The Sine Rule, Cosine Rule and Triangle Area

In short: The sine and cosine rules solve triangles that have no right angle. Use the sine rule $\frac{a}{sin A} = \frac{b}{sin B}$ when you know a side and its opposite angle, the cosine rule $c^{2} = a^{2} + b^{2} - 2 ab cos C$ otherwise, and find the area with $\frac{1}{2} ab sin C$ .

Starting point: Years 9 and 10

Label a triangle with capital letters $A$ , $B$ , $C$ for the angles and small letters $a$ , $b$ , $c$ for the sides. Side $a$ is opposite angle $A$ , and so on.

Each side is named after the angle across from it.

The sine rule

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

Use it when you know a side and its opposite angle, plus one more side or angle.

The cosine rule

$c^{2} = a^{2} + b^{2} - 2 ab cos C$

Use it when you know two sides and the angle between them (to find the third side), or all three sides (to find an angle). For an angle, rearrange:

$cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$

Area of any triangle

$Area = \frac{1}{2} ab sin C$

Two sides and the angle between them. For the triangle above, area $= \frac{1}{2} (7) (9) sin 4 0^{\circ} \approx 20.2 cm^{2}$ .

Non-right triangles

Labels

Side a opposite angle A

Sine rule

a / sin A = b / sin B
Needs a matching pair

Cosine rule

c² = a² + b² - 2ab cos C
Two sides and included angle
Three sides for an angle

Area

½ ab sin C

Ambiguous case

Two possible triangles

Year 11Obtuse angles and the ambiguous case

Sine is positive for angles between $9 0^{\circ}$ and $18 0^{\circ}$ too: $sin 15 0^{\circ} = sin 3 0^{\circ} = \frac{1}{2}$ . In general $sin (18 0^{\circ} - θ) = sin θ$ .

So when the sine rule gives an angle, there may be two answers: $θ$ and $18 0^{\circ} - θ$ . This is the ambiguous case.

Example. $a = 8$ , $b = 10$ , $A = 4 0^{\circ}$ . Then $sin B = \frac{10 sin 4 0 ^{\circ}}{8} \approx 0.803$ , so $B \approx 53. 5^{\circ}$ or $B \approx 126. 5^{\circ}$ . Both work, because $4 0^{\circ} + 126. 5^{\circ} < 18 0^{\circ}$ . Two different triangles fit the data.

Side a = 8 can swing to two places on the base, shown by the dashed arc. B₁ gives an obtuse angle (126.5°) and B₂ an acute one (53.5°). They add to 180°, because sin(180° − θ) = sin θ.

Cosine is negative for obtuse angles, so the cosine rule never gives two answers.

Exact answers. Use exact values in the rules. A triangle with sides 5 and 8 and the angle $6 0^{\circ}$ between them has area $\frac{1}{2} (5) (8) sin 6 0^{\circ} = 20 \times \frac{3}{2} = 103$ . For an obtuse angle, $sin 12 0^{\circ} = \frac{3}{2}$ , $sin 15 0^{\circ} = \frac{1}{2}$ and $cos 12 0^{\circ} = - \frac{1}{2}$ .

The height is the slanted side times the sine of the angle: 5 sin 60°. So the area is ½ × 8 × 5 sin 60° = 10√3.

Working back from an area. If two equal sides $x$ meet at $15 0^{\circ}$ and the area is 16: $\frac{1}{2} x^{2} sin 15 0^{\circ} = \frac{x ^{2}}{4} = 16$ , so $x = 8$ .

With an obtuse angle the height drops outside the triangle, onto the base extended. It is still the slanted side times the sine of the angle: x sin 150° = x ÷ 2, so the area is ½ × x × x ÷ 2 = x² ÷ 4.

Year 11Bearings with non-right triangles

A ship sails 12 km on a bearing of $05 0^{\circ}$ , then 8 km on a bearing of $14 0^{\circ}$ . The turn between the two legs is $9 0^{\circ}$ here, so Pythagoras gives the distance from the start: $144 + 64 \approx 14.4$ km. When the angle is not $9 0^{\circ}$ , find the angle inside the triangle using parallel north lines, then use the cosine rule.

Draw a north line at every turn. The first leg points back on a bearing of 230° from the turn, and 230° − 140° = 90°, so the triangle has a right angle at the turn.

Worked example. Two ships leave port together, one on a bearing of $04 0^{\circ}$ at 10 km/h and the other on $15 0^{\circ}$ at 12 km/h. After 2 hours they are 20 km and 24 km from port, and the angle between their paths is $15 0^{\circ} - 4 0^{\circ} = 11 0^{\circ}$ . So the distance between them is $d^{2} = 2 0^{2} + 2 4^{2} - 2 (20) (24) cos 11 0^{\circ} \approx 1304 \Rightarrow d \approx 36.1 km$

Both bearings are measured clockwise from the same north line, so the angle between the paths is 150° − 40° = 110°. Two sides and the angle between them: use the cosine rule.

To find a bearing at the end, find the angle in the triangle with the sine rule, then add or subtract it from a north or south line.

Quick summary

Side $a$ is opposite angle $A$ .
Sine rule: $\frac{a}{sin A} = \frac{b}{sin B}$ . Needs a side with its opposite angle.
Cosine rule: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ . Two sides and the angle between, or three sides.
Area $= \frac{1}{2} ab sin C$ .
Finding an angle with the sine rule may give two answers: $θ$ and $18 0^{\circ} - θ$ .

Quiz

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In triangle $A B C$ , side $b$ is opposite:
1. A angle $A$
2. B angle $B$
3. C angle $C$
4. D the right angle
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B: angle $B$
Each side is named after the angle across from it.
Two angles of a triangle are $5 0^{\circ}$ and $7 0^{\circ}$ . The third angle is:
1. A $4 0^{\circ}$
2. B $6 0^{\circ}$
3. C $12 0^{\circ}$
4. D $7 0^{\circ}$
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B: $6 0^{\circ}$
Angles add to $18 0^{\circ}$ .
You know two sides and the angle between them and want the third side. Use:
1. A the sine rule
2. B the cosine rule
3. C SOH CAH TOA
4. D the area formula
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B: the cosine rule
Two sides and the included angle means cosine rule.