IGCSE Mathematics: Trigonometry

Worked Example 1.1 — Finding the Hypotenuse

A right-angled triangle has legs of 7 cm and 24 cm. Find the hypotenuse.

Step 1 Apply Pythagoras' theorem.

$$c = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}$$

Worked Example 1.2 — Finding a Shorter Side

A right-angled triangle has hypotenuse 17 cm and one leg 8 cm. Find the other leg.

Step 1

$$a = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 \text{ cm}$$

Worked Example 1.3 — Space Diagonal of a Cuboid Extended

A box measures 6 cm by 4 cm by 3 cm. Find the length of the longest diagonal.

$$d = \sqrt{6^2 + 4^2 + 3^2} = \sqrt{36 + 16 + 9} = \sqrt{61} \approx 7.81 \text{ cm}$$

Worked Example 2.1 — Finding a Side

In a right-angled triangle, the hypotenuse is 12 cm and one angle is 35°. Find the side opposite the 35° angle.

Step 1 We know the hypotenuse and want the opposite side, so use $\sin$.

$$\sin 35° = \frac{\text{Opposite}}{12}$$

Step 2 Opposite $= 12\sin 35° = 12 \times 0.5736 \approx 6.88$ cm.

Worked Example 2.2 — Finding an Angle

A right-angled triangle has adjacent side 5 cm and hypotenuse 11 cm (adjacent to angle $\theta$). Find $\theta$.

Step 1 We know adjacent and hypotenuse, so use $\cos$.

$$\cos\theta = \frac{5}{11}$$

Step 2 $\theta = \cos^{-1}\!\left(\dfrac{5}{11}\right) \approx 62.96° \approx 63.0°$.

Worked Example 2.3 — Exact Values

Without a calculator, find the exact value of $\sin 30° + \cos 60° + \tan 45°$.

$$= \frac{1}{2} + \frac{1}{2} + 1 = 2$$

Worked Example 3.1 — Finding a Side with the Sine Rule

In triangle $ABC$, $\angle A = 50°$, $\angle B = 70°$, and $b = 15$ cm. Find side $a$.

Step 1 Apply the sine rule.

$$\frac{a}{\sin 50°} = \frac{15}{\sin 70°}$$

Step 2 Solve.

$$a = \frac{15 \sin 50°}{\sin 70°} = \frac{15 \times 0.7660}{0.9397} \approx 12.2 \text{ cm}$$

Worked Example 3.2 — Finding an Angle with the Sine Rule

In triangle $PQR$, $p = 10$ cm, $q = 14$ cm, and $\angle P = 42°$. Find $\angle Q$.

Step 1

$$\frac{\sin Q}{14} = \frac{\sin 42°}{10}$$

Step 2

$$\sin Q = \frac{14 \sin 42°}{10} = \frac{14 \times 0.6691}{10} = 0.9367$$

Step 3

$$\angle Q = \sin^{-1}(0.9367) \approx 69.4° \quad \text{(or } 110.6° \text{ — ambiguous case, both valid here)}$$

Worked Example 4.1 — Finding a Side with the Cosine Rule

In triangle $ABC$, $b = 8$ cm, $c = 11$ cm, and $\angle A = 65°$. Find side $a$.

Step 1

$$a^2 = 8^2 + 11^2 - 2(8)(11)\cos 65° = 64 + 121 - 176 \times 0.4226 = 185 - 74.38 = 110.62$$

Step 2 $a = \sqrt{110.62} \approx 10.5$ cm.

Worked Example 4.2 — Finding an Angle with the Cosine Rule

A triangle has sides $a = 7$ cm, $b = 9$ cm, $c = 12$ cm. Find angle $C$.

Step 1

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{49 + 81 - 144}{2(7)(9)} = \frac{-14}{126} = -0.1111$$

Step 2 $C = \cos^{-1}(-0.1111) \approx 96.4°$.

Worked Example 5.1 — Area Formula

Find the area of a triangle with sides 10 cm and 7 cm and an included angle of 55°.

$$A = \frac{1}{2}(10)(7)\sin 55° = 35\sin 55° = 35 \times 0.8192 \approx 28.7 \text{ cm}^2$$

Worked Example 5.2 — Finding an Angle from the Area

A triangle has sides 8 cm and 12 cm and an area of 36 cm². Find the possible values of the included angle $C$.

Step 1

$$36 = \frac{1}{2}(8)(12)\sin C = 48\sin C \implies \sin C = \frac{36}{48} = 0.75$$

Step 2 $C = \sin^{-1}(0.75) \approx 48.6°$ or $C = 180° - 48.6° = 131.4°$. Both are valid geometrically.

