Ext Maths · 6 Trigonometry

6.5 Non-right-angled triangles

Before you start

SOH CAH TOA and the accuracy rules: 3 s.f. for lengths, 1 d.p. for angles (6.2).
The graphs of $y = \sin x$ and $y = \cos x$ , and that $\sin (180 ° - x) = \sin x$ (6.4).
Bearings, and alternate angles between parallel north lines.
Angles in a triangle add up to $180 °$ .

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 = ½ab sin C.

Part 1 of 7

What the formula page gives you, and the labelling it expects

SOH CAH TOA needs a right angle. For any other triangle you use the sine rule, the cosine rule and the area formula $\frac{1}{2} a b \sin C$ .
All three are given in the List of formulas on page 2 of the paper. You don't need to memorise them. You do need to choose the right one and substitute into it correctly.
Every one of these formulas uses the same labelling: capital letters for angles, and the same lower-case letter for the side opposite.
Side a faces angle A, side b faces angle B, side c faces angle C.
Side $a$ is opposite angle $A$ , so it does not touch $A$ . The same goes for $b$ with $B$ and $c$ with $C$ .
On the paper the triangle might be $P Q R$ . Then the side opposite angle $P$ is $Q R$ . Relabel before you substitute, or write the formula directly in the question's letters, such as $\frac{Q R}{\sin P} = \frac{P R}{\sin Q}$ .

Part 2 of 7

Obtuse angles: what sine and cosine do past 90°

A triangle without a right angle can have one obtuse angle, between $90 °$ and $180 °$ . So you need sine and cosine for those angles too.
Sine is positive for obtuse angles, and it repeats the acute values: $\sin (180 ° - x) = \sin x$ . For example, $\sin 150 ° = \sin 30 ° = \frac{1}{2}$ .
That means a sine value alone cannot tell you whether an angle is acute or obtuse. Keep that in mind; it causes the ambiguous case below.
Cosine is negative for obtuse angles. For example, $\cos 120 ° = - \frac{1}{2}$ . That is the cosine graph dipping below the axis after $90 °$ .
So a cosine value does tell you: positive means acute, negative means obtuse, zero means exactly $90 °$ .

Part 3 of 7

The sine rule

The sine rule pairs each side with the angle opposite it.
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
Use it when you know a matching pair (a side and its opposite angle), plus one more side or angle.
You only ever use two of the three fractions: the pair you know completely, and the pair containing your unknown.
If you know two angles, find the third first, because angles in a triangle add to $180 °$ . Often that third angle is the one opposite the side you were given.
To find a side, use the rule as printed, with sides on top. Then multiply across.
To find an angle, turn both fractions upside down first, so the unknown is on top.
$\frac{\sin A}{a} = \frac{\sin B}{b}$
Sense check: in any triangle the longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.

Part 4 of 7

The ambiguous case

When you use the sine rule to find an angle, the inverse key gives only the acute answer, $θ$ . But $180 ° - θ$ has exactly the same sine, so it might also be correct.
This happens when you are given two sides and an angle that is not between them. Those facts can sometimes fit two different triangles.
The 7 cm side can swing to two positions, so the same three facts make an acute-angled and an obtuse-angled triangle.
Picture it: side $A B$ and the angle at $A$ are fixed, and the $7$ cm side hangs from $B$ . It can swing to meet the base line at two points, one near $A$ and one further away.
The test: work out $180 ° - θ$ . Add it to the angle you were given. If the total is less than $180 °$ , both triangles exist and both answers are valid.
If the total is $180 °$ or more, the obtuse option is impossible, and only $θ$ is correct.
For example, with $A B = 10$ cm, angle $A = 40 °$ and $B C = 12$ cm, the sine rule gives $\sin C = 0.5357$ , so $C = 32.4 °$ or $147.6 °$ . But $40 ° + 147.6 ° = 187.6 °$ , which is too much, so only $C = 32.4 °$ works.
A quick predictor: if the side opposite the given angle is the longer of the two given sides (as $12 > 10$ above), there is only one triangle.
If the question tells you the angle is obtuse, or asks for 'the two possible values', that tells you which answer to give.

