Sine Rule, Cosine Rule and Area of a Triangle

1 Why We Need These Rules

SOH CAH TOA only works in a right-angled triangle. The sine and cosine rules work in any triangle.

Labelling convention

Capital letters $A$, $B$, $C$ for the angles
Lower-case $a$, $b$, $c$ for the sides opposite them

Sine rule

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
(or upside down, $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$, when finding an angle)

Cosine rule

$a^2 = b^2 + c^2 - 2bc\cos A$
rearranged for an angle: $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$

Area of any triangle

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

2 Choosing the Right Rule

This is the most important decision, and it takes only a few seconds.

What you are given	Use
A right angle	SOH CAH TOA (much quicker)
A matching pair (a side and its opposite angle) plus one more piece	Sine rule
Two sides and the included angle (SAS)	Cosine rule — to find the third side
All three sides (SSS)	Cosine rule — to find an angle
Two sides and the included angle, and you want the area	Area $= \tfrac{1}{2}ab\sin C$

The single fastest test: look for a matching pair — a side with its opposite angle both known. If you can see one, use the sine rule. If you cannot, use the cosine rule.

The angle must be opposite the side. In the sine rule, $a$ and $A$ must be a genuine opposite pair — the side $a$ facing the angle $A$ across the triangle.

3 The Sine Rule

Label the triangle: angles $A$, $B$, $C$ and opposite sides $a$, $b$, $c$.
Identify the complete pair and the incomplete pair.
Write the two relevant fractions equal to each other.
If finding a side, put the sides on top. If finding an angle, put the sines on top.
Cross-multiply and solve.

Worked Example 1 — Finding a side

In triangle $ABC$, $A = 52^\circ$, $B = 71^\circ$ and $a = 9$ cm. Find $b$, to 2 d.p.

①The complete pair is $a$ and $A$; we want $b$, whose angle $B$ is known.

②$\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$, that is $\dfrac{b}{\sin 71^\circ} = \dfrac{9}{\sin 52^\circ}$

③$b = \dfrac{9 \sin 71^\circ}{\sin 52^\circ} = \dfrac{9 \times 0.94552}{0.78801}$

④$b = \dfrac{8.5097}{0.78801} = 10.80$ cm

Worked Example 2 — Finding an angle

In triangle $PQR$, $p = 12$ cm, $q = 8$ cm and $P = 78^\circ$. Find $Q$, to 1 d.p.

①Finding an angle, so put the sines on top:

②$\dfrac{\sin Q}{8} = \dfrac{\sin 78^\circ}{12}$

③$\sin Q = \dfrac{8 \sin 78^\circ}{12} = \dfrac{8 \times 0.97815}{12} = 0.65210$

④$Q = \sin^{-1}(0.65210) = 40.7^\circ$

The ambiguous case. $\sin\theta = \sin(180 - \theta)$, so an equation like $\sin Q = 0.652$ has two solutions between $0^\circ$ and $180^\circ$: $40.7^\circ$ and $139.3^\circ$. Check whether the second one is possible — if the angles would then total more than $180^\circ$, discard it. In Worked Example 2, $78 + 139.3 = 217.3 \gt 180$, so only $40.7^\circ$ works.

4 The Cosine Rule

How to write it correctly: the letter on the left of the formula, and the angle used, must be the same letter. If you want side $c$, write $c^2 = a^2 + b^2 - 2ab\cos C$.

Worked Example 3 — Finding a side (SAS)

A triangle has sides $7$ cm and $10$ cm with an angle of $46^\circ$ between them. Find the third side, to 2 d.p.

①Two sides and the included angle → cosine rule.

②$a^2 = 7^2 + 10^2 - 2(7)(10)\cos 46^\circ$

③$= 49 + 100 - 140 \times 0.69466$

④$= 149 - 97.252 = 51.748$

⑤$a = \sqrt{51.748} = 7.19$ cm

Work out the whole right-hand side before square rooting. Square rooting too early is a classic slip.

Worked Example 4 — Finding an angle (SSS)

A triangle has sides $5$ cm, $7$ cm and $9$ cm. Find the largest angle, to 1 d.p.

①The largest angle is opposite the longest side, so it faces the $9$ cm side. Let $a = 9$, $b = 5$, $c = 7$.

②$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{25 + 49 - 81}{2 \times 5 \times 7}$

③$= \dfrac{-7}{70} = -0.1$

④$A = \cos^{-1}(-0.1) = 95.7^\circ$

A negative cosine means the angle is obtuse — a useful check. If the cosine comes out positive, the angle is acute.

The cosine rule contains Pythagoras. If $A = 90^\circ$ then $\cos A = 0$, and the formula collapses to $a^2 = b^2 + c^2$.

