Trigonometry

Work in a group. Look at the two right-angled triangles below, measure all their sides and angles, and talk about these questions: Which angles are equal? Are the sides equal? Are △ABC and △PQR similar?

Then find the ratios $\dfrac{AB}{BC}$, $\dfrac{BC}{AC}$ and $\dfrac{AB}{AC}$ in △ABC, and $\dfrac{PQ}{QR}$, $\dfrac{QR}{PR}$ and $\dfrac{PQ}{PR}$ in △PQR. Which of them are equal?

That last line is the seed of the whole unit. Two right-angled triangles with one more angle the same are similar, and in similar triangles the sides are in the same proportions. So in a right-angled triangle, the ratio of two sides depends only on the angle, not on how big the triangle is.

The sides of a right-angled triangle have names that depend on the angle you look from. Seen from the angle at A:

• the hypotenuse h is the longest side, opposite the right angle: AC;
• the perpendicular p is the side opposite the angle at A: BC;
• the base b is the third side, the one next to the angle at A: AB.

So the three ratios above are $\dfrac{b}{p}$, $\dfrac{p}{h}$ and $\dfrac{b}{h}$, the same in both triangles.

These fixed ratios are the trigonometric ratios. Trigonometry (from the Greek for “triangle measuring”) studies how the sides and angles of triangles are related. It lets you find a height, a length or an angle you can’t measure directly: how tall a tree is, how wide a river is, how steep a road climbs. Surveyors, builders, engineers, pilots and astronomers use it every day, and physics and engineering couldn’t do without it.

Naming the sides

In the right-angled triangle ABC, with the right angle at B, take ∠ACB = θ as the angle we are looking from. It is called the reference angle. From θ:

• AB, opposite θ, is the perpendicular p (also called the opposite side);
• BC, next to θ, is the base b (the adjacent side);
• AC, opposite the right angle, is the hypotenuse h.

Now the three ratios get their names:

$$\sin\theta = \frac{AB}{AC} = \frac{p}{h} = \frac{\text{perpendicular}}{\text{hypotenuse}}$$

$$\cos\theta = \frac{BC}{AC} = \frac{b}{h} = \frac{\text{base}}{\text{hypotenuse}}$$

$$\tan\theta = \frac{AB}{BC} = \frac{p}{b} = \frac{\text{perpendicular}}{\text{base}}$$

They are read “sine theta”, “cosine theta” and “tangent theta”. These three, sin θ, cos θ and tan θ, are the fundamental trigonometric ratios.

Turn each one upside down and you get the other three, the reciprocal ratios:

$$\operatorname{cosec}\theta = \frac{1}{\sin\theta} = \frac{h}{p}, \quad \sec\theta = \frac{1}{\cos\theta} = \frac{h}{b}$$

$$\cot\theta = \frac{1}{\tan\theta} = \frac{b}{p}$$

read “cosecant”, “secant” and “cotangent”. So $\sin\theta \times \operatorname{cosec}\theta = 1$, $\cos\theta \times \sec\theta = 1$ and $\tan\theta \times \cot\theta = 1$.

Angles are usually named with Greek letters. These are the ones you will meet:

Letter	Name	Say it
α	alpha	AL-fuh
β	beta	BAY-tuh (or BEE-tuh)
γ	gamma	GAM-uh
δ	delta	DEL-tuh
θ	theta	THAY-tuh (or THEE-tuh)
φ	phi	fy (or fee)
π	pi	pie
ψ	psi	sy (or psy)
λ	lambda	LAM-duh

Examples

Two rules that always hold

A second rule comes from dividing two ratios:

$$\frac{\sin\theta}{\cos\theta} = \frac{p}{h} \div \frac{b}{h} = \frac{p}{h} \times \frac{h}{b} = \frac{p}{b} = \tan\theta$$

So $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, and turned upside down, $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$.

Question 1

In the right-angled triangle ABC, $\angle ACB = \varphi$ and ∠BAC = β.

Question 2

In the right-angled triangle PQR, PQ = 6 and QR = 8.