Part 5 of 7

The cosine rule

The cosine rule works when there is no matching pair.
$a^{2} = b^{2} + c^{2} - 2 b c \cos A$
Use it with two sides and the angle between them (to find the third side), or with all three sides (to find an angle).
Look at its shape: the side on the left, $a$ , is opposite the angle in the cosine, $A$ . The other two sides, $b$ and $c$ , are the ones that meet at $A$ .
To find an angle, substitute the three sides and rearrange. Or use this rearranged form, where $a$ must be the side opposite the angle you want:
$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
If $\cos A$ comes out negative, the angle is obtuse, and $\cos^{- 1}$ gives the obtuse angle directly. The cosine rule is never ambiguous.
The obtuse-angle trap when finding a side: if $A$ is obtuse, $\cos A$ is negative, so $- 2 b c \cos A$ is added. If you work out $2 b c \cos A$ separately and forget its minus sign, your side comes out far too short.
The last step for a side is a square root. The rule gives $a^{2}$ .
If $A = 90 °$ then $\cos A = 0$ and the rule becomes $a^{2} = b^{2} + c^{2}$ , which is Pythagoras. The cosine rule is Pythagoras with a correction for angles that are not right angles.

Part 6 of 7

Choosing the rule

Write down what you know as sides and angles, then run down this list and stop at the first match.
Is there a right angle? Use SOH CAH TOA or Pythagoras. They are quicker.
Two angles known? Find the third from $180 °$ , then use the sine rule.
A side and its opposite angle known? Sine rule.
Two sides and the angle between them? Cosine rule for the third side, or $\frac{1}{2} a b \sin C$ for the area.
Three sides? Cosine rule for an angle.
Once you have found one new value, you usually have a matching pair and can switch to the quicker sine rule.
When you do switch, find the angle opposite the shortest side with the sine rule. It is the smallest angle, so it must be acute, and the ambiguous case cannot trip you. Find the largest angle with the cosine rule or from $180 °$ .

Part 7 of 7

Area of any triangle: ½ab sin C

The formula needs two sides and the angle between them, the included angle.
$Area = \frac{1}{2} a b \sin C$
Why it works: take $a$ as the base. The height of the triangle, measured from the base up to the opposite vertex, is $b \sin C$ (SOH in the right-angled triangle that the height makes). So $\frac{1}{2} \times base \times height = \frac{1}{2} \times a \times b \sin C$ .
That also shows why the angle must be the one between the two sides: it is the angle that sets the height.
If you only have an angle that isn't between your two sides, find the included angle first, using $180 °$ or the sine rule.
Give areas in square units, such as cm², to 3 s.f.
Working backwards. If the area is given and you want the angle, rearrange to $\sin C = \frac{2 \times Area}{a b}$ . Because it is a sine, both $C$ and $180 ° - C$ give the same area. There are two possible angles unless the question rules one out.

Worked examples

Example 1. In triangle $P Q R$ , angle $P = 48 °$ , angle $Q = 67 °$ and $P Q = 12.5$ cm. Calculate the length of $Q R$ .

Show the working

Two angles are known, so find the third first.
$R = 180 ° - 48 ° - 67 ° = 65 °$
Now $P Q = 12.5$ cm is opposite angle $R = 65 °$ , a matching pair. $Q R$ is opposite angle $P = 48 °$ . That is the sine rule.
Write the two fractions with sides on top, since you want a side.
$\frac{Q R}{\sin 48 °} = \frac{12.5}{\sin 65 °}$
Multiply both sides by $\sin 48 °$ .
$Q R = \frac{12.5 \sin 48 °}{\sin 65 °} = 10.249 \dots$
Round to 3 s.f.: $10.2$ cm. Check: $48 °$ is the smallest angle, so $Q R$ should be the shortest side. It is shorter than $12.5$ cm.

Answer: $Q R = 10.2$ cm (3 s.f.).

Example 2. A triangle has sides of $6$ cm, $9$ cm and $13$ cm. Calculate its largest angle.

Show the working

Three sides and no angles means the cosine rule. The largest angle is opposite the longest side, so $a = 13$ , with $b = 6$ and $c = 9$ .
Use the rearranged form.
$\cos A = \frac{6^{2} + 9^{2} - 13^{2}}{2 \times 6 \times 9} = \frac{36 + 81 - 169}{108} = \frac{- 52}{108}$
The cosine is negative, so the angle is obtuse. The inverse key handles that correctly, with no second answer to consider.
$A = \cos^{- 1} (\frac{- 52}{108}) = 118.782 \dots °$
Round to 1 d.p.: $118.8 °$ .

Answer: $118.8 °$ .

Example 3. In triangle $A B C$ , $A B = 10$ cm, $B C = 7$ cm and angle $B A C = 40 °$ . Calculate the two possible sizes of angle $A C B$ .

Show the working

$B C = 7$ cm is opposite the known angle $A = 40 °$ , a matching pair. $A B = 10$ cm is opposite the angle $C$ you want. Sine rule, turned upside down to find an angle.
$\frac{\sin C}{10} = \frac{\sin 40 °}{7}$
Multiply both sides by $10$ .
$\sin C = \frac{10 \sin 40 °}{7} = 0.9182 \dots$
The inverse key gives the acute answer.
$C = \sin^{- 1} (0.9182 \dots) = 66.674 \dots °$
The obtuse partner has the same sine.
$180 ° - 66.674 \dots ° = 113.325 \dots °$
Test both. $40 ° + 66.7 ° = 106.7 °$ and $40 ° + 113.3 ° = 153.3 °$ are each less than $180 °$ , so both triangles exist.
This matches the predictor: the side opposite the $40 °$ angle ( $7$ cm) is the shorter of the two given sides, so two triangles were possible. They really are different: $A C$ comes out as $10.4$ cm in one and $4.89$ cm in the other.

Answer: Angle $A C B = 66.7 °$ or $113.3 °$ .

Example 4. A triangle has two sides of $8$ cm and $11$ cm. Its area is $30$ cm². Calculate the two possible sizes of the angle between these two sides.

Show the working

Two sides, the angle between them and the area are linked by the area formula.
$30 = \frac{1}{2} \times 8 \times 11 \times \sin C$
Simplify: $\frac{1}{2} \times 8 \times 11 = 44$ , so divide by $44$ .
$\sin C = \frac{30}{44} = 0.6818 \dots$
Inverse sine gives the acute angle.
$C = \sin^{- 1} (0.6818 \dots) = 42.985 \dots °$
The obtuse angle $180 ° - 42.985 \dots ° = 137.014 \dots °$ has the same sine, so it gives the same area. Nothing in the question rules it out.
Round each to 1 d.p.

Answer: $43.0 °$ or $137.0 °$ .

Where people go wrong

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

Instead: Use the sine rule or the cosine rule.

Opposite, adjacent and hypotenuse only exist relative to a right angle.

Mistake: Giving only the calculator's acute angle when the sine rule finds an angle.

Instead: Also test $180 ° - θ$ : it is valid if it still leaves the angle sum below $180 °$ .

$\sin θ = \sin (180 ° - θ)$ , so two different triangles can fit the same data.

Mistake: Losing the minus sign when the included angle is obtuse, so $- 2 b c \cos A$ is subtracted instead of added.

Instead: Type the whole expression into the calculator in one go, or remember that a negative cosine makes that term positive.

For $b = 7.5$ , $c = 11$ , $A = 124 °$ the correct side is $16.4$ cm. Dropping the sign gives $\sqrt{177.25 - 92.27} = 9.22$ cm, which is shorter than the $11$ cm side even though it faces the obtuse angle, so it can't be right.

Mistake: Using an angle that is not between the two sides in $\frac{1}{2} a b \sin C$ .

Instead: Use the included angle, where the two sides meet.

$b \sin C$ is the height only when $C$ is the angle between $a$ and $b$ . Any other angle gives a wrong height and so a wrong area.

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

Instead: Label every side from the angle it faces before substituting.

All three formulas assume that convention, and a mislabelled substitution gives a believable wrong answer.

Check yourself

You know two angles and one side of a triangle. Which rule finds the other sides?
Answer
The sine rule, after finding the third angle from $180 °$ if you need it.
You know all three sides. Which rule finds an angle, and which angle is safest to find first?
Answer
The cosine rule. Find the largest angle, opposite the longest side, first: the cosine rule handles an obtuse angle, and the remaining angles are then acute.
Calculate the third side of a triangle with sides $7.5$ cm and $11$ cm and an included angle of $124 °$ .
Answer
$a^{2} = {7.5}^{2} + 11^{2} - 2 (7.5) (11) \cos 124 ° = 177.25 + 92.27 = 269.52$ , so $a = 16.4$ cm (3 s.f.).
Without a calculator, write down the exact values of $\sin 150 °$ and $\cos 120 °$ .
Answer
$\sin 150 ° = \sin 30 ° = \frac{1}{2}$ and $\cos 120 ° = - \frac{1}{2}$ .
The cosine rule gives $\cos A = - 0.25$ . Is angle $A$ acute or obtuse? Explain.
Answer
Obtuse, because the cosine of an angle between $90 °$ and $180 °$ is negative.
When can the sine rule give two valid answers for an angle?
Answer
When you are given two sides and an angle that is not between them, and the obtuse option $180 ° - θ$ added to the given angle is still less than $180 °$ .
Without a calculator, work out the area of a triangle with sides $7$ cm and $10$ cm and an included angle of $30 °$ .
Answer
$\frac{1}{2} \times 7 \times 10 \times \sin 30 ° = 35 \times \frac{1}{2} = 17.5$ cm².

Exam questions

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

Question 1. A boat sails from $A$ to $B$ , a distance of $14$ km on a bearing of $050 °$ . It then sails from $B$ to $C$ , a distance of $9$ km on a bearing of $160 °$ .

(a) Show that angle $A B C = 70 °$ . 2 marks No calculator

Mark scheme

B1 bearing of A from B is 230°, or angle between south at B and BA is 50° (alternate angles) oe
B1 230 − 160 = 70, or 50 + 20 = 70, with the 20 from 180 − 160

(b) Calculate the distance $A C$ . 3 marks

Mark scheme

M1 14² + 9² − 2 × 14 × 9 × cos 70° oe
M1 √(their 190.8) ; dep on first M1
A1 13.8 awrt (km)

(c) Calculate angle $B A C$ . 3 marks

Mark scheme

M1 sin BAC / 9 = sin 70° / their AC oe
M1 sin BAC = 0.612 awrt, or 9 sin 70° ÷ their AC seen
A1 37.8 awrt; FT their (b)

(d) Calculate the bearing of $C$ from $A$ . 2 marks

Mark scheme

M1 50 + their (c)
A1 087.8 awrt; FT their (c)

Question 2. In triangle $P Q R$ , $P Q = 9$ cm, $Q R = 6.5$ cm and angle $Q P R = 35 °$ .

(a) Calculate the two possible values of angle $P R Q$ . 4 marks

Mark scheme

M1 sin PRQ = 9 sin 35° ÷ 6.5 oe
A1 52.6 awrt
M1 180 − their acute angle
A1 127.4 awrt

(b) Explain why angle $P R Q$ would have only one possible value if $Q R$ were $10$ cm instead. 2 marks

Mark scheme

B1 finds sin PRQ = 0.516 and the obtuse option 148.9 awrt, or states that PRQ must be smaller than QPR because QR is longer than PQ
B1 the obtuse option would make the angle sum more than 180° (35 + 148.9 = 183.9) so it is impossible oe

Question 3. Triangle $X Y Z$ has $X Y = 7.2$ cm, $X Z = 9.5$ cm and angle $Y X Z = 47 °$ .

(a) Calculate the area of triangle $X Y Z$ . 2 marks

Mark scheme

M1 ½ × 7.2 × 9.5 × sin 47° oe
A1 25.0 awrt (cm²)

(b) A different triangle also has two sides of $7.2$ cm and $9.5$ cm and the same area as triangle $X Y Z$ . Find the angle between these two sides. 2 marks No calculator

Mark scheme

M1 recognises sin θ = sin 47°, so θ = 180° − 47° oe
A1 133 cao

How this is examined

A bearings journey turned into a triangle, solved with the cosine rule then the sine rule: A long multi-part question: find an angle inside the triangle from two bearings, then a distance with the cosine rule, then an angle with the sine rule, then a final bearing. A reason for the interior angle, such as the back bearing $230 °$ minus $160 °$ . A correct cosine-rule substitution with a square root. The sine rule with their distance, which follow-through protects. A bearing given with three figures.
Finding an angle, including an obtuse one, from three sides: A 3- or 4-mark Calculate, often 'the largest angle' of a triangle or a field. The side opposite the wanted angle placed correctly in $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$ (or the rule as printed, rearranged). The value of $\cos A$ seen. The angle to 1 d.p.
The ambiguous case: both possible angles from the sine rule: A Calculate asking for 'the two possible values' of an angle, sometimes followed by an Explain. The sine rule set up for the angle, $\sin C$ correct to at least 3 s.f., the acute angle, then $180 ° -$ that angle. An explanation needs $\sin θ = \sin (180 ° - θ)$ and the angle-sum check.
Area with the included angle, forwards and backwards: A 2-mark Calculate for an area, or a 2- or 3-mark part giving the area and asking for the angle. $\frac{1}{2} \times$ side $\times$ side $\times \sin (included angle)$ with the numbers in, then the area in square units. Backwards, the rearrangement to $\sin C$ , and the obtuse option if asked.

Words to know

Sine rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ ; given in the List of formulas. Use it when a side and its opposite angle are known.
Cosine rule: $a^{2} = b^{2} + c^{2} - 2 b c \cos A$ ; given in the List of formulas. Use it with two sides and the included angle, or with three sides.
Included angle: The angle between two named sides, at the vertex where they meet.
Obtuse angle: An angle between $90 °$ and $180 °$ . Its sine is positive and its cosine is negative.
Ambiguous case: When the sine rule finds an angle, both $θ$ and $180 ° - θ$ may fit the data, giving two possible triangles.

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