Maths · 7 Trigonometry

7.5 Non-right-angled triangles

Before you start

Trigonometry in right-angled triangles, and the accuracy rules for angles (7.2).
The exact values of sine and cosine at $30 °$ , $45 °$ and $60 °$ (7.3).
Rearranging a formula where the unknown appears inside a fraction (2.5).

Syllabus objectives covered (2)

Use the sine and cosine rules in calculations involving lengths and angles for any triangle.
Use the formula area of triangle = ½ ab sin C.

Part 1 of 4

Labelling, and what is given to you

SOH CAH TOA only works in right-angled triangles. For any other triangle you need the sine rule or the cosine rule.
Good news from the syllabus: the sine rule, the cosine rule and the area formula are all given on the formula list for Papers 2 and 4. You do not have to memorise them.
What you do have to know is which one to use, and that is entirely about matching the rule to what the question gives you.
First, the labelling convention, which every one of these formulas depends on. Capital letters for angles, lower-case for the sides opposite them.
So side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , side $c$ is opposite angle $C$ . Label your diagram this way before you write anything down.
Capital for the angle, lower-case for the side facing it.
Get this labelling wrong and the right formula gives the wrong answer, which is much harder to spot than using the wrong formula.

Part 2 of 4

The sine rule

The sine rule links a side with the angle opposite it.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Use it when you have a matching pair — a side and its opposite angle — plus one more piece of information.
That covers two situations: two angles and any side, or two sides and an angle opposite one of them.
You only ever use two of the three fractions at a time. Pick the pair that contains your unknown and the pair you know.
Finding a side, use it as printed. Finding an angle, turn it upside down first — it saves a rearrangement:
$\frac{\sin A}{a} = \frac{\sin B}{b}$
The ambiguous case. The syllabus names this explicitly, so it will be examined.
When you use the sine rule to find an angle, there may be two possible answers, because $\sin θ = \sin (180 ° - θ)$ . The calculator only gives you the acute one.
So if a question mentions an obtuse angle, or asks for two possible triangles, work out $180 ° - (your answer)$ as well.
Discard any second answer that would make the angles sum to more than $180 °$ . That check usually settles which of the two is intended.

Part 3 of 4

The cosine rule

The cosine rule handles the cases the sine rule cannot — when you have no matching side-and-opposite-angle pair.
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$
Use it when you have two sides and the angle between them (to find the third side), or all three sides (to find any angle).
Notice the structure: the side you want, $a$ , sits opposite the angle you know, $A$ . The two known sides are $b$ and $c$ .
'The angle between them' is the key phrase. If the given angle is not between the two given sides, the cosine rule will not help and you need the sine rule instead.
To find an angle from three sides, rearrange it:
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
This rearranged form is not on the formula list — only the first version is — so either learn it or be ready to rearrange under pressure.
A useful feature: if the answer comes out with $\cos A$ negative, the angle is obtuse, and the calculator handles it correctly with no ambiguity. Unlike the sine rule, the cosine rule never has two answers.
That is a reason to prefer the cosine rule when both would work: it cannot give you the ambiguous case.
Sanity check: if $A = 90 °$ then $\cos A = 0$ and the formula collapses to $a^{2} = b^{2} + c^{2}$ — Pythagoras. The cosine rule is its generalisation.

Part 4 of 4

Area of a triangle

The familiar $\frac{1}{2} \times base \times height$ needs a perpendicular height, which a non-right-angled triangle rarely gives you. This formula does not.
$Area = \frac{1}{2} a b \sin C$
It needs two sides and the angle between them — the same information the cosine rule uses to find a third side.
The angle must be the included angle, the one between the two sides you are using. This is where marks are lost: using an angle that is not between them gives a plausible but wrong area.
The letters say it: sides $a$ and $b$ , angle $C$ — and $C$ is the angle where sides $a$ and $b$ meet.
Put more usefully: whichever two sides you multiply, the angle you need is the one where those two sides meet.
If the question gives you an angle that is not between the two sides, find the one you need first — using the fact that angles in a triangle sum to $180 °$ , or the sine rule.
Areas are lengths squared, so give the units as cm², m² and so on, and use 3 significant figures unless told otherwise.

Worked examples

Example 1. In triangle $A B C$ , $A = 52 °$ , $B = 71 °$ and $a = 9.4$ cm. Calculate the length of side $b$ .