Question 3

Question 4

Find the value of each, using the trigonometric ratios.

Question 5

Prove, using the trigonometric ratios:

Question 6

An inclined plane 29 ft long is built to carry goods up to the top of a building 20 ft high.

Question 7

In the figure, AB is a lamp post and C is the foot of a tree. AC is the distance from the foot of the tree to the top of the lamp post, and BC the distance between the post and the tree.

Most angles have untidy ratios ($\sin 37^\circ$ is about 0.6018…), but five angles, 0∘°, 30∘°, 45∘°, 60∘° and 90∘°, have exact values you can work out with nothing but Pythagoras’ rule. Three of them, 30∘°, 45∘° and 60∘°, are the angles of your geometry box’s two set squares, and exam questions are built on all five.

a. The ratios of 45°

In △OQP, $\angle POQ = 45^\circ$ and $\angle OQP = 90^\circ$, so the third angle is 45∘° too, and the triangle is isosceles: OQ = PQ. Call each x. By Pythagoras’ rule,

$$OP^2 = x^2 + x^2 = 2x^2, \quad \text{so } OP = \sqrt2\, x$$

From the 45∘° angle at O: p = PQ = x, b = OQ = x, $h = OP = \sqrt2\, x$. So

$$\sin 45^\circ = \frac{x}{\sqrt2\, x} = \frac{1}{\sqrt2} \approx 0.707$$

$$\cos 45^\circ = \frac{x}{\sqrt2\, x} = \frac{1}{\sqrt2} \approx 0.707$$

$$\tan 45^\circ = \frac{x}{x} = 1$$

and upside down, $\operatorname{cosec} 45^\circ = \sqrt2$, $\sec 45^\circ = \sqrt2$, $\cot 45^\circ = 1$. The x always cancels: any 45∘° right-angled triangle, a set square or a field cut corner to corner, gives the same values.

b. The ratios of 30° and 60°

For 30∘° and 60∘° we need a right-angled triangle with those angles. Take an equilateral triangle PQR: all its angles are 60∘° and all its sides are equal, say x. Draw $PM \perp QR$. It cuts the base into two equal halves, $QM = MR = \dfrac{x}{2}$, and the angle at P into two halves of 30∘°.

In the right-angled △PMR, $\angle R = 60^\circ$, $\angle M = 90^\circ$, so $\angle MPR = 30^\circ$. By Pythagoras’ rule:

$$\begin{aligned} PM^2 &= PR^2 - MR^2 \\ &= x^2 - \frac{x^2}{4} = \frac{3x^2}{4} \\ PM &= \frac{\sqrt3}{2}\, x \end{aligned}$$

From 30∘° (at P): $p = MR = \dfrac{x}{2}$, $b = PM = \dfrac{\sqrt3}{2}x$, h = PR = x.

$$\sin 30^\circ = \frac{x/2}{x} = \frac12 = 0.5$$

$$\cos 30^\circ = \frac{\sqrt3 x/2}{x} = \frac{\sqrt3}{2} \approx 0.866$$

$$\tan 30^\circ = \frac{x/2}{\sqrt3 x/2} = \frac{1}{\sqrt3} \approx 0.577$$

and upside down, $\operatorname{cosec} 30^\circ = 2$, $\sec 30^\circ = \dfrac{2}{\sqrt3}$, $\cot 30^\circ = \sqrt3$.

From 60∘° (at R), the perpendicular and the base swap: $p = PM = \dfrac{\sqrt3}{2}x$, $b = MR = \dfrac{x}{2}$, h = x.

$$\sin 60^\circ = \frac{\sqrt3}{2} \approx 0.866, \quad \cos 60^\circ = \frac12 = 0.5$$

$$\tan 60^\circ = \frac{\sqrt3 x/2}{x/2} = \sqrt3 \approx 1.732$$

and upside down, $\operatorname{cosec} 60^\circ = \dfrac{2}{\sqrt3}$, $\sec 60^\circ = 2$, $\cot 60^\circ = \dfrac{1}{\sqrt3}$.

c. The ratios of 0° and 90°

A right-angled triangle can’t really have a second angle of 0∘° or 90∘°, but we can see what the ratios approach as the angle gets close to them. Take △PQR, right-angled at Q, with θ = ∠PRQ.

