Introduction to trigonometry - Year 9 Mathematics

What you will learn

identify the hypotenuse, opposite, and adjacent sides relative to a given angle in a right-angled triangle,
define the three trigonometric ratios: sine, cosine, and tangent,
use SOHCAHTOA to remember the ratios,
find an unknown side when given an angle and one side,
find an unknown angle when given two sides (using inverse trig functions).

Worked example 0 Real-world example: height of a tree

You stand $12$ m from the base of a tree. Your clinometer reads an angle of elevation of $40°$ to the top. Your eye height is $1.6$ m. How tall is the tree?

The distance along the ground ( $12$ m) is adjacent to the $40°$ angle.
The height above eye level is opposite the angle.
$\tan 40° = \dfrac{\text{opposite}}{12}$ , so opposite $= 12 \times \tan 40° \approx 12 \times 0.8391 \approx 10.07$ m.
Total tree height $\approx 10.07 + 1.6 = 11.67$ m.

Key idea: choose the ratio that links the angle to the sides you know and need.

1. Naming the sides

In a right-angled triangle, the sides are named relative to a chosen acute angle $\theta$ :

Hypotenuse (H): the longest side, opposite the right angle. Always the same regardless of which acute angle you pick.
Opposite (O): the side directly across from angle $\theta$ .
Adjacent (A): the side next to angle $\theta$ (that is not the hypotenuse).

A right-angled triangle with sides labelled O (opposite), A (adjacent), and H (hypotenuse) relative to angle theta.

2. The three ratios: SOHCAHTOA

Trigonometric ratios

Sine

\sin \theta = \frac{O}{H} = \frac{\text{Opposite}}{\text{Hypotenuse}}

Cosine

\cos \theta = \frac{A}{H} = \frac{\text{Adjacent}}{\text{Hypotenuse}}

Tangent

\tan \theta = \frac{O}{A} = \frac{\text{Opposite}}{\text{Adjacent}}

Worked example 1 Identifying ratios

In a right-angled triangle, $\theta = 35°$ , the opposite side is $7$ cm, and the hypotenuse is $12$ cm. Write the values of $\sin 35°$ , $\cos 35°$ , and $\tan 35°$ as fractions.

We need the adjacent side first. By Pythagoras: $A = \sqrt{12^2 - 7^2} = \sqrt{144 - 49} = \sqrt{95} \approx 9.75$ cm.
$\sin 35° = \dfrac{7}{12}$ , $\cos 35° = \dfrac{9.75}{12} \approx 0.8124$ , $\tan 35° = \dfrac{7}{9.75} \approx 0.7179$ .

3. Finding a side given an angle

Method: label the sides O, A, H relative to the known angle; choose the ratio that involves the unknown side and the known side; solve for the unknown.

Worked example 2 Finding the opposite side

In a right-angled triangle, $\theta = 50°$ and the hypotenuse is $15$ cm. Find the opposite side.

We know H and want O, so use $\sin$ : $\sin 50° = \dfrac{O}{15}$ .
$O = 15 \times \sin 50° = 15 \times 0.7660 \approx 11.49$ cm.

Worked example 3 Finding the adjacent side

In a right-angled triangle, $\theta = 28°$ and the opposite side is $9$ cm. Find the adjacent side.

We know O and want A, so use $\tan$ : $\tan 28° = \dfrac{9}{A}$ .
Rearranging: $A = \dfrac{9}{\tan 28°} = \dfrac{9}{0.5317} \approx 16.93$ cm.

Worked example 4 Finding the hypotenuse

In a right-angled triangle, $\theta = 42°$ and the adjacent side is $20$ cm. Find the hypotenuse.

We know A and want H, so use $\cos$ : $\cos 42° = \dfrac{20}{H}$ .
Rearranging: $H = \dfrac{20}{\cos 42°} = \dfrac{20}{0.7431} \approx 26.91$ cm.

