Sine, Cosine and Tangent

In a right-angled triangle, the ratio of two sides depends only on the angle. Name the sides from an angle, write its sine, cosine and tangent as fractions, and find their values on a calculator.

Before you start

In a right-angled triangle the side opposite the right angle is the hypotenuse, the longest side. Pythagoras' theorem (chapter 9): in $\triangle ABC$ with $\angle C = 90^{\circ}$ , $AB^2 = AC^2 + BC^2$ .
The angles of a triangle add up to $180^{\circ}$ , so the two acute angles of a right-angled triangle add up to $90^{\circ}$ .
Similar triangles (chapter 8) have equal angles, and their corresponding sides are in the same ratio.

Naming the Sides

Trigonometry links the angles of a right-angled triangle with the lengths of its sides. It begins by naming the three sides from one of the two acute angles. That angle is often called $\theta$ (the Greek letter theta).

The names of the sides

The hypotenuse is the side opposite the right angle. It is the longest side.

The opposite side is the side opposite the angle $\theta$ . It does not touch $\theta$ .

The adjacent side is the side next to $\theta$ that is not the hypotenuse.

In $\triangle ABC$ above, with $\theta$ at $A$ : the hypotenuse is $AB$ , the opposite side is $BC$ and the adjacent side is $AC$ .

Which side is opposite and which is adjacent depends on the angle you look from. From the other acute angle they change places. The hypotenuse never changes.

Example 1: Naming the sides from each acute angle

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ . Name the hypotenuse, and the sides opposite and adjacent to (a) $\angle P$ , (b) $\angle R$ .

Solution

Find the right angle first: it is at $Q$ , so the hypotenuse is the side opposite $Q$ , which is $PR$ .

(a) From $\angle P$ : the side that does not touch $P$ is $QR$ , so $QR$ is opposite. $PQ$ touches $P$ and is not the hypotenuse, so $PQ$ is adjacent.

(b) From $\angle R$ : $PQ$ does not touch $R$ , so $PQ$ is opposite, and $QR$ is adjacent.

Hypotenuse $PR$ . (a) For $\angle P$ : opposite $QR$ , adjacent $PQ$ . (b) For $\angle R$ : opposite $PQ$ , adjacent $QR$ .

The Ratios Stay the Same

Here are three right-angled triangles of different sizes, $\triangle ABC$ , $\triangle ADE$ and $\triangle AFG$ , which share the angle of $30^{\circ}$ at $A$ . Their hypotenuses are $AB = 4$ cm, $AD = 8$ cm and $AF = 12$ cm, and the sides opposite the $30^{\circ}$ angle are $BC = 2$ cm, $DE = 4$ cm and $FG = 6$ cm.

The third side of each comes from Pythagoras' theorem: in $\triangle ABC$ , $AC = \sqrt{4^2 - 2^2} = \sqrt{12} = 3.464\ldots$ cm. Now divide one side by another in each triangle:

Triangle	$ABC$	$ADE$	$AFG$
hypotenuse (cm)	$4$	$8$	$12$
opposite (cm)	$2$	$4$	$6$
adjacent (cm)	$3.464\ldots$	$6.928\ldots$	$10.392\ldots$
opposite ÷ hypotenuse	$0.5$	$0.5$	$0.5$
adjacent ÷ hypotenuse	$0.866$	$0.866$	$0.866$
opposite ÷ adjacent	$0.577$	$0.577$	$0.577$

Each ratio is the same in all three triangles (the last two are given to 3 decimal places). This is no accident. The three triangles have the same angles, $30^{\circ}$ , $60^{\circ}$ and $90^{\circ}$ , so they are similar, and the sides of similar triangles are in the same ratio.

So for a given angle, the ratio of two sides of a right-angled triangle is fixed, however large or small the triangle is. These ratios are given names.

Sine, Cosine and Tangent

The trigonometric ratios

For an acute angle $\theta$ in a right-angled triangle:

$\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$

$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

They are read "sine theta", "cosine theta" and "tangent theta". $\sin \theta$ is one number, fixed by the angle: it is not $\sin$ times $\theta$ .

SOH-CAH-TOA

The first letters help you remember which sides go in each ratio: Sin = Opposite ÷ Hypotenuse, Cos = Adjacent ÷ Hypotenuse, Tan = Opposite ÷ Adjacent.

