Year 1 Pure · Chapter 8

Triangle Geometry

Specification: Edexcel Pure 5 — Trigonometry · AQA section E

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In this chapter

1. Labelling and right-angled triangles
2. The sine rule
3. The cosine rule
4. Area of a triangle
Practice

Key facts — this chapter

Labelling: side $a$ is opposite angle $A$ , side $b$ opposite $B$ , side $c$ opposite $C$ .
Right-angled triangles: SOH-CAH-TOA — $\sin = \frac{\text{opp}}{\text{hyp}}$ , $\cos = \frac{\text{adj}}{\text{hyp}}$ , $\tan = \frac{\text{opp}}{\text{adj}}$ .
Sine rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ — use it with a side and its opposite angle.
Cosine rule: $a^2 = b^2 + c^2 - 2bc\cos A$ — use it for SAS (find a side) or SSS (find an angle).
Area: $\frac{1}{2}ab\sin C$ — two sides and the angle between them.
The sine rule can be ambiguous: $\sin\theta = \sin(180^\circ - \theta)$ , so a “find the angle” sine-rule problem can have two answers.

Once a triangle stops being right-angled, SOH-CAH-TOA alone can’t finish the job. The sine and cosine rules pick up where it leaves off — between them they solve any triangle, given enough information.

1. Labelling and right-angled triangles

First, the convention that makes every formula in this chapter readable: each side is named by the angle it faces.

Key result

In triangle $ABC$ , side $a$ is opposite angle $A$ , side $b$ opposite $B$ , and side $c$ opposite $C$ .

A triangle with vertices A, B, C and sides a, b, c, where side a is opposite angle A, side b opposite B and side c opposite C — Each side faces the angle of the same letter — the key to reading the rules.

For a right-angled triangle you already have everything you need: SOH-CAH-TOA relates an angle to the opposite, adjacent and hypotenuse. Keep it in mind — the sine and cosine rules are both derived from it.

Worked example 8.1 · Right-angled refresher

A right-angled triangle has hypotenuse $10$ and one angle of $35^\circ$ . Find the side opposite the $35^\circ$ angle.

The opposite side and the hypotenuse call for sine (SOH):

\sin 35^\circ = \frac{\text{opp}}{10} \;\;\Longrightarrow\;\; \text{opp} = 10\sin 35^\circ = 5.74 \;\,(3\text{ s.f.})

2. The sine rule

The sine rule links each side to the sine of its opposite angle. Reach for it whenever you have (or want) a side paired with the angle across from it.

Key result

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Turn it upside down — $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ — when you’re finding an angle.

Why it works

Drop a perpendicular of height $h$ from $C$ onto $AB$ . It splits the triangle into two right-angled triangles, and in each one $h$ is the opposite side: $h = b\sin A$ from the left, $h = a\sin B$ from the right. Setting them equal, $b\sin A = a\sin B$ , which rearranges to $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ .

A triangle with an altitude h dropped from C to the base AB, showing h equals b sin A equals a sin B — The shared height h gives b sin A = a sin B in one step.

Worked example 8.2 · Finding a side with the sine rule

In triangle $ABC$ , $A = 50^\circ$ , $B = 60^\circ$ and $a = 10$ . Find $b$ .

Pair each side with its opposite angle:
$\frac{b}{\sin 60^\circ} = \frac{10}{\sin 50^\circ}$
Rearrange and evaluate:
$b = \frac{10\sin 60^\circ}{\sin 50^\circ} = 11.3 \;\,(3\text{ s.f.})$

The ambiguous case

Finding an angle with the sine rule comes with a catch: $\sin\theta$ and $\sin(180^\circ - \theta)$ are equal, so the calculator’s answer might not be the only one.

One fixed side b from A and a fixed angle at A; a second side a swung from C meets the base at two points B1 and B2, giving two possible triangles — With two sides and a non-included angle, the third vertex can land in two places.

Worked example 8.3 · Two possible triangles

In triangle $ABC$ , $A = 40^\circ$ , $a = 9$ and $b = 12$ . Find the two possible values of angle $B$ .