5 Area of Any Triangle

Area formula

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

The two sides $a$ and $b$ must have the angle $C$ between them. In other words, this is the SAS formula for area.

Worked Example 5 — Area

A triangle has sides $11$ cm and $14$ cm with an angle of $63^\circ$ between them. Find its area, to 1 d.p.

①Area $= \tfrac{1}{2} \times 11 \times 14 \times \sin 63^\circ$

②$= 77 \times 0.89101$

③$= 68.6\text{ cm}^2$

Worked Example 6 — Working backwards from the area

A triangle has sides $9$ cm and $12$ cm and area $40\text{ cm}^2$. Find the acute angle between them, to 1 d.p.

①$\tfrac{1}{2} \times 9 \times 12 \times \sin C = 40$

②$54\sin C = 40$

③$\sin C = \dfrac{40}{54} = 0.74074$

④$C = \sin^{-1}(0.74074) = 47.8^\circ$

(The obtuse alternative $132.2^\circ$ would also give this area, which is why the question specifies "acute".)

6 Multi-Step Problems

Worked Example 7 — Cosine rule then area

A triangular field has sides $80$ m, $110$ m and $150$ m. Find its area, to the nearest m².

①Three sides but no angles — find an angle first with the cosine rule.

②Let $a = 150$, $b = 80$, $c = 110$.

③$\cos A = \dfrac{80^2 + 110^2 - 150^2}{2 \times 80 \times 110} = \dfrac{6400 + 12100 - 22500}{17600}$

④$= \dfrac{-4000}{17600} = -0.227273$, so $A = 103.14^\circ$.

⑤Area $= \tfrac{1}{2} \times 80 \times 110 \times \sin 103.14^\circ = 4400 \times 0.97383$

⑥$= 4284.8$, so about $\mathbf{4285}\text{ m}^2$.

Worked Example 8 — Bearings with the cosine rule

A ship sails $60$ km on a bearing of $040^\circ$, then $90$ km on a bearing of $130^\circ$. Find its distance from the start.

①The turn between the two bearings is $130 - 40 = 90^\circ$, so the interior angle of the triangle is $180 - 90 = 90^\circ$.

②With a right angle, Pythagoras is quickest: $d = \sqrt{60^2 + 90^2} = \sqrt{3600 + 8100}$

③$= \sqrt{11700} = 108.2$ km (1 d.p.)

Check with the cosine rule: $d^2 = 60^2 + 90^2 - 2(60)(90)\cos 90^\circ = 11700 - 0$ ✓

7 Quick Reference

Labelling

Capital angle, matching lower-case side opposite it.

Which rule?

A matching pair → sine rule. No matching pair → cosine rule.

Sine rule (side)

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ — sides on top.

Sine rule (angle)

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$ — sines on top.

Cosine rule (side)

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

Cosine rule (angle)

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$.

Area

$\tfrac{1}{2}ab\sin C$, with $C$ between $a$ and $b$.

Negative cosine

Means the angle is obtuse — a handy check.

Ambiguous case

$\sin\theta = \sin(180-\theta)$; check whether both are possible.

Right angle?

Use SOH CAH TOA instead — it is far quicker.

8 Practice Questions

Question 1

In triangle $ABC$, $A = 40^\circ$, $B = 65^\circ$ and $a = 8$ cm. Find $b$, to 2 d.p.

▶ Show solution

$\dfrac{b}{\sin 65^\circ} = \dfrac{8}{\sin 40^\circ}$

$b = \dfrac{8 \times 0.90631}{0.64279} = \dfrac{7.2505}{0.64279} = 11.28$ cm

Question 2

A triangle has sides $6$ cm and $9$ cm with an angle of $55^\circ$ between them. Find the third side, to 2 d.p.

▶ Show solution

$a^2 = 6^2 + 9^2 - 2(6)(9)\cos 55^\circ$

$= 36 + 81 - 108 \times 0.57358 = 117 - 61.947 = 55.053$

$a = \sqrt{55.053} = 7.42$ cm

Question 3

Find the area of a triangle with sides $10$ cm and $15$ cm and an included angle of $40^\circ$, to 1 d.p.

▶ Show solution

Area $= \tfrac{1}{2} \times 10 \times 15 \times \sin 40^\circ$

$= 75 \times 0.64279 = 48.2\text{ cm}^2$

Question 4

A triangle has sides $4$ cm, $6$ cm and $8$ cm. Find the largest angle, to 1 d.p.

▶ Show solution

The largest angle is opposite the $8$ cm side.