Example 2: The ratios of both acute angles

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $BC = 3$ cm, $AC = 4$ cm and $AB = 5$ cm. Write down $\sin A$ , $\cos A$ , $\tan A$ , $\sin B$ , $\cos B$ and $\tan B$ as fractions.

Solution

The right angle is at $C$ , so the hypotenuse is $AB = 5$ cm.

From $\angle A$ : opposite $BC = 3$ , adjacent $AC = 4$ .

$\sin A = \dfrac{3}{5}$ , $\cos A = \dfrac{4}{5}$ , $\tan A = \dfrac{3}{4}$

From $\angle B$ the opposite and adjacent change places: opposite $AC = 4$ , adjacent $BC = 3$ .

$\sin B = \dfrac{4}{5}$ , $\cos B = \dfrac{3}{5}$ , $\tan B = \dfrac{4}{3}$

$\sin A = \frac{3}{5}$ , $\cos A = \frac{4}{5}$ , $\tan A = \frac{3}{4}$ ; $\sin B = \frac{4}{5}$ , $\cos B = \frac{3}{5}$ , $\tan B = \frac{4}{3}$ .

Notice that $\sin A = \cos B$ and $\cos A = \sin B$ : the side opposite one acute angle is adjacent to the other. A tangent can be greater than $1$ (here $\tan B = \frac{4}{3}$ ), but a sine or a cosine never is, because the hypotenuse is the longest side.

Example 3: Sides given by letters

In $\triangle XYZ$ , $\angle Z = 90^{\circ}$ , $XY = c$ m, $YZ = a$ m and $XZ = b$ m. Write down $\sin X$ , $\cos X$ , $\tan X$ and $\tan Y$ .

Solution

The hypotenuse is $XY = c$ , opposite the right angle at $Z$ .

From $\angle X$ : $YZ = a$ does not touch $X$ , so it is opposite; $XZ = b$ is adjacent.

$\sin X = \dfrac{a}{c}$ , $\cos X = \dfrac{b}{c}$ , $\tan X = \dfrac{a}{b}$

From $\angle Y$ : opposite $XZ = b$ , adjacent $YZ = a$ , so $\tan Y = \dfrac{b}{a}$ .

$\sin X = \frac{a}{c}$ , $\cos X = \frac{b}{c}$ , $\tan X = \frac{a}{b}$ and $\tan Y = \frac{b}{a}$ .

Finding the Third Side First

When only two sides are given, find the third by Pythagoras' theorem, then write the ratios.

Example 4: Two sides given

In $\triangle ABC$ , $\angle B = 90^{\circ}$ , $AB = 15$ cm and $BC = 8$ cm. Find $\sin A$ , $\cos A$ and $\tan A$ .

Solution

The hypotenuse is $AC$ , opposite the right angle at $B$ . By Pythagoras' theorem:

$AC^2 = AB^2 + BC^2 = 15^2 + 8^2 = 225 + 64 = 289\text{,}$ so $AC = \sqrt{289} = 17$ cm.

From $\angle A$ : opposite $BC = 8$ , adjacent $AB = 15$ , hypotenuse $AC = 17$ .

$\sin A = \dfrac{8}{17}$ , $\cos A = \dfrac{15}{17}$ , $\tan A = \dfrac{8}{15}$

$\sin A = \frac{8}{17}$ , $\cos A = \frac{15}{17}$ , $\tan A = \frac{8}{15}$

One ratio is enough to find the other two. A ratio fixes the shape of the triangle, so you may draw any triangle with that ratio and use its sides.

Example 5: From one ratio to the others

In $\triangle ABC$ , $\angle C = 90^{\circ}$ and $\tan A = \frac{5}{12}$ . Find $\sin A$ and $\cos A$ as fractions, without a calculator.

Solution

$\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}\text{,}$ so take $BC = 5$ and $AC = 12$ (any lengths in the ratio $5 : 12$ give the same answer).

$AB^2 = 12^2 + 5^2 = 144 + 25 = 169\text{,}$ so $AB = 13$ .