4. Finding an angle given two sides

When you know two sides and need the angle, use the inverse trig functions on your calculator: $\sin^{-1}$ , $\cos^{-1}$ , or $\tan^{-1}$ .

Inverse trigonometric functions

\theta = \sin^{-1}\!\left(\frac{O}{H}\right), \qquad \theta = \cos^{-1}\!\left(\frac{A}{H}\right), \qquad \theta = \tan^{-1}\!\left(\frac{O}{A}\right).

Worked example 5 Finding an angle from two sides

A right-angled triangle has opposite side $8$ cm and adjacent side $11$ cm. Find angle $\theta$ .

We have O and A, so use $\tan^{-1}$ : $\theta = \tan^{-1}\!\left(\dfrac{8}{11}\right)$ .
$\theta = \tan^{-1}(0.7273) \approx 36.0°$ .

Worked example 6 Finding an angle using sine

A ladder $5$ m long leans against a wall. The foot of the ladder is $2$ m from the wall. Find the angle between the ladder and the ground.

Hypotenuse $= 5$ m (the ladder), adjacent $= 2$ m (ground distance).
$\cos \theta = \dfrac{2}{5} = 0.4$ .
$\theta = \cos^{-1}(0.4) \approx 66.4°$ .

5. Choosing the right ratio

Worked example 7 Full problem: ramp angle

A wheelchair ramp rises $0.8$ m over a horizontal distance of $6$ m. Find (a) the angle of inclination, and (b) the length of the ramp surface.

(a) The rise ( $0.8$ m) is opposite the angle; the run ( $6$ m) is adjacent. $\theta = \tan^{-1}\!\left(\dfrac{0.8}{6}\right) = \tan^{-1}(0.1333) \approx 7.6°$ .

(b) The ramp surface is the hypotenuse. $H = \dfrac{0.8}{\sin 7.6°} = \dfrac{0.8}{0.1323} \approx 6.05$ m.

Alternatively: $H = \sqrt{0.8^2 + 6^2} = \sqrt{36.64} \approx 6.05$ m (Pythagoras check).

Practice

Fluency

Tier 1: basic skills

In a right-angled triangle, angle $A = 30°$ . The opposite side is $5$ cm and the hypotenuse is $10$ cm. Find $\sin 30°$ .
In a right-angled triangle with $\theta = 45°$ , the adjacent and opposite sides are both $7$ cm. Find $\tan 45°$ .
Find the opposite side if $\theta = 40°$ and $H = 20$ cm.
Find the adjacent side if $\theta = 55°$ and $H = 18$ cm.
Find the hypotenuse if $\theta = 33°$ and $O = 10$ cm.
Find $\theta$ if $O = 6$ cm and $H = 13$ cm.
Find $\theta$ if $A = 9$ cm and $H = 15$ cm.
Find $\theta$ if $O = 12$ cm and $A = 5$ cm.
A kite string is $50$ m long and makes an angle of $60°$ with the ground. How high is the kite?
A $3$ m ladder leans against a wall at $72°$ to the ground. How far up the wall does it reach?

Reasoning

Tier 2: mixed practice

From the top of a $25$ m cliff, the angle of depression to a boat is $38°$ . How far is the boat from the base of the cliff?
A road rises $1$ m for every $8$ m of horizontal distance. Find the angle of incline.
A rectangular gate is $1.8$ m wide and $1.2$ m tall. Find the angle its diagonal makes with the bottom edge.
An isosceles triangle has equal sides of $10$ cm and a base of $12$ cm. Find the base angles. (Hint: split it into two right-angled triangles.)
A ship sails $15$ km due east then $9$ km due north. Find the bearing from the starting point to the ship’s final position.
Two buildings are $30$ m apart. From the roof of the shorter building ( $20$ m tall), the angle of elevation to the top of the taller building is $35°$ . Find the height of the taller building.