$\cos A = \dfrac{4^2 + 6^2 - 8^2}{2 \times 4 \times 6} = \dfrac{16 + 36 - 64}{48} = \dfrac{-12}{48} = -0.25$

$A = \cos^{-1}(-0.25) = 104.5^\circ$

Question 5

In triangle $XYZ$, $x = 14$ cm, $y = 9$ cm and $X = 95^\circ$. Find $Y$, to 1 d.p.

▶ Show solution

$\dfrac{\sin Y}{9} = \dfrac{\sin 95^\circ}{14}$

$\sin Y = \dfrac{9 \times 0.99619}{14} = 0.64041$

$Y = \sin^{-1}(0.64041) = 39.8^\circ$

(The obtuse alternative $140.2^\circ$ is impossible, since $95 + 140.2 \gt 180$.)

Question 6

A triangle has sides $7$ cm and $11$ cm and area $30\text{ cm}^2$. Find the acute angle between them, to 1 d.p.

▶ Show solution

$\tfrac{1}{2} \times 7 \times 11 \times \sin C = 30$

$38.5\sin C = 30$

$\sin C = 0.77922$

$C = \sin^{-1}(0.77922) = 51.2^\circ$

Question 7

In triangle $ABC$, $A = 38^\circ$, $C = 71^\circ$ and $b = 15$ cm. Find $a$, to 2 d.p.

▶ Show solution

First find the missing angle: $B = 180 - 38 - 71 = 71^\circ$.

Now there is a matching pair ($b$ and $B$):

$\dfrac{a}{\sin 38^\circ} = \dfrac{15}{\sin 71^\circ}$

$a = \dfrac{15 \times 0.61566}{0.94552} = \dfrac{9.2349}{0.94552} = 9.77$ cm

Question 8

Two ships leave a port at the same time. One sails on a bearing of $050^\circ$ at $18$ km/h; the other on a bearing of $110^\circ$ at $24$ km/h. How far apart are they after $2$ hours, to 1 d.p.?

▶ Show solution

Distances after $2$ hours: $36$ km and $48$ km.

Angle at the port $= 110 - 50 = 60^\circ$.

Two sides and the included angle → cosine rule.

$d^2 = 36^2 + 48^2 - 2(36)(48)\cos 60^\circ$

$= 1296 + 2304 - 3456 \times 0.5 = 3600 - 1728 = 1872$

$d = \sqrt{1872} = 43.3$ km

Question 9

A triangular plot has sides $25$ m, $32$ m and $40$ m.

(a) Find the largest angle, to 1 d.p. (b) Find the area, to the nearest m².

▶ Show solution

(a) Largest angle is opposite the $40$ m side.

$\cos A = \dfrac{25^2 + 32^2 - 40^2}{2 \times 25 \times 32} = \dfrac{625 + 1024 - 1600}{1600} = \dfrac{49}{1600} = 0.030625$

$A = \cos^{-1}(0.030625) = 88.2^\circ$

(b) Area $= \tfrac{1}{2} \times 25 \times 32 \times \sin 88.2^\circ = 400 \times 0.99954$

$= 399.8$, so about $\mathbf{400}\text{ m}^2$.

(The angle is very close to $90^\circ$, so the area is very close to $\tfrac{1}{2} \times 25 \times 32 = 400$.)

Question 10

In triangle $ABC$, $AB = 12$ cm, $BC = 15$ cm and angle $ABC = 68^\circ$.

(a) Find $AC$, to 2 d.p. (b) Find angle $BAC$, to 1 d.p. (c) Find the area, to 1 d.p. (d) Find the perpendicular distance from $B$ to $AC$, to 2 d.p.

▶ Show solution

(a) Two sides and the included angle → cosine rule.

$AC^2 = 12^2 + 15^2 - 2(12)(15)\cos 68^\circ$

$= 144 + 225 - 360 \times 0.37461 = 369 - 134.86 = 234.14$

$AC = \sqrt{234.14} = \mathbf{15.30}$ cm

(b) Sine rule, using the pair $BC = 15$ opposite $A$, and $AC = 15.30$ opposite $B$:

$\dfrac{\sin A}{15} = \dfrac{\sin 68^\circ}{15.30}$

$\sin A = \dfrac{15 \times 0.92718}{15.30} = 0.90900$

$A = \sin^{-1}(0.90900) = \mathbf{65.4^\circ}$

(c) Area $= \tfrac{1}{2} \times 12 \times 15 \times \sin 68^\circ = 90 \times 0.92718 = \mathbf{83.4\text{ cm}^2}$

(d) Using $AC$ as the base: Area $= \tfrac{1}{2} \times AC \times h$

$83.45 = \tfrac{1}{2} \times 15.30 \times h$

$h = \dfrac{2 \times 83.45}{15.30} = \mathbf{10.91}$ cm

Sine Rule, Cosine Rule & Area (G22) · GCSE Maths Revision · Created with MathJax