Worked Example 6.1 — Angle in a Pyramid

A square-based pyramid has a base of side 10 cm and vertical height 12 cm. Find the angle that a slant edge makes with the base.

Step 1 The slant edge runs from an apex $V$ to a base corner $A$. The foot of the vertical from $V$ is the centre $O$ of the base.

Step 2 The distance from the centre of the base to a corner is half the diagonal: $OA = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$ cm.

Step 3 In right-angled triangle $VOA$: $VO = 12$ (vertical height), $OA = 5\sqrt{2}$.

$$\tan(\angle VAO) = \frac{VO}{OA} = \frac{12}{5\sqrt{2}} = \frac{12}{7.071} \approx 1.697$$

$$\angle VAO = \tan^{-1}(1.697) \approx 59.5°$$

Worked Example 6.2 — Angle in a Cuboid

A cuboid has dimensions 8 cm by 6 cm by 5 cm. Find the angle that the space diagonal makes with the base.

Step 1 Diagonal of the base: $d = \sqrt{8^2 + 6^2} = \sqrt{100} = 10$ cm.

Step 2 The space diagonal has horizontal component 10 cm and vertical component 5 cm.

$$\theta = \tan^{-1}\!\left(\frac{5}{10}\right) = \tan^{-1}(0.5) \approx 26.6°$$

Worked Example 7.1 — Reading from a Trig Graph

The graph of $y = 3\sin(2x)$ is drawn for $0° \le x \le 360°$. State (a) its amplitude; (b) its period; (c) the number of complete cycles shown.

(a) Amplitude $= 3$ (the coefficient of $\sin$).

(b) Period $= \dfrac{360°}{2} = 180°$.

(c) Number of complete cycles $= \dfrac{360°}{180°} = 2$.

Worked Example 7.2 — Solving Using a Graph

From the graph of $y = \cos x$, write down all solutions of $\cos x = 0.5$ for $0° \le x \le 360°$.

$x = \cos^{-1}(0.5) = 60°$. The cosine graph is symmetric, so the second solution is $x = 360° - 60° = 300°$.

Solutions: $x = 60°$ and $x = 300°$.

Worked Example 8.1 — Finding Distance Using Bearings and Trig

A ship sails 15 km on a bearing of 040° from port $A$. How far north and how far east of $A$ is it?

Step 1 The bearing of 040° means the angle from North is 40° towards the East.

Northward component $= 15\cos 40° \approx 15 \times 0.766 = 11.5$ km.

Eastward component $= 15\sin 40° \approx 15 \times 0.643 = 9.64$ km.

Worked Example 8.2 — Bearing Between Two Points

Point $B$ is 8 km east and 6 km north of point $A$. Find the bearing of $B$ from $A$.

Step 1 Draw a sketch. The angle from North towards East satisfies:

$$\tan\theta = \frac{\text{East}}{\text{North}} = \frac{8}{6} = 1.\overline{3}$$

Step 2 $\theta = \tan^{-1}(1.\overline{3}) \approx 53.1°$. Bearing $= 053°$ (to the nearest degree).

Worked Example 8.3 — Using the Cosine Rule with Bearings

A boat travels 12 km on a bearing of 070° from $A$ to $B$, then 9 km on a bearing of 145° from $B$ to $C$. Find the distance $AC$ and the bearing of $C$ from $A$.

Step 1 — Angle at B The bearing from $A$ to $B$ is 070°; the bearing from $B$ to $C$ is 145°. The interior angle at $B$ between $BA$ and $BC$:

Back bearing of $B$ from $A$: $070° + 180° = 250°$. Angle $\angle ABC = 250° - 145° = 105°$.

Step 2 — Cosine Rule

$$AC^2 = 12^2 + 9^2 - 2(12)(9)\cos 105° = 144 + 81 - 216(-0.2588) = 225 + 55.9 = 280.9$$

$$AC = \sqrt{280.9} \approx 16.8 \text{ km}$$

Step 3 — Bearing Use the sine rule to find the angle at $A$, then add to the bearing 070°.

$\dfrac{\sin(\angle BAC)}{9} = \dfrac{\sin 105°}{16.8} \implies \sin(\angle BAC) = \dfrac{9 \times 0.9659}{16.8} \approx 0.5178 \implies \angle BAC \approx 31.2°$.

Bearing of $C$ from $A = 070° + 31.2° \approx 101°$.