SOH CAH TOA only works in a right-angled triangle. The sine rule is what happens when you drop a perpendicular into any triangle and refuse to give up.
Today's hook, Almost no real triangle is right-angled. Surveyors, navigators and engineers work with the sine and cosine rules far more often than with SOH CAH TOA, and both come from the right-angled case you already know.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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In a triangle, the largest angle sits opposite the longest side. Why must that be true? Then guess: if you know two angles and one side, is the triangle fully determined?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Drop a perpendicular and the right-angled case reappears
Work through the core explanation before applying it.
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Drop a perpendicular and the right-angled case reappears
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Drop a perpendicular of height $h$ from $B$ to side $AC$. In the left triangle $h = c\sin A$, and in the right triangle $h = a\sin C$. Setting them equal gives $c\sin A = a\sin C$, which rearranges to $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$. That is the sine rule.
Side $a$ is opposite angle $A$. Every sine rule ratio pairs a side with the angle facing it, never with the angle next to it.
Use it when you have a matching pair
The sine rule needs one complete side-and-opposite-angle pair. Without that pair, reach for the cosine rule instead.
The area formula needs the included angle
$\tfrac{1}{2}ab\sin C$ works only when $C$ is the angle between sides $a$ and $b$. Any other angle gives the wrong answer.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
The sine rule is $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ in any triangle.
It follows from dropping a perpendicular and writing its height two ways.
The area of a triangle is $\tfrac{1}{2}ab\sin C$, where $C$ is the angle between $a$ and $b$.
Angles may be given in degrees, or in degrees and minutes, where 60 minutes make one degree.
Understand
Concepts
Why the sine rule is the right-angled case applied twice, not a new fact.
Why you need a complete side-and-opposite-angle pair before the sine rule is usable.
Why the area formula requires the angle included between the two sides.
Can do
Skills
Examine and reproduce the proof of the sine rule and the area formula.
Find an unknown side or angle in a non-right-angled triangle.
Work with angles in degrees and minutes, and find a triangle area from two sides and the included angle.
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Key terms
Sine ruleA relationship that holds in every triangle, pairing each side with the sine of the angle opposite it. Like this: if $a = 8$, $A = 40^\circ$ and $B = 60^\circ$, then $\frac{8}{\sin 40^\circ} = \frac{b}{\sin 60^\circ}$.
Opposite sideThe side facing an angle, labelled with the matching lower-case letter. Like this: side $a$ is opposite angle $A$, so in triangle $ABC$ side $a$ is the side $BC$.
Included angleThe angle sitting between two named sides, which is the one the area formula needs. Like this: in Area $= \frac{1}{2}ab\sin C$, angle $C$ is between sides $a$ and $b$.
Degrees and minutesA way of writing part of a degree, where 60 minutes make one degree. Like this: $37^\circ 30'$ is $37.5^\circ$, because 30 minutes is half a degree.
Non-right-angled triangleA triangle with no $90^\circ$ angle, so SOH CAH TOA cannot be used directly. Like this: a triangle with angles $50^\circ$, $60^\circ$ and $70^\circ$.
Perpendicular heightA line dropped from a vertex at right angles to the opposite side, used to split a triangle into two right-angled triangles. Like this: dropping $h$ from $B$ onto $AC$ creates two right triangles sharing $h$.
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Proving the sine rule
Work through the core explanation before applying it.
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Proving the sine rule
core concept
Take any triangle $ABC$ and drop a perpendicular of height $h$ from vertex $B$ to side $AC$. This splits it into two right-angled triangles that share the side $h$.
In the left-hand triangle, $\sin A = \dfrac{h}{c}$, so $h = c\sin A$. In the right-hand triangle, $\sin C = \dfrac{h}{a}$, so $h = a\sin C$.
The same $h$ cannot equal two different things, so $c\sin A = a\sin C$. Dividing both sides by $\sin A \sin C$ gives $\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$. Repeating with a perpendicular from a different vertex brings in $\dfrac{b}{\sin B}$.
The syllabus asks you to examine the proof. You may be asked to reproduce it, so learn the two-expressions-for-$h$ step rather than only the final formula.
Quick check: in the proof, why can we write $c\sin A = a\sin C$?
Drop a perpendicular $h$ from a vertex. Writing $h$ from each of the two right-angled triangles gives $h = c\sin A$ and $h = a\sin C$. Equating them and dividing gives the sine rule. It is the right-angled case used twice.
Pause, copy the proof in three lines: drop $h$, write $h = c\sin A$ and $h = a\sin C$, equate and divide, into your book.
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Finding a side or an angle
core concept
We just saw that the sine rule comes from writing one perpendicular height two different ways. That raises a question: what do you actually need to know before you can use it? This card answers it → one complete side-and-opposite-angle pair, plus one more piece of information.
