Finding an Unknown Side

When one side and one acute angle of a right-angled triangle are known, choose the ratio that links the side you know with the side you want, and multiply.

Before you start

From an acute angle $\theta$ (lesson 10.1): the hypotenuse is opposite the right angle, the opposite side does not touch $\theta$ , and the adjacent side is next to $\theta$ and is not the hypotenuse.
$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}\text{,}$ $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\text{,}$ $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ (SOH-CAH-TOA).
The two acute angles of a right-angled triangle add up to $90^{\circ}$ .

Choosing the Ratio

Each ