Sine Rule: One of the Key GCSE Maths Topics You Must Know
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Sine Rule — How to find missing sides and missing angles in non-right-angled triangles
This blog explains how to use the sine rule to find missing sides and missing angles in non-right-angled triangles. You’ll learn the exact formulas, when to use them, and see detailed, step-by-step worked examples.
There are also sine rule worksheets based on GCSE exam questions that mirror the kinds of problems shown here — ideal for practising the techniques and avoiding common exam pitfalls.
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What is the Sine Rule?
The sine rule (also called the law of sines) connects the ratios of side lengths to the sines of their opposite angles in any triangle. For a triangle \(ABC\) with sides \(a,b,c\) opposite angles \(A,B,C\) respectively, the sine rule states:
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\]
From this relationship you can rearrange to solve for either a missing side or a missing angle:
- To find a missing side (given its opposite angle and another side and its opposite angle), we can put the sides on top ( \(a, b, c\)):
\frac{a}{\sin A}=\frac{b}{\sin B}\quad\Rightarrow\quad a = b\cdot\frac{\sin A}{\sin B}
\]
- To find a missing angle (given its opposite sides and another side and its opposite angle), we can put the angles on top (\(\sin A, \sin B, \sin C\)):
\frac{a}{\sin A}=\frac{b}{\sin B}\quad\Rightarrow\quad \sin A = \frac{a}{b}\sin B \quad\Rightarrow\quad A = \sin^{-1}\!\Big(\frac{a}{b}\sin B\Big)
\]
When to use the Sine Rule
- AAS / ASA (two angles and one side) → find a missing side (straightforward).
- SSA (two sides and a non-included angle) → find a missing angle (ambiguous case may arise).
*If you have SAS (two sides and the included angle), use the cosine rule instead.
Finding a Missing Side
Worked Example 1 — AAS case
Problem: In triangle \(ABC\) you are given \(A=38^\circ\), \(B=64^\circ\) and side \(b=9.2\text{ cm}\). Find the missing side \(a\) (to 3 significant figures).
Step 1: Check what you know:
We have two angles and one side (AAS). Use the sine rule:
\frac{a}{\sin A}=\frac{b}{\sin B}
\]
Step 2 — Substitute values:
\frac{a}{\sin 38^\circ} = \frac{9.2}{\sin 64^\circ}
\]
Step 3 — Rearrange and solve the equation (calculator in degree mode)
a = 9.2 \times \frac{\sin 38^\circ}{\sin 64^\circ}=\frac{9.2 \cdot \sin 38^\circ}{\sin 64^\circ}
\]
Step 4 — Round as requested
a \approx 6.3018726…
\]
Final answer: \(a \approx 6.30\text{ cm}\) (3 s.f.).
Worked Example 2 — ASA case
Problem: In triangle \(ABC\) you are given \(A=46^\circ\), \(C=71^\circ\) and the included side \(b=12.8\text{ cm}\) (between \(A\) and \(C\)). Find the missing side \(a\) (to 3 s.f.).
Step 1 — Check what you know
We have two angles and one included side. We need to find the angle which is opposite to side \(b\) first.
B = 180^\circ – (A + C) = 180^\circ – (46^\circ + 71^\circ) = 63^\circ
\]
Now use the sine rule:
\frac{a}{\sin A}=\frac{b}{\sin B}
\]
Step 2 — Substitute values
\frac{a}{\sin 46^\circ} = \frac{12.8}{\sin 63^\circ}
\]
Step 3 — Rearrange and solve the equation (calculator in degree mode)
a = \frac{12.8}{\sin 63^\circ} \cdot \sin 46^\circ = \frac{12.8 \cdot \sin 46^\circ}{\sin 63^\circ}
\]
Step 4 — Round as requested
a \approx 10.33387433…
\]
Final answer: \(a \approx 10.3\text{ cm}\) (3 s.f.).
Finding a Missing Angle
Worked Example — SSA (ambiguous case)
Problem: In triangle \(ABC\) you are given side \(a=26\text{ cm}\), side \(c=45\text{ cm}\) and angle \(A=33^\circ\). Find angle \(C\) (nearest degree).