Use the angle form of the sine rule:
$\frac{\sin B}{12} = \frac{\sin 40^\circ}{9} \;\;\Longrightarrow\;\; \sin B = \frac{12\sin 40^\circ}{9} = 0.857$
The calculator gives $B = 59.0^\circ$ . But $\sin B = \sin(180^\circ - B)$ , so $B = 121.0^\circ$ is also possible.
Check both fit ( $A + B < 180^\circ$ ): $40 + 59 = 99^\circ$ ✓ and $40 + 121 = 161^\circ$ ✓ — both are valid triangles.

So $B = 59.0^\circ$ or $B = 121.0^\circ$ .

Common mistake

Don’t trust the calculator’s single answer when finding an angle. Always ask whether the obtuse partner $180^\circ - \theta$ also fits (i.e. whether the angles still sum to less than $180^\circ$ ). If it does, the problem has two solutions.

3. The cosine rule

When the sine rule can’t start — because you don’t have a side-and-its-opposite-angle pair — the cosine rule steps in. It needs either two sides and the angle between them, or all three sides.

Key result

a^2 = b^2 + c^2 - 2bc\cos A

Here $a$ is the side opposite the angle $A$ , and $b$ , $c$ are the two sides that enclose $A$ . Use it to find a side from SAS (two sides and the included angle). Rearranged for the angle,

\cos A = \frac{b^2 + c^2 - a^2}{2bc},

it finds an angle from SSS (all three sides). The letters just rotate — to find $B$ instead, use $\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}$ , and similarly for $C$ .

Worked example 8.4 · Finding a side (SAS)

In triangle $ABC$ , $b = 8$ , $c = 5$ and the included angle $A = 60^\circ$ . Find $a$ .

Substitute into the cosine rule — $a$ is opposite the known angle $A$ :
$a^2 = 8^2 + 5^2 - 2(8)(5)\cos 60^\circ$
Evaluate, using $\cos 60^\circ = \tfrac{1}{2}$ :
$a^2 = 64 + 25 - 80 \times \tfrac{1}{2} = 89 - 40 = 49$

So $a = 7$ .

Worked example 8.5 · Finding an angle (SSS)

A triangle has sides $5$ , $7$ and $8$ . Find the size of the angle opposite the side of length $7$ .

Put the wanted angle as $A$ , so its opposite side is $a = 7$ , with the other two $b = 5$ and $c = 8$ . Rearrange the cosine rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{25 + 64 - 49}{2(5)(8)} = \frac{40}{80} = \tfrac{1}{2}$
So $A = 60^\circ$ .

Common mistake

The angle in the cosine rule must be the one opposite the side you’re finding (equivalently, the angle between the two known sides). In $a^2 = b^2 + c^2 - 2bc\cos A$ , the angle $A$ sits between sides $b$ and $c$ and faces side $a$ — never one of the angles at the other corners.

4. Area of a triangle

The familiar $\frac{1}{2}\times\text{base}\times\text{height}$ becomes far more useful once you can write the height in terms of a side and an angle.

Key result

\text{Area} = \tfrac{1}{2}ab\sin C

— two sides and the angle between them. (Any pair of sides works, as long as you use the angle they enclose.)

Why it works

Take base $a$ . The perpendicular height onto it, dropped from the opposite vertex, is $b\sin C$ — the side $b$ times the sine of the angle $C$ it makes with the base. So $\text{Area} = \frac{1}{2} \times a \times b\sin C$ .

A triangle with base a and the perpendicular height from the opposite vertex marked as b sin C, giving area one half a b sin C — The height onto base a is b sin C, so the area is ½ ab sin C.

Set two sides and the angle between them. The triangle is drawn to scale, and the opposite side (cosine rule) and the area (½ bc sin A) update as you go.

b and c enclose angle A · the opposite side a comes from the cosine rule

Worked example 8.6 · Area from two sides and the included angle

Find the area of the triangle with $a = 8$ , $b = 5$ and included angle $C = 60^\circ$ .

\text{Area} = \tfrac{1}{2}(8)(5)\sin 60^\circ = 20\sin 60^\circ = 17.3 \;\,(3\text{ s.f.})

(This is the same triangle as in Example 8.4 — sides $5$ , $7$ , $8$ — so its area is $10\sqrt{3}$ in exact form.)

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers. Give non-exact answers to 3 s.f.