Show the working

Check what you have: a side ( $a$ ) with its opposite angle ( $A$ ), plus another angle. That is a matching pair, so use the sine rule.
Write the two fractions you need — the known pair and the unknown pair.
$\frac{a}{\sin A} = \frac{b}{\sin B}$
Substitute the values.
$\frac{9.4}{\sin 52 °} = \frac{b}{\sin 71 °}$
Multiply both sides by $\sin 71 °$ .
$b = \frac{9.4 \sin 71 °}{\sin 52 °}$
Evaluate in one go, without rounding partway: $b = 11.3$ cm.
Sanity check: $B$ is a bigger angle than $A$ , so $b$ should be longer than $a$ . It is.

Answer: $b = 11.3$ cm (3 s.f.).

Example 2. A triangle has sides $b = 7$ cm and $c = 10$ cm with an included angle $A = 43 °$ . Calculate the length of side $a$ .

Show the working

Two sides and the angle between them, with no matching side-and-opposite-angle pair. That is the cosine rule.
Write the formula as given.
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$
Substitute.
$a^{2} = 7^{2} + 10^{2} - 2 (7) (10) \cos 43 °$
Evaluate the pieces: $49 + 100 = 149$ , and $140 \cos 43 ° = 102.4$ .
$a^{2} = 149 - 102.4 = 46.6$
Take the square root — the step most often forgotten here, because the formula gives $a^{2}$ .
$a = \sqrt{46.6} = 6.83 cm$
Check the size: $a$ is opposite the smallest angle, so it should be the shortest side. At $6.83$ cm it is.

Answer: $a = 6.83$ cm (3 s.f.).

Example 3. A triangle has sides $a = 8$ cm, $b = 5$ cm and $c = 11$ cm. Calculate angle $A$ , correct to one decimal place.

Show the working

Three sides and no angles, so this is the cosine rule rearranged to find an angle.
Write the rearranged form.
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
Substitute, taking care that $a$ is the side opposite the angle you want.
$\cos A = \frac{25 + 121 - 64}{2 (5) (11)} = \frac{82}{110}$
That gives $\cos A = 0.7455$ , so use the inverse cosine.
$A = \cos^{- 1} (0.7455) = 41.8 °$
Give the angle to one decimal place, as the syllabus requires. Note $\cos A$ came out positive, so the angle is acute — consistent with $a$ not being the longest side.

Answer: $A = 41.8 °$ .

Example 4. A triangle has sides of $12$ cm and $9$ cm with an angle of $58 °$ between them. Calculate its area.

Show the working

Two sides and the included angle is exactly what the area formula needs.
Write the formula, with the two sides as $a$ and $b$ and the included angle as $C$ .
$Area = \frac{1}{2} a b \sin C$
Substitute.
$Area = \frac{1}{2} (12) (9) \sin 58 °$
Evaluate: $\frac{1}{2} (108) = 54$ , and $54 \sin 58 ° = 45.8$ .
$Area = 45.8 {cm}^{2}$
Give the units as cm², since an area is a length squared. Check for plausibility: the maximum possible area with these sides is $54$ cm² at $90 °$ , and $45.8$ is sensibly below it.

Answer: $45.8 {cm}^{2}$ (3 s.f.).

Where people go wrong

Mistake: Using SOH CAH TOA in a triangle with no right angle.

Instead: Use the sine or cosine rule instead.

The three ratios are defined by reference to a right angle. Without one they are meaningless.

Mistake: Forgetting to square-root at the end of a cosine rule calculation.

Instead: The formula gives $a^{2}$ , not $a$ .

An answer of $46.6$ cm where $6.83$ cm was expected is a missing root, and a quick size check catches it.

Mistake: Using the sine rule when the given angle is not opposite a given side.

Instead: The sine rule needs a matching pair. With two sides and the angle between them, use the cosine rule.

Every sine-rule fraction pairs a side with its own opposite angle, so without such a pair there is nothing to substitute.

Mistake: Using an angle that is not between the two sides in the area formula.

Instead: It must be the included angle, where the two sides meet.

The formula derives from $\frac{1}{2} \times base \times height$ with the height found from the included angle. Any other angle gives a wrong height.

Mistake: Giving only the acute answer when the sine rule is used to find an angle.

