Sine and Cosine Rules – A Level Maths Year 1

1 Beyond Right-Angled Triangles

SOH CAH TOA only works in right-angled triangles. The sine rule and cosine rule solve any triangle. You met them at Higher GCSE; at A Level you must also handle the ambiguous case, area problems and multi-step contexts such as bearings. This page is part of DfE content area E · Trigonometry.

The labelling convention. Angles are capital letters $A$, $B$, $C$ at the vertices. The side opposite each angle takes the matching lower-case letter: $a$ is opposite $A$, and so on.

2 The Sine Rule

The sine rule

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

When to use it: you know a side and its opposite angle, plus one more side or angle. Put the unknown on top: sides on top to find a side, sines on top to find an angle.

Worked Example 1 — Finding a side

In triangle $ABC$, $A = 40°$, $B = 65°$ and $a = 12$ cm. Find $b$.

①$\dfrac{b}{\sin 65°} = \dfrac{12}{\sin 40°}$

②$b = \dfrac{12 \sin 65°}{\sin 40°} = 16.919\ldots$

③$b = 16.9$ cm (3 s.f.)

Worked Example 2 — Finding an angle (only one answer)

In triangle $ABC$, $a = 10$ cm, $b = 7$ cm and $A = 50°$. Find $B$.

①$\dfrac{\sin B}{7} = \dfrac{\sin 50°}{10}$, so $\sin B = \dfrac{7\sin 50°}{10} = 0.5362\ldots$

②$B = 32.4°$ — or possibly $180° - 32.4° = 147.6°$?

③$50° + 147.6° > 180°$, so the obtuse answer is impossible. $B = 32.4°$ (1 d.p.).

Here $b < a$, so $B$ must be smaller than $A$ — it cannot be obtuse. The longest side is always opposite the largest angle.

3 The Ambiguous Case

$\sin\theta = \sin(180° - \theta)$. So when the sine rule gives $\sin B = k$, there are two candidate angles: $B$ and $180° - B$. If both fit inside a triangle with the given angle, there are two different triangles matching the information.

Side $CB$ of length $8$ can "swing" to hit the base line in two places, $B_1$ and $B_2$. Triangle $AB_2C$ has an acute angle at $B_2$, and triangle $AB_1C$ has the obtuse supplement at $B_1$.

Worked Example 3 — Two possible triangles

In triangle $ABC$, $A = 35°$, $a = 8$ cm and $b = 11$ cm. Find the two possible values of $B$, and the corresponding values of $C$.

①$\sin B = \dfrac{11 \sin 35°}{8} = 0.7887\ldots$

②$B = 52.1°$ or $B = 180° - 52.1° = 127.9°$

③Check: $35° + 127.9° = 162.9° < 180°$, so both are valid.

④$C = 180° - 35° - 52.1° = 92.9°$ or $C = 180° - 35° - 127.9° = 17.1°$

When does the ambiguous case happen?

Given angle $A$ (acute), side $a$ opposite it and another side $b$: two triangles exist when $b\sin A < a < b$

Calculators only give the acute angle. $\sin^{-1}$ always returns an answer between $-90°$ and $90°$. You must think about $180° - \theta$ yourself every time you find an angle with the sine rule.

4 The Cosine Rule

The cosine rule

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

When to use it: you know two sides and the angle between them (to find the third side), or all three sides (to find an angle). If $A = 90°$ it reduces to Pythagoras, because $\cos 90° = 0$.

Worked Example 4 — Finding a side

In triangle $ABC$, $b = 7$ cm, $c = 9$ cm and $A = 52°$. Find $a$.

①$a^2 = 7^2 + 9^2 - 2(7)(9)\cos 52°$

②$a^2 = 130 - 126\cos 52° = 52.427\ldots$

③$a = 7.24$ cm (3 s.f.)

Worked Example 5 — Finding an angle

A triangle has sides $5$ cm, $7$ cm and $9$ cm. Find its largest angle.

①The largest angle is opposite the longest side, $9$ cm.

②$\cos\theta = \dfrac{5^2 + 7^2 - 9^2}{2(5)(7)} = \dfrac{-7}{70} = -0.1$

③$\theta = \cos^{-1}(-0.1) = 95.7°$ (1 d.p.)

A negative cosine means an obtuse angle — and unlike $\sin^{-1}$, the $\cos^{-1}$ button gets it right automatically. That is why it is safer to find the largest angle with the cosine rule first; the others must then be acute.