Step 1 — Check what you know
We know two sides and a non-included angle (SSA). Use the sine rule to find \(\sin A\):
\frac{a}{\sin A}=\frac{c}{\sin C}
\]
Step 2 — Substitute values
\frac{26}{\sin 33^\circ} = \frac{45}{\sin C}
\]
Step 3 — Find principal value and consider ambiguity
\sin C = \frac{\sin 33^\circ}{26} \cdot 45 = 0.9426444837\ldots
\]
\[
C = \sin^{-1}(0.9426444837) = 70.5^\circ
\]
Principal value \( C_1 = 70.5^\circ \), but there is a second possible angle \( C_2 = 180^\circ – 70.5^\circ = 109.5^\circ \)
Step 4 — Check triangle angle sum and feasibility
We already have \( A = 33^\circ \).
- If \( C_1 = 70.5^\circ \): then \( C_1 + A = 70.5^\circ + 33^\circ = 103.5^\circ \). Remaining angle \( C = 180^\circ – 103.5^\circ = 76.5^\circ \). All angles positive — feasible.
- If \( C_2 = 109.5^\circ \): then \( C_2 + A = 109.5^\circ + 33^\circ = 142.5^\circ < 180^\circ \).This is possible (angles do not exceed \( 180^\circ \)). So \( C_2 \) is accepted.
Step 5 — Round as requested
Final answer: \(C \approx 71^\circ\) or \(109^\circ\) (nearest degree).
*Key point: In the SSA case always check whether the supplementary angle is feasible by using the angle sum — many ambiguous cases are ruled out that way.
The Ambiguous Case (SSA) — Short Summary
When using the sine rule with SSA (two sides and a non-included angle), one of three situations can occur:
- No triangle — the computed sin value is greater than 1 (no solution).
- One triangle — either the principal value is valid and the supplementary is
impossible (angles sum \(> 180^\circ\)), or the principal and supplementary collapse (rare). - Two distinct triangles — both the acute and the obtuse solutions of the inverse sine yield valid triangles
(both fit with \( B \) and leave a positive third angle). In this case give both possible sets of answers.
Always test both possible angles \( A \) and \( 180^\circ – A \) against the known angle(s) to see which are geometrically valid.
Common Misconceptions & Exam Traps
Here are some common misconceptions and exam traps students often encounter when using the sine rule:
- Longer side ↔ larger angle — True, but students sometimes draw a triangle with the wrong orientation and then match sides to the wrong angles.
Always label sides and angles clearly before applying formulas. - Using the wrong formula (sine vs cosine) — If you have SAS (two sides and included angle), use the cosine rule;
if you have AAS/ASA or SSA (with caution), use the sine rule. - Rounding too early — Intermediate rounding can change which ambiguous solution appears valid.
Keep 4–6 significant figures in intermediate steps and round only the final answer. - Forgetting calculator mode — degrees vs radians: GCSE problems use degrees unless specified.
- Cancelling sine function incorrectly — Students sometimes “cancel” sines across a proportion incorrectly.
For example, treating sin like a plain number without considering units/angles:\[
a = \frac{12.8}{\sin 63^\circ} \cdot \sin 46^\circ = \frac{12.8 \sin 46^\circ}{\sin 63^\circ} = \frac{12.8 \times 46}{63} = 9.35
\]which is a wrong calculation.
- Missing the ambiguous case — When using \( \sin^{-1} \), always remember the supplementary angle (\( 180^\circ – \text{angle} \)) and check both; don’t assume the calculator’s principal value is the only answer.
Practice & Next Steps
Quick Summary
- Sine rule:\[
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\] - Use for AAS/ASA (to find sides) and SSA (to find angles — watch for ambiguity).
- In SSA, always consider \(A\) and \((180^\circ – A)\), then check triangle feasibility.
- Label clearly, keep precision until the end, use the correct calculator mode, and practise with exam-style questions.
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