Reasoning

Tier 3: explain and apply

Explain why $\sin \theta$ can never be greater than $1$ .
Show that for any acute angle $\theta$ , $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ .
A surveyor needs to find the width of a river. She stands at point A on one bank and sights a tree at point B directly opposite. She then walks $40$ m along the bank to point C and measures angle ACB as $56°$ . Find the width of the river.
Without a calculator, explain why $\sin 30° = \cos 60°$ . Use a diagram if helpful.
A ski slope is $200$ m long (measured along the slope) and drops $80$ m vertically. Find (a) the angle of the slope, and (b) the horizontal distance covered.

Challenge

Reasoning

Harder reasoning

From a point on level ground, the angle of elevation to the top of a building is $28°$ . Moving $20$ m closer, the angle becomes $42°$ . Find the height of the building.
A regular hexagon has side length $8$ cm. Using trigonometry, find (a) the distance from the centre to a vertex, and (b) the area of the hexagon.
Prove that in any right-angled triangle, $\sin^2\theta + \cos^2\theta = 1$ . (Hint: use the definitions and Pythagoras’ theorem.)
A plane takes off at an angle of $15°$ to the horizontal and climbs at a constant speed of $250$ km/h. After $2$ minutes, find (a) the plane’s altitude, and (b) the horizontal distance it has covered from the runway.

Interactive

Try it yourself: build the right triangle

Work through 3 examples on one triangle. Drag the pink vertex.

Example 1 (easy). Drag the pink vertex to make a triangle where sin θ = 0.5. Then read off θ (should be 30°).

Opposite: 4
Adjacent: 3
Hypotenuse: 5.00
θ: 53.1°
sin θ: 0.800
cos θ: 0.600
tan θ: 1.333

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$\sin 30° = \dfrac{5}{10} = 0.5$ .
$\tan 45° = \dfrac{7}{7} = 1$ .
$O = 20 \sin 40° \approx 20 \times 0.6428 = 12.86$ cm.
$A = 18 \cos 55° \approx 18 \times 0.5736 = 10.32$ cm.
$H = \dfrac{10}{\sin 33°} = \dfrac{10}{0.5446} \approx 18.36$ cm.
$\theta = \sin^{-1}\!\left(\dfrac{6}{13}\right) \approx 27.5°$ .
$\theta = \cos^{-1}\!\left(\dfrac{9}{15}\right) = \cos^{-1}(0.6) \approx 53.1°$ .
$\theta = \tan^{-1}\!\left(\dfrac{12}{5}\right) = \tan^{-1}(2.4) \approx 67.4°$ .
Height $= 50 \sin 60° = 50 \times 0.8660 \approx 43.3$ m.
Height up wall $= 3 \sin 72° \approx 3 \times 0.9511 = 2.85$ m.

Tier 2

$\approx 32.0$ m. Method: the angle of depression equals the angle at the boat. $\tan 38° = \dfrac{25}{d}$ ; $d = \dfrac{25}{\tan 38°} = \dfrac{25}{0.7813} \approx 32.0$ .
$\approx 7.1°$ . Method: $\tan \theta = \dfrac{1}{8} = 0.125$ ; $\theta = \tan^{-1}(0.125)$ .
$\approx 33.7°$ . Method: $\tan \theta = \dfrac{1.2}{1.8} = 0.6\overline{6}$ ; $\theta = \tan^{-1}(0.667)$ .
$\approx 53.1°$ . Method: half-base $= 6$ cm; $\cos \theta = \dfrac{6}{10} = 0.6$ ; $\theta = \cos^{-1}(0.6)$ .
Bearing $\approx 031°$ . Method: $\tan \alpha = \dfrac{9}{15} = 0.6$ ; $\alpha = \tan^{-1}(0.6) \approx 31.0°$ . Bearing from east is $90° - 31° = 59°$ … Correction: bearing is measured clockwise from north. The ship is east then north, so the angle from north $= 90° - \tan^{-1}(\tfrac{9}{15})$ . Actually: from start, east $= 15$ , north $= 9$ . Bearing $= \tan^{-1}(\tfrac{15}{9}) = \tan^{-1}(1.667) \approx 59.0°$ . Bearing $\approx 059°$ .
$41.0$ m. Method: let extra height above the shorter building $= x$ . $\tan 35° = \dfrac{x}{30}$ ; $x = 30 \tan 35° \approx 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