5 The Area of a Triangle

Area from two sides and the included angle

Area $= \tfrac12 ab\sin C$

Why it works. Take $a$ as the base. The perpendicular height from $A$ is $b\sin C$, so the area is $\tfrac12 \times a \times b\sin C$.

Worked Example 6 — Finding an area

Find the area of the triangle in Worked Example 4 ($b = 7$, $c = 9$, $A = 52°$).

①The angle between $b$ and $c$ is $A$: Area $= \tfrac12 bc\sin A$

②$= \tfrac12 (7)(9)\sin 52° = 24.8$ cm$^2$ (3 s.f.)

Worked Example 7 — Working backwards from the area

A triangle has sides $8$ cm and $10$ cm and area $24$ cm$^2$. Find the possible sizes of the angle between these sides.

①$\tfrac12(8)(10)\sin\theta = 24 \;\Rightarrow\; \sin\theta = 0.6$

②$\theta = 36.9°$ or $\theta = 180° - 36.9° = 143.1°$

③Both are possible: two different triangles have the same area.

6 Bearings Problems

Bearings are measured clockwise from north and written with three figures. The hard part is finding the angle inside the triangle; use a sketch with a north line at every point and the fact that the back bearing differs by $180°$.

Worked Example 8 — A two-leg journey

A ship sails $15$ km from port $O$ on a bearing of $060°$ to $A$, then $9$ km on a bearing of $170°$ to $B$. Find the distance $OB$ and the bearing of $B$ from $O$.

①At $A$, the bearing back to $O$ is $060° + 180° = 240°$. The ship leaves $A$ on $170°$. The angle between these directions is $\angle OAB = 240° - 170° = 70°$.

②Cosine rule: $OB^2 = 15^2 + 9^2 - 2(15)(9)\cos 70° = 306 - 270\cos 70° = 213.65\ldots$

③$OB = 14.6$ km (3 s.f.)

④Sine rule for $\angle AOB$: $\sin\angle AOB = \dfrac{9\sin 70°}{14.616\ldots} = 0.5786\ldots$, so $\angle AOB = 35.4°$ (it is opposite the shortest side, so acute).

⑤Bearing of $B$ from $O$: $060° + 35.4° = 095.4°$, so about $095°$.

Keep the unrounded value of $OB$ in your calculator for step ④. Using $14.6$ instead of $14.616\ldots$ is a common source of lost accuracy marks.

7 Quick Reference

Sine rule

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

Cosine rule

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

Cosine rule (angle)

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

Area

$\tfrac12 ab\sin C$, $C$ between $a$ and $b$.

Which rule?

Opposite pair known → sine. Otherwise → cosine.

Ambiguous case

Sine rule angle: also test $180° - \theta$.

Obtuse check

Angles must total $180°$.

Largest angle

Opposite the longest side; find it first with cosine rule.

Bearings

Clockwise from north; back bearing $\pm 180°$.

Accuracy

Store unrounded values; give 3 s.f. / 1 d.p.

8 Practice Questions

Question 1

In triangle $PQR$, $P = 48°$, $Q = 71°$ and $p = 9$ cm. Find $q$.

▶ Show solution

$\dfrac{q}{\sin 71°} = \dfrac{9}{\sin 48°}$

$q = \dfrac{9\sin 71°}{\sin 48°} = 11.5$ cm (3 s.f.)

Question 2

In triangle $ABC$, $b = 5$ cm, $c = 8$ cm and $A = 120°$. Find the exact value of $a$, and give it to 3 s.f.

▶ Show solution

$a^2 = 25 + 64 - 2(5)(8)\cos 120° = 89 - 80\left(-\tfrac12\right) = 89 + 40 = 129$

$a = \sqrt{129} = 11.4$ cm (3 s.f.)

Question 3

A triangle has sides $4$ cm, $6$ cm and $7$ cm. Find its largest angle.

▶ Show solution

Opposite the $7$ cm side: $\cos\theta = \dfrac{16 + 36 - 49}{2(4)(6)} = \dfrac{3}{48} = 0.0625$

$\theta = 86.4°$ (1 d.p.)

Question 4

Find the exact area of a triangle with sides $5$ cm and $8$ cm and an included angle of $30°$.

▶ Show solution

Area $= \tfrac12(5)(8)\sin 30° = 20 \times \tfrac12 = 10$ cm$^2$.

Question 5

In this question you must show detailed reasoning. A triangle has sides $3$, $5$ and $7$. Show that its largest angle is exactly $120°$.

▶ Show solution

The largest angle is opposite the side of length $7$.