To find a **side**, put the unknown on top: $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$, so $b = \dfrac{a\sin B}{\sin A}$. You need the pair $a$ and $A$, plus the angle $B$ facing the side you want.
To find an **angle**, put the sines on top: $\dfrac{\sin B}{b} = \dfrac{\sin A}{a}$, so $\sin B = \dfrac{b\sin A}{a}$. Then take the inverse sine.
If the angles are given in degrees and minutes, convert to decimal degrees before calculating: $37^\circ 30'$ is $37.5^\circ$, because $30 \div 60 = 0.5$.
Angle sum first. If you are given two angles, find the third by subtracting from $180^\circ$ before doing anything else. It often turns out to be the pair you were missing.
Fill the blank: $45^\circ 30'$ written in decimal degrees is $45.$ $^\circ$.
Sine rule for a side: unknown on top, $b = \frac{a\sin B}{\sin A}$. For an angle: sines on top, $\sin B = \frac{b\sin A}{a}$, then inverse sine. Convert degrees and minutes to decimal degrees first, dividing the minutes by 60.
Pause, copy both rearrangements of the sine rule, the degrees-and-minutes conversion, and the reminder to find the third angle from the $180^\circ$ sum first, into your book.
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The area of a triangle from two sides and the included angle
core concept
We just saw how the sine rule finds missing sides and angles. That raises a question: the familiar area formula needs a perpendicular height, so what do you do when you only have two sides and the angle between them? This card answers it → the same perpendicular that proved the sine rule gives the area formula.
Start from Area $= \tfrac{1}{2} \times \text{base} \times \text{height}$. Take $b$ as the base, and drop the perpendicular $h$ as before, where $h = a\sin C$.
Substituting gives Area $= \tfrac{1}{2} b \times a\sin C = \tfrac{1}{2}ab\sin C$. So the formula is the standard one with the height rewritten.
The angle must be the one **between** the two sides you use. With sides $a$ and $b$ the angle is $C$; with $b$ and $c$ it is $A$.
The angle is not optional. $\tfrac{1}{2}ab\sin C$ with a non-included angle is simply wrong, not an approximation. Label the diagram before substituting.
Which is NOT a correct area formula for triangle $ABC$?
Area $= \frac{1}{2}ab\sin C$ comes from the standard base-times-height formula with $h = a\sin C$. The angle must be included between the two sides used, so $\frac{1}{2}ab\sin A$ is wrong because $A$ is not between $a$ and $b$.
Pause, copy the derivation from $\frac{1}{2}bh$ with $h = a\sin C$, and the three valid versions of the formula, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · FINDING A SIDE
In triangle $ABC$, $A = 40^\circ$, $B = 75^\circ$ and $a = 12$ cm. Find $b$ correct to one decimal place.
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Pair available: $a = 12$ with $A = 40^\circ$
The sine rule needs one complete side-and-opposite-angle pair.
Find the area of a triangle with sides 6 cm and 9 cm and an included angle of $40^\circ$.
Write $28^\circ 45'$ in decimal degrees.
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Revisit the largest angle
Run the quick drill and copy the summary into your book.
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Revisit the largest angle
At the start you argued that the largest angle sits opposite the longest side. Explain how the sine rule makes that obvious in one line, and say what extra information two angles and one side actually give you.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
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Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. In triangle $PQR$, $P = 48^\circ$, $Q = 67^\circ$ and $p = 14$ cm. Find $q$ correct to one decimal place. (3 marks)
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ApplyBand 54 marks
Q2. A triangular block has sides 23 m and 31 m with an included angle of $71^\circ 24'$. Find its area to the nearest square metre, showing your conversion of the angle. (4 marks)
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UnderstandBand 43 marks
Q3. Prove the sine rule for an acute-angled triangle $ABC$ by dropping a perpendicular from $B$ to $AC$. (3 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $a = \frac{15\sin 35^\circ}{\sin 80^\circ} \approx 8.7$. Practice 2: $\sin B = \frac{8\sin 50^\circ}{10} \approx 0.6128$, so $B \approx 37.8^\circ$. Practice 3: $\frac{1}{2}(6)(9)\sin 40^\circ \approx 17.4$ cm$^2$. Practice 4: $28.75^\circ$.
Q3 (3 marks): Drop a perpendicular $h$ from $B$ to $AC$, meeting it at $H$ [1]. In triangle $ABH$, $\sin A = \frac{h}{c}$ so $h = c\sin A$; in triangle $CBH$, $\sin C = \frac{h}{a}$ so $h = a\sin C$ [1]. Therefore $c\sin A = a\sin C$, and dividing by $\sin A \sin C$ gives $\frac{a}{\sin A} = \frac{c}{\sin C}$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Sine Rule Sprint
earn bronze · silver · gold
Find missing sides, angles and areas in non-right-angled triangles at speed. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.