Tier 3

$\sin \theta = \dfrac{O}{H}$ . In a right-angled triangle, the opposite side is always shorter than the hypotenuse (the hypotenuse is the longest side). Therefore $O < H$ , so $\dfrac{O}{H} < 1$ , meaning $\sin \theta < 1$ for acute angles.
$\dfrac{\sin \theta}{\cos \theta} = \dfrac{O/H}{A/H} = \dfrac{O}{H} \times \dfrac{H}{A} = \dfrac{O}{A} = \tan \theta$ .
$\approx 59.3$ m. Method: the river width $w$ is opposite the $56°$ angle; the $40$ m walk is adjacent. $\tan 56° = \dfrac{w}{40}$ ; $w = 40 \tan 56° \approx 59.3$ m.
In a 30-60-90 triangle, the side opposite $30°$ is half the hypotenuse, and the side adjacent to $30°$ is $\dfrac{\sqrt{3}}{2}$ times the hypotenuse. Now $\sin 30° = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . For $60°$ , the opposite and adjacent sides swap: $\cos 60° = \dfrac{\text{adj}_{60}}{H} = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . So $\sin 30° = \cos 60°$ .
(a) $\sin \theta = \dfrac{80}{200} = 0.4$ ; $\theta = \sin^{-1}(0.4) \approx 23.6°$ . (b) Horizontal distance $= \sqrt{200^2 - 80^2} = \sqrt{33600} \approx 183.3$ m. Or: $200 \cos 23.6° \approx 183.3$ m.

Challenge

$\approx 24.9$ m. Method: let $h$ = height, $d$ = original distance. From the first position: $\tan 28° = \dfrac{h}{d}$ , so $d = \dfrac{h}{\tan 28°}$ . From the closer position: $\tan 42° = \dfrac{h}{d - 20}$ , so $d - 20 = \dfrac{h}{\tan 42°}$ . Substituting: $\dfrac{h}{\tan 28°} - 20 = \dfrac{h}{\tan 42°}$ ; $h\!\left(\dfrac{1}{\tan 28°} - \dfrac{1}{\tan 42°}\right) = 20$ ; $h\!\left(\dfrac{1}{0.5317} - \dfrac{1}{0.9004}\right) = 20$ ; $h(1.8807 - 1.1106) = 20$ ; $h \times 0.7701 = 20$ ; $h \approx 26.0$ m. (Accept $25-26$ m depending on rounding.)
(a) In a regular hexagon, the distance from centre to vertex equals the side length, so $8$ cm. (b) The hexagon splits into $6$ equilateral triangles of side $8$ cm. Area of each $= \dfrac{\sqrt{3}}{4}(8^2) = 16\sqrt{3}$ . Total $= 96\sqrt{3} \approx 166.3$ cm $^2$ .
$\sin^2\theta + \cos^2\theta = \left(\dfrac{O}{H}\right)^2 + \left(\dfrac{A}{H}\right)^2 = \dfrac{O^2 + A^2}{H^2}$ . By Pythagoras, $O^2 + A^2 = H^2$ . Therefore $\dfrac{H^2}{H^2} = 1$ .
Distance travelled in $2$ min $= 250 \times \dfrac{2}{60} = 8.33$ km. (a) Altitude $= 8.33 \sin 15° \approx 8.33 \times 0.2588 \approx 2.16$ km. (b) Horizontal distance $= 8.33 \cos 15° \approx 8.33 \times 0.9659 \approx 8.05$ km.

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