknown $b$ pairs with $B = 75°$.
Both halves of the rule are complete, so no third angle is needed.
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PROBLEM 2 · THIRD ANGLE FIRST

In triangle $ABC$, $A = 52°$, $B = 61°$ and $c = 20$. Find $a$, correct to 2 decimal places.

1
The known side is $c = 20$, but $C$ is not given. There is no matched pair yet.
Spot this before substituting. It is the reason the question feels stuck.
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PROBLEM 3 · UNKNOWN ANGLE, ACUTE THEN OBTUSE

(a) In triangle $ABC$, $a = 9$, $A = 35°$ and $b = 13$, and $B$ is acute. Find $B$ to the nearest minute.   (b) In a different triangle, $a = 7$, $A = 28°$, $b = 13$, and $B$ is known to be obtuse. Find $B$ to the nearest minute.

1
(a) $\sin B = \dfrac{13\sin 35°}{9} = \dfrac{13 \times 0.573576}{9} = 0.828498$
Flip the rule so the sines are on top, because the unknown is an angle.
1 of 4

True or false: $\sin 150°$ and $\sin 30°$ are equal.

4

Avoid the three traps

Spot the wrong pairing, the missing third angle and the ignored obtuse solution.

Common errors · the 3 traps that cost marks

Trap 01
Pairing a side with the wrong angle
The sine rule only works side-with-opposite-angle. Using $a$ with $B$ because they appear together in the question produces a confident, wrong number with no warning sign. Label the diagram before you substitute anything.
Trap 02
Substituting when there is no matched pair
If the known side has no known opposite angle, the rule cannot start. Find the third angle with $180° - A - B$ first. Worked example 2 is exactly this case, and it is the most common reason a sine rule question stalls.
Trap 03
Taking the calculator's angle when the question said obtuse
Because $\sin\theta = \sin(180° - \theta)$, the inverse sine always returns the acute value. If the question states the angle is obtuse, subtract from $180°$. Then check the three angles still add to less than $180°$, which catches the error if you subtracted when you should not have.

Fill the gaps: With $A = 40°$, $B = 75°$ and $a = 12$: $\sin A =$ (4 dp), $\sin B =$ (4 dp), so $b =$ (2 dp).

5

Drill the sine rule

Run the quick-fire calculations until the arrangement choice is automatic.

Quick-fire practice · 5 calculations

1

$A = 50°$, $B = 60°$, $a = 10$. Find $b$ to 2 decimal places.

2

$A = 35°$, $C = 95°$, $a = 8$. Find $c$ to 2 decimal places.

3

$a = 10$, $A = 45°$, $b = 12$. Find the acute angle $B$ to 2 decimal places.

4

$A = 110°$, $B = 25°$, $b = 6$. Find $a$ to 2 decimal places.

5

$a = 5$, $A = 20°$, $b = 12$, and $B$ is obtuse. Find $B$ to 2 decimal places, then check the triangle closes.

Match each situation to the first move:

  • Two angles and the side between them
  • A side, its opposite angle, and a second side
  • A side, its opposite angle, and a second angle
  • The question says the unknown angle is obtuse
  • Sine rule straight away, unknown side on top
  • Find the acute value, then subtract it from 180 degrees
  • Find the third angle first, then use the sine rule
  • Flip the rule so the sines are on top, then inverse sine
6

Revisit your thinking

Return to your opening answer and name what has changed.

10
Revisit your thinking

Back to the surveyor. She has a baseline $PQ = 500\text{ m}$ and measures the angle at $P$ to be $60°$ and the angle at $Q$ to be $85°$.

Third angle: $R = 180° - 60° - 85° = 35°$, which pairs with the known baseline.

Then $QR = \dfrac{500\sin 60°}{\sin 35°} = \dfrac{500 \times 0.866025}{0.573576} \approx \mathbf{755\text{ m}}$.

She never crossed the river, and there was never a right angle. One matched pair was enough.

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7

Practise the sine rule

Answer the question bank, then write full short-answer responses.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 33 marks

Q1. In triangle $ABC$, $A = 63°$, $C = 47°$ and $b = 24$. Find $c$, correct to 2 decimal places. (3 marks)

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ApplyBand 44 marks

Q2. In triangle $ABC$, $a = 15$, $A = 42°$ and $b = 20$. Find the acute angle $B$, correct to the nearest minute. (4 marks)

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AnalyseBand 54 marks

Q3. From point $P$, a surveyor sights $Q$ on a bearing of $040°$T and $R$ on a bearing of $100°$T. She measures $PQ = 500\text{ m}$ and the angle $PQR = 85°$. Find $QR$, correct to the nearest metre. (4 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $b = \tfrac{10\sin 60°}{\sin 50°} = 11.31$  ·  Drill 2: $c = \tfrac{8\sin 95°}{\sin 35°} = 13.89$

Drill 3: $\sin B = \tfrac{12\sin 45°}{10} = 0.848528$, so $B = 58.05°$

Drill 4: $a = \tfrac{6\sin 110°}{\sin 25°} = 13.34$

Drill 5: $\sin B = \tfrac{12\sin 20°}{5} = 0.820848$; acute value $55.17°$, so $B = 180° - 55.17° = 124.83°$. Check: $20° + 124.83° = 144.83° < 180°$, so the triangle closes.

Q1 (3 marks): The known side $b = 24$ has no known opposite angle yet, so there is no matched pair. Find it first: $B = 180° - 63° - 47° = 70°$ [1]. Now $b = 24$ pairs with $B = 70°$, and the unknown $c$ pairs with $C = 47°$, so $c = \tfrac{24\sin 47°}{\sin 70°} = \tfrac{24 \times 0.731354}{0.939693} = 18.68$ [2]. Check: $18.68 \times \sin 70° = 17.55$ and $24 \times \sin 47° = 17.55$.

Q2 (4 marks): $\sin B = \tfrac{20\sin 42°}{15} = \tfrac{20 \times 0.669131}{15} = 0.892175$ [2]. $B = \sin^{-1}(0.892175) = 63.1477°$ [1]. In minutes: $0.1477 \times 60 = 8.9$, so $B = 63°\,9'$ [1].

Q3 (4 marks): Angle $QPR = 100° - 40° = 60°$ [1]. Angle $PRQ = 180° - 60° - 85° = 35°$ [1]. The matched pair is $PQ = 500$ with $R = 35°$, so $QR = \tfrac{500\sin 60°}{\sin 35°} = \tfrac{500 \times 0.866025}{0.573576} = 754.93$ [1], which is $\mathbf{755\text{ m}}$ to the nearest metre [1].