Shrink θ to 0∘°. Turn RP down towards RQ. The perpendicular PQ gets shorter and shorter, until P lands on Q: then PQ = 0, and PR lies along QR, so PR = QR. Call that length x. So

$$\begin{aligned} \sin 0^\circ &= \frac{0}{x} = 0 \\ \cos 0^\circ &= \frac{x}{x} = 1 \\ \tan 0^\circ &= \frac{0}{x} = 0 \end{aligned}$$

Grow θ to 90∘°. Now keep P where it is and slide R towards Q. The angle at R grows and the base QR gets shorter, until at 90∘° it is 0 and PR lies along PQ: PR = PQ = x. So

$$\sin 90^\circ = \frac{x}{x} = 1, \quad \cos 90^\circ = \frac{0}{x} = 0$$

and $\tan 90^\circ = \dfrac{x}{0}$, which is not defined: no number times 0 gives x.

Here are all the values in one table:

θ	0∘°	30∘°	45∘°	60∘°	90∘°
sin θ	0	$\dfrac12$	$\dfrac{1}{\sqrt2}$	$\dfrac{\sqrt3}{2}$	1
cos θ	1	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{\sqrt2}$	$\dfrac12$	0
tan θ	0	$\dfrac{1}{\sqrt3}$	1	$\sqrt3$	not defined

Examples

Question 1

Find the value of each trigonometric ratio.

Question 2

Find the reference angle of each right-angled triangle.

Question 3

Find the remaining sides of each right-angled triangle.

Question 4

Find the value of each expression.

Question 5

Prove that:

Question 6

Yes: Sunmaya’s 6 m ladder reaches 5.81 m at the safe angle, 0.81 m above the leak.

The rule puts the foot a quarter of the ladder’s length from the wall: $\dfrac{6}{4} = 1.5$ m. The ladder is the hypotenuse and the foot’s distance the base, so

$$\cos\theta = \frac{b}{h} = \frac{1.5}{6} = \frac14$$

That is the safe angle for every ladder, whatever its length: about 75.5∘° (a calculator’s cos⁻¹ button gives it). The height on the wall is the perpendicular:

$$p = \sqrt{6^2 - 1.5^2} = \sqrt{33.75} \approx 5.81 \text{ m}$$

and $\sin\theta = \dfrac{5.81}{6} \approx 0.968$: at the safe angle, a ladder reaches about 97% of its length up the wall. That is the number the opening widget shows as “height ÷ ladder” at 75∘° or 76∘°, for the 3 m ladder as much as the 6 m one.

What if they lean it at 60∘°, so it feels less steep? Then the top reaches only $6 \sin 60^\circ = 3\sqrt3 \approx 5.20$ m, still above the leak, but the foot is $6\cos 60^\circ = 3$ m out, twice the safe distance, and much more likely to slide.

Grade 9’s final exam is set to CDC’s specification grid for Compulsory Mathematics, Grades 9–10. There, Trigonometry is 4 marks in one question; in the grid’s model paper it is question 16:

Part	Level	Marks	What it asks
a.	knowledge	1	which ratio a given pair of sides makes
b.	understanding	1	a value from Pythagoras’ rule, such as $\sin^2\theta + \cos^2\theta$
c.	application	1	a side from one ratio and one side
d.	higher ability	1	an angle from how two sides compare

Here is one of our own in that shape.

Question 1

In the figure, △CAB is right-angled at A, and ∠B = θ.

Question 2

In the right-angled triangle XYZ, $\angle Y = 90^\circ$, $\angle X = 60^\circ$ and YZ = 8 cm.

Question 3

From a point on level ground 30 ft from the foot of a bamboo, the angle of elevation of its top (the angle the line to the top makes with the ground) is 30∘°.