$\sin A = \dfrac{BC}{AB} = \dfrac{5}{13}$ and $\cos A = \dfrac{AC}{AB} = \dfrac{12}{13}$

$\sin A = \frac{5}{13}$ and $\cos A = \frac{12}{13}$

Values on a Calculator

A scientific calculator gives the sine, cosine and tangent of any angle, so there is no need to draw and measure. Angles in this chapter are in degrees, so the calculator must be in degree mode: most show a small D or DEG on the screen. As a check, $\sin 30^{\circ}$ must come out as $0.5$ .

Give a value to 3 decimal places unless the question says otherwise, and round only the final answer.

Example 6: Reading values from the calculator

Find, correct to 3 decimal places, (a) $\tan 47^{\circ}$ , (b) $\sin 75.3^{\circ}$ .

Solution

(a) Key tan 47 =. The screen shows $1.072368\ldots$ , so $\tan 47^{\circ} = 1.072$ (3 d.p.).

(b) Key sin 75.3 =. The screen shows $0.967268\ldots$ , so $\sin 75.3^{\circ} = 0.967$ (3 d.p.).

(a) $\tan 47^{\circ} = 1.072$ (b) $\sin 75.3^{\circ} = 0.967$

Example 7: Keying a whole expression

Find the value of $2 \cos 42.3^{\circ} + 3 \sin 16.8^{\circ}\text{,}$ correct to 3 decimal places.

Solution

Key the whole expression in one go: 2 cos 42.3 ) + 3 sin 16.8 ) =. (On most calculators a bracket opens after cos and sin; close it after the angle.)

The screen shows $2.346358\ldots$

Do not round each part first: $2(0.740) + 3(0.289) = 2.347\text{,}$ which is wrong in the last place.

$2 \cos 42.3^{\circ} + 3 \sin 16.8^{\circ} = 2.346$ (3 d.p.)

Common Mistakes

Using the same opposite and adjacent sides for both acute angles. In Example 2, $\tan A = \frac{3}{4}$ but $\tan B = \frac{4}{3}$ : name the sides again from each angle.
Naming the sides by how the triangle is drawn (the bottom side as adjacent, the slanting side as hypotenuse). Find the right angle first: the side opposite it is the hypotenuse, whatever way the triangle is turned; then name the other two from the marked angle.
Leaving the calculator in radian mode: it then treats $30$ as a different unit of angle (radians, met in later years), not degrees, and gives $\sin 30 = -0.988\ldots$ instead of $0.5$ .
Rounding values before adding or multiplying them. Key the whole expression and round once at the end.
Writing $\sin$ on its own. Sine is always the sine of an angle: $\sin 40^{\circ}$ , $\sin A$ , $\sin \theta$ .

Practice Questions

Give values to 3 decimal places unless the question says otherwise. Try each question before opening its solution.

Question 1

In $\triangle LMN$ , $\angle M = 90^{\circ}$ . Name the hypotenuse, and the sides opposite and adjacent to (a) $\angle L$ , (b) $\angle N$ .

ShowHide solution

The right angle is at $M$ , so the hypotenuse is $LN$ .

(a) $MN$ does not touch $L$ : it is opposite. $LM$ is adjacent.

(b) $LM$ does not touch $N$ : it is opposite. $MN$ is adjacent.

Hypotenuse $LN$ . (a) Opposite $MN$ , adjacent $LM$ . (b) Opposite $LM$ , adjacent $MN$ .

Question 2

In $\triangle PQR$ , $\angle R = 90^{\circ}$ , $PR = x$ cm, $QR = y$ cm and $PQ = z$ cm. Write down $\sin P$ , $\cos P$ , $\tan P$ and $\sin Q$ .

ShowHide solution

The hypotenuse is $PQ = z$ . From $\angle P$ : opposite $QR = y$ , adjacent $PR = x$ .

$\sin P = \dfrac{y}{z}$ , $\cos P = \dfrac{x}{z}$ , $\tan P = \dfrac{y}{x}$

From $\angle Q$ the opposite side is $PR = x$ , so $\sin Q = \dfrac{x}{z}$ .

$\sin P = \frac{y}{z}$ , $\cos P = \frac{x}{z}$ , $\tan P = \frac{y}{x}$ , $\sin Q = \frac{x}{z}$

Question 3

In $\triangle DEF$ , $\angle E = 90^{\circ}$ , $DE = 7$ cm, $EF = 24$ cm and $DF = 25$ cm. Write down $\sin D$ , $\cos D$ , $\tan D$ , $\sin F$ and $\tan F$ .