$\cos\theta = \dfrac{3^2 + 5^2 - 7^2}{2(3)(5)} = \dfrac{9 + 25 - 49}{30} = \dfrac{-15}{30} = -\dfrac12$

$\cos 120° = -\tfrac12$ and $0° < \theta < 180°$, so $\theta = 120°$.

Question 6

In triangle $ABC$, $AB = 9$ cm, $AC = 7$ cm and $\angle ABC = 40°$. Find the two possible sizes of $\angle ACB$.

▶ Show solution

$AB = c = 9$ is opposite $C$ and $AC = b = 7$ is opposite $B$.

$\dfrac{\sin C}{9} = \dfrac{\sin 40°}{7} \;\Rightarrow\; \sin C = \dfrac{9\sin 40°}{7} = 0.8264\ldots$

$C = 55.7°$ or $C = 180° - 55.7° = 124.3°$.

Check: $40° + 124.3° = 164.3° < 180°$, so both are possible.

Question 7

A triangle has area $20$ cm$^2$. Two of its sides are $8$ cm and $x$ cm, with an angle of $30°$ between them. Find $x$.

▶ Show solution

$\tfrac12(8)(x)\sin 30° = 20$

$4x \times \tfrac12 = 20 \;\Rightarrow\; 2x = 20 \;\Rightarrow\; x = 10$.

Question 8

In triangle $ABC$, $a = 6$, $b = 10$ and $A = 25°$. Show that there are two possible triangles, and find the difference between the two possible values of $c$.

▶ Show solution

$\sin B = \dfrac{10\sin 25°}{6} = 0.7044\ldots$, so $B = 44.8°$ or $135.2°$.

$25° + 135.2° < 180°$, so both triangles exist.

Case 1: $C = 180° - 25° - 44.78° = 110.22°$, $c = \dfrac{6\sin 110.22°}{\sin 25°} = 13.32\ldots$

Case 2: $C = 180° - 25° - 135.22° = 19.78°$, $c = \dfrac{6\sin 19.78°}{\sin 25°} = 4.80\ldots$

Difference $= 13.322\ldots - 4.804\ldots = 8.52$ (3 s.f.).

Check: the cosine rule $36 = 100 + c^2 - 20c\cos 25°$ is the quadratic $c^2 - (20\cos 25°)c + 64 = 0$, whose two roots multiply to $64$: $13.32 \times 4.80 \approx 64$ ✓

Question 9

A yacht leaves harbour $H$ and sails $20$ km on a bearing of $040°$ to a buoy $A$. It then sails $12$ km on a bearing of $120°$ to a point $B$. Find the distance $HB$ and the bearing of $B$ from $H$.

▶ Show solution

At $A$ the back bearing to $H$ is $220°$; the new course is $120°$, so $\angle HAB = 220° - 120° = 100°$.

$HB^2 = 20^2 + 12^2 - 2(20)(12)\cos 100° = 544 - 480\cos 100° = 627.35\ldots$

$HB = 25.0$ km (3 s.f.)

$\sin\angle AHB = \dfrac{12\sin 100°}{25.047\ldots} = 0.4718\ldots$, so $\angle AHB = 28.2°$ (acute, as $100°$ is already the obtuse angle).

Bearing of $B$ from $H$: $040° + 28.2° = 068.2°$, about $068°$.

Question 10

A plot of land is the quadrilateral $ABCD$. $AB = 6$ m, $BC = 9$ m, $\angle ABC = 110°$, $CD = 7$ m and $DA = 8$ m.

(a) Find the length of the diagonal $AC$. (b) Find the area of triangle $ABC$. (c) Find $\angle ADC$. (d) Find the total area of the plot, to the nearest square metre. (e) State one assumption made in modelling the plot this way.

▶ Show solution

(a) $AC^2 = 6^2 + 9^2 - 2(6)(9)\cos 110° = 117 - 108\cos 110° = 153.94\ldots$

$AC = 12.4$ m (3 s.f.)

(b) Area $ABC = \tfrac12(6)(9)\sin 110° = 25.37\ldots$ m$^2$

(c) $\cos\angle ADC = \dfrac{8^2 + 7^2 - AC^2}{2(8)(7)} = \dfrac{113 - 153.94\ldots}{112} = -0.3655\ldots$

$\angle ADC = 111.4°$ (1 d.p.)

(d) Area $ACD = \tfrac12(8)(7)\sin 111.44° = 26.06\ldots$ m$^2$

Total $= 25.37 + 26.06 = 51.43\ldots \approx 51$ m$^2$.

(e) For example: the ground is flat (the plot lies in a plane), and the boundaries are straight lines.

Sine and Cosine Rules · A Level Maths Year 1 · Created with MathJax