Instead: Consider $180 ° - θ$ as well — the ambiguous case.

$\sin θ = \sin (180 ° - θ)$ , so two triangles can fit the same data. The syllabus names this case explicitly.

Mistake: Mislabelling, so that $a$ is not opposite $A$ .

Instead: Capital letters for angles, lower-case for the side opposite each.

Every one of these formulas assumes that convention, and breaking it produces a wrong answer that looks entirely reasonable.

Check yourself

State the sine rule and say what information you need in order to use it.
Answer
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ . You need a side with its opposite angle, plus one more piece of information.
State the cosine rule and the two situations in which it is used.
Answer
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$ . Use it with two sides and the included angle, or with all three sides.
A triangle has $a = 6$ , $b = 9$ and $C = 40 °$ . Which rule applies, and what would it find?
Answer
The cosine rule, since $C$ is between sides $a$ and $b$ . It would find side $c$ .
What is the ambiguous case, and which rule does it affect?
Answer
When the sine rule is used to find an angle, both $θ$ and $180 ° - θ$ may be valid, because they have the same sine. It affects the sine rule only.
Find the area of a triangle with sides $7$ cm and $10$ cm and an included angle of $30 °$ .
Answer
$\frac{1}{2} (7) (10) \sin 30 ° = 35 \times 0.5 = 17.5 {cm}^{2}$ .
How does the cosine rule relate to Pythagoras' theorem?
Answer
When $A = 90 °$ , $\cos A = 0$ and the formula becomes $a^{2} = b^{2} + c^{2}$ . Pythagoras is the special case.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. In triangle $P Q R$ , $P Q = 8.2$ cm, $Q R = 6.5$ cm and angle $P Q R = 112 °$ .

(a) Calculate the length of $P R$ . 3 marks

Mark scheme

M1 correct use of the cosine rule with the included angle 112°
M1 8.2² + 6.5² - 2(8.2)(6.5)cos112° evaluated oe
A1 12.3 awrt (cm)

(b) Calculate angle $Q P R$ , correct to one decimal place. 3 marks

Mark scheme

M1 sine rule with sin(QPR)/6.5 = sin112°/their PR oe
M1 sin(QPR) = 0.4899 awrt seen; FT their (a)
A1 29.3° awrt; FT

(c) Calculate the area of triangle $P Q R$ . 2 marks

Mark scheme

M1 ½ × 8.2 × 6.5 × sin112° oe
A1 24.7 awrt (cm²); unit required

Question 2. A triangular field has sides of $140$ m, $180$ m and $250$ m.

(a) Calculate the largest angle of the field, correct to one decimal place. 4 marks

Mark scheme

B1 identifies the largest angle as the one opposite the 250 m side
M1 cos A = (140² + 180² - 250²)/(2 × 140 × 180) oe
M1 cos A = -0.1746 awrt seen
A1 100.1° awrt

(b) Calculate the area of the field. 2 marks

Mark scheme

M1 ½ × 140 × 180 × sin(their largest angle) oe
A1 12 400 awrt (m²); FT their (a)

(c) Explain how the sign of $\cos A$ in part (a) tells you the angle is obtuse. 1 mark No calculator

Mark scheme

B1 cosine is negative for angles between 90° and 180°, so a negative value means an obtuse angle oe

Question 3. In triangle $A B C$ , $a = 11$ cm, $b = 14$ cm and angle $A = 38 °$ .

(a) Calculate the two possible values of angle $B$ , correct to one decimal place. 4 marks

Mark scheme

M1 sinB/14 = sin38°/11 oe
M1 sin B = 0.7834 awrt seen
A1 51.6° awrt
A1 128.4° awrt; dep on the first value

(b) Explain why two answers are possible. 2 marks No calculator

Mark scheme

B1 sin θ = sin(180° - θ), so two angles have the same sine
B1 both give a valid triangle because the angles still sum to less than 180° oe

Words to know

Sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ , used when a side and its opposite angle are both known.
Cosine rule: $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ , used with two sides and the included angle, or with three sides.
Included angle: The angle between two named sides, where those sides meet.
Ambiguous case: When the sine rule gives an angle, both $θ$ and $180 ° - θ$ may satisfy the data.
Obtuse angle: An angle between $90 °$ and $180 °$ , whose cosine is negative.

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