ShowHide solution

The hypotenuse is $DF = 25$ . From $\angle D$ : opposite $EF = 24$ , adjacent $DE = 7$ .

$\sin D = \dfrac{24}{25}$ , $\cos D = \dfrac{7}{25}$ , $\tan D = \dfrac{24}{7}$

From $\angle F$ : opposite $DE = 7$ , adjacent $EF = 24$ , so $\sin F = \dfrac{7}{25}$ and $\tan F = \dfrac{7}{24}$ .

$\sin D = \frac{24}{25}$ , $\cos D = \frac{7}{25}$ , $\tan D = \frac{24}{7}$ , $\sin F = \frac{7}{25}$ , $\tan F = \frac{7}{24}$

Question 4

In $\triangle KLM$ , $\angle L = 90^{\circ}$ , $KM = 29$ cm and $LM = 20$ cm. Find $KL$ , then write down $\sin K$ , $\cos K$ and $\tan K$ .

ShowHide solution

The hypotenuse is $KM = 29$ cm. $KL^2 = 29^2 - 20^2 = 841 - 400 = 441\text{,}$ so $KL = \sqrt{441} = 21$ cm.

From $\angle K$ : opposite $LM = 20$ , adjacent $KL = 21$ .

$\sin K = \dfrac{20}{29}$ , $\cos K = \dfrac{21}{29}$ , $\tan K = \dfrac{20}{21}$

$KL = 21$ cm; $\sin K = \frac{20}{29}$ , $\cos K = \frac{21}{29}$ , $\tan K = \frac{20}{21}$

Question 5

In $\triangle ABC$ , $\angle C = 90^{\circ}$ and $\cos B = \frac{8}{17}$ . Without a calculator, find $\sin B$ , $\tan B$ and $\sin A$ .

ShowHide solution

$\cos B = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{8}{17}\text{,}$ so take $BC = 8$ and $AB = 17$ .

$AC^2 = 17^2 - 8^2 = 289 - 64 = 225\text{,}$ so $AC = 15$ .

From $\angle B$ : opposite $AC = 15$ . $\sin B = \dfrac{15}{17}$ and $\tan B = \dfrac{15}{8}$ .

From $\angle A$ : opposite $BC = 8$ , so $\sin A = \dfrac{8}{17}$ (the same as $\cos B$ ).

$\sin B = \frac{15}{17}$ , $\tan B = \frac{15}{8}$ , $\sin A = \frac{8}{17}$

Question 6

Use a calculator to find, correct to 3 decimal places, (a) $\cos 28^{\circ}$ , (b) $\tan 63.5^{\circ}$ , (c) $\sin 8.2^{\circ}$ .

ShowHide solution

(a) cos 28 = gives $0.882948\ldots$

(b) tan 63.5 = gives $2.005690\ldots$

(a) $0.883$ (b) $2.006$ (c) $0.143$

Question 7

Find, correct to 3 decimal places, (a) $\sin 35^{\circ} + \cos 49^{\circ}$ , (b) $\tan 71.6^{\circ} \times \tan 16.7^{\circ}\text{,}$ (c) $\dfrac{3 \cos 20^{\circ}}{\sin 70^{\circ}}$ . What do you notice about (c)?

ShowHide solution

(a) Keyed in one go: $1.229635\ldots$

(b) $0.901876\ldots$

(c) $3$ exactly. In a right-angled triangle with angles $20^{\circ}$ and $70^{\circ}$ , the side adjacent to the $20^{\circ}$ angle is opposite the $70^{\circ}$ angle, so $\cos 20^{\circ} = \sin 70^{\circ}$ and the fraction is $3$ .

(a) $1.230$ (b) $0.902$ (c) $3.000$ , because $\cos 20^{\circ} = \sin 70^{\circ}$ .

Question 8

Find the error in each. (a) In $\triangle ABC$ of Example 2 ( $\angle C = 90^{\circ}$ , $BC = 3$ cm, $AC = 4$ cm), Wei Ming writes $\tan B = \frac{3}{4}$ . (b) Priya keys cos 60 = and gets $-0.952$ .