Warm-up

1 [3 marks]

In triangle $ABC$ , $A = 45^\circ$ , $B = 65^\circ$ and $a = 12$ . Find $b$ .

Show solution

Sine rule, pairing each side with its opposite angle:

\frac{b}{\sin 65^\circ} = \frac{12}{\sin 45^\circ} \;\;\Longrightarrow\;\; b = \frac{12\sin 65^\circ}{\sin 45^\circ} = 15.4 \;\,(3\text{ s.f.})

2 [3 marks]

In triangle $ABC$ , $b = 6$ , $c = 9$ and the included angle $A = 40^\circ$ . Find $a$ .

Show solution

Cosine rule, with $a$ opposite the known angle $A$ :

a^2 = 6^2 + 9^2 - 2(6)(9)\cos 40^\circ = 36 + 81 - 108\cos 40^\circ = 34.267\ldots

So $a = \sqrt{34.267\ldots} = 5.85 \;\,(3\text{ s.f.})$ .

3 [3 marks]

Find the area of the triangle with sides $a = 7$ and $b = 10$ enclosing an angle $C = 55^\circ$ .

Show solution

\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(7)(10)\sin 55^\circ = 35\sin 55^\circ = 28.7 \;\,(3\text{ s.f.})

4 [2 marks]

A triangle has sides $4$ , $6$ and $9$ . State, with a reason, whether the angle opposite the side of length $9$ is acute or obtuse.

Show solution

By the cosine rule the cosine of that angle is

\cos\theta = \frac{4^2 + 6^2 - 9^2}{2(4)(6)} = \frac{16 + 36 - 81}{48} = \frac{-29}{48} < 0

A negative cosine means the angle is obtuse.

Standard

5 [4 marks]

In triangle $ABC$ , $a = 10$ , $b = 7$ and $c = 8$ . Find angle $A$ .

Show solution

$A$ is opposite the longest side, so use the cosine rule rearranged:

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{49 + 64 - 100}{2(7)(8)} = \frac{13}{112} = 0.11607\ldots

A = \cos^{-1}(0.11607\ldots) = 83.3^\circ \;\,(3\text{ s.f.})

6 [4 marks]

In triangle $ABC$ , $A = 35^\circ$ , $a = 9$ and $b = 13$ . Find the two possible values of angle $B$ .

Show solution

Angle form of the sine rule:

\frac{\sin B}{13} = \frac{\sin 35^\circ}{9} \;\;\Longrightarrow\;\; \sin B = \frac{13\sin 35^\circ}{9} = 0.82849\ldots

The calculator gives $B = 55.9^\circ$ . Its obtuse partner $B = 180^\circ - 55.9^\circ = 124.1^\circ$ also fits, since $35 + 124.1 = 159.1^\circ < 180^\circ$ .

So $B = 55.9^\circ$ or $B = 124^\circ \;\,(3\text{ s.f.})$ .

7 [5 marks]

In triangle $ABC$ , $b = 11$ , $c = 6$ and $A = 72^\circ$ .

Find $a$ . [3]
Find the area of the triangle. [2]

Show solution

(a) Cosine rule (SAS, angle $A$ between sides $b$ and $c$ ):

\begin{aligned} a^2 &= 11^2 + 6^2 - 2(11)(6)\cos 72^\circ \\[0.3em] &= 121 + 36 - 132\cos 72^\circ = 116.2\ldots \end{aligned}

a = \sqrt{116.2\ldots} = 10.8 \;\,(3\text{ s.f.})

(b) Area $= \tfrac{1}{2}bc\sin A = \tfrac{1}{2}(11)(6)\sin 72^\circ = 33\sin 72^\circ = 31.4 \;\,(3\text{ s.f.})$ .

8 [5 marks]

In triangle $ABC$ , $A = 48^\circ$ , $B = 61^\circ$ and $b = 14$ .

Find angle $C$ . [1]
Find side $c$ . [2]
Find the area of the triangle. [2]

Show solution

(a) Angles sum to $180^\circ$ : $C = 180^\circ - 48^\circ - 61^\circ = 71^\circ$ .