ShowHide solution

(a) From $\angle B$ the opposite side is $AC = 4$ and the adjacent side is $BC = 3$ . Wei Ming used the sides for $\angle A$ . $\tan B = \frac{4}{3}$ .

(b) The calculator is in radian mode, so it treated $60$ as a different unit of angle (radians, met in later years), not degrees. In degree mode, $\cos 60^{\circ} = 0.5$ .

(a) $\tan B = \frac{4}{3}$ (b) Radian mode; $\cos 60^{\circ} = 0.5$ .

Question 9

In $\triangle MNO$ , $\angle O = 90^{\circ}$ , $OM = b$ cm and $ON = c$ cm. Given that $\tan M = \tan N$ , show that $b = c$ . What kind of triangle is $MNO$ , and what are $\angle M$ and $\angle N$ ?

ShowHide solution

From $\angle M$ : opposite $ON = c$ , adjacent $OM = b$ , so $\tan M = \dfrac{c}{b}$ .

From $\angle N$ : opposite $OM = b$ , adjacent $ON = c$ , so $\tan N = \dfrac{b}{c}$ .

$\dfrac{c}{b} = \dfrac{b}{c}$ gives $c^2 = b^2$ , and lengths are positive, so $b = c$ .

Two sides are equal, so the triangle is isosceles, and its base angles are equal: $\angle M = \angle N = \dfrac{90^{\circ}}{2} = 45^{\circ}\text{.}$

$b = c$ . $MNO$ is an isosceles right-angled triangle, with $\angle M = \angle N = 45^{\circ}$ .

Question 10

A ladder $2.5$ m long leans against a vertical wall of a block of flats. Its foot is on level ground, $0.7$ m from the wall. The ladder makes an angle $\theta$ with the ground. Find the height, $h$ m, that the ladder reaches up the wall, then write down $\sin \theta$ , $\cos \theta$ and $\tan \theta$ as fractions.

ShowHide solution

The wall is vertical and the ground level, so $\angle W = 90^{\circ}$ and the ladder $FT$ is the hypotenuse.

$h^2 = 2.5^2 - 0.7^2 = 6.25 - 0.49 = 5.76\text{,}$ so $h = \sqrt{5.76} = 2.4$ m.

From $\theta$ at $F$ : opposite $WT = 2.4$ , adjacent $FW = 0.7$ , hypotenuse $2.5$ .

$\sin \theta = \dfrac{2.4}{2.5} = \dfrac{24}{25}\text{,}$ $\cos \theta = \dfrac{0.7}{2.5} = \dfrac{7}{25}\text{,}$ $\tan \theta = \dfrac{2.4}{0.7} = \dfrac{24}{7}$

The ladder reaches $2.4$ m up the wall; $\sin \theta = \frac{24}{25}$ , $\cos \theta = \frac{7}{25}$ , $\tan \theta = \frac{24}{7}$ .

Question 11

A straight ramp at a void deck rises $0.5$ m over a horizontal distance of $6$ m. The ramp makes an angle $\theta$ with the ground. (a) Write down $\tan \theta$ as a fraction. (b) Find the length of the ramp, correct to 3 significant figures. (c) Find $\sin \theta$ , correct to 3 significant figures.

ShowHide solution

(a) Opposite $0.5$ , adjacent $6$ : $\tan \theta = \dfrac{0.5}{6} = \dfrac{1}{12}\text{.}$

(b) The ramp is the hypotenuse: $\sqrt{6^2 + 0.5^2} = \sqrt{36.25} = 6.020797\ldots = 6.02$ m (3 s.f.).

(a) $\frac{1}{12}$ (b) $6.02$ m (c) $0.0830$

Key Points

Find the right angle first: the side opposite it is the hypotenuse, the longest side.
From an acute angle $\theta$ , the opposite side does not touch $\theta$ , and the adjacent side touches $\theta$ and is not the hypotenuse. From the other acute angle they change places.
For a given angle the ratio of two sides is the same in every right-angled triangle, because such triangles are similar.
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\text{,}$ $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\text{,}$ $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ (SOH-CAH-TOA).
When a side is missing, find it by Pythagoras' theorem first. One ratio is enough to find the other two.
Set the calculator in degree mode, key whole expressions, and round only at the end.