(b) Sine rule:

c = \frac{b\sin C}{\sin B} = \frac{14\sin 71^\circ}{\sin 61^\circ} = 15.13\ldots = 15.1 \;\,(3\text{ s.f.})

(c) Two sides and the included angle — use $b$ , $c$ and angle $A$ , keeping the unrounded $c$ from part (b):

\text{Area} = \tfrac{1}{2}bc\sin A = \tfrac{1}{2}(14)(15.13\ldots)\sin 48^\circ = 78.7 \;\,(3\text{ s.f.})

Rounding $c$ to $15.1$ first would give $78.6$ : carry full accuracy through, and round only the final answer.

9 [4 marks]

The area of triangle $ABC$ is $40\text{ cm}^2$ . Given $a = 12$ cm and $b = 9$ cm, find the two possible values of the included angle $C$ .

Show solution

\tfrac{1}{2}(12)(9)\sin C = 40 \;\;\Longrightarrow\;\; 54\sin C = 40 \;\;\Longrightarrow\;\; \sin C = 0.7407\ldots

The calculator gives $C = 47.8^\circ$ ; the obtuse partner $C = 132.2^\circ$ also gives this area. So $C = 47.8^\circ$ or $C = 132^\circ \;\,(3\text{ s.f.})$ .

Exam style

10 [6 marks]

A ship sails $8$ km on a bearing of $030^\circ$ from $P$ to $Q$ , then $11$ km on a bearing of $100^\circ$ from $Q$ to $R$ .

Show that angle $PQR = 110^\circ$ . [2]
Find the distance $PR$ . [3]
Find the area of triangle $PQR$ . [1]

Show solution

(a) The bearing of $P$ from $Q$ is $030^\circ + 180^\circ = 210^\circ$ . The angle between $QP$ ( $210^\circ$ ) and $QR$ ( $100^\circ$ ) is $210^\circ - 100^\circ = 110^\circ$ . ✓

(b) Cosine rule with the $110^\circ$ angle between $PQ = 8$ and $QR = 11$ :

\begin{aligned} PR^2 &= 8^2 + 11^2 - 2(8)(11)\cos 110^\circ \\[0.3em] &= 64 + 121 - 176\cos 110^\circ = 245.19\ldots \end{aligned}

PR = 15.7 \text{ km} \;\,(3\text{ s.f.})

(c) Area $= \tfrac{1}{2}(8)(11)\sin 110^\circ = 44\sin 110^\circ = 41.3 \text{ km}^2 \;\,(3\text{ s.f.})$ .

11 [5 marks]

In triangle $ABC$ , $a = 9$ , $b = 10$ and $c = 11$ .

Find the largest angle of the triangle. [3]
Hence find the area of the triangle. [2]

Show solution

(a) The largest angle faces the longest side $c = 11$ , so find $C$ :

\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{81 + 100 - 121}{2(9)(10)} = \frac{60}{180} = \tfrac{1}{3}

C = \cos^{-1}\left(\tfrac{1}{3}\right) = 70.5^\circ \;\,(3\text{ s.f.})

(b) Area $= \tfrac{1}{2}ab\sin C = \tfrac{1}{2}(9)(10)\sin 70.5^\circ = 45\sin 70.5^\circ = 42.4 \;\,(3\text{ s.f.})$ .

12 [6 marks]

Points $A$ , $B$ and $C$ lie on a circle of radius $7$ , with centre $O$ . The angle $AOB$ (at the centre) is $80^\circ$ .

Find the length of the chord $AB$ . [3]
Find the area of triangle $AOB$ . [1]
Hence find the area of the minor segment cut off by the chord $AB$ . [2]

Show solution

(a) Triangle $AOB$ has $OA = OB = 7$ (radii) with the $80^\circ$ angle between them. Cosine rule:

AB^2 = 7^2 + 7^2 - 2(7)(7)\cos 80^\circ = 98 - 98\cos 80^\circ = 80.982\ldots

AB = 9.00 \;\,(3\text{ s.f.})

(b) Area of triangle $AOB = \tfrac{1}{2}(7)(7)\sin 80^\circ = 24.5\sin 80^\circ = 24.1$ .

(c) The sector $AOB$ has area $\tfrac{80}{360}\times\pi\times 7^2 = 34.2$ . The segment is the sector minus the triangle:

34.2 - 24.1 = 10.1 \;\,(3\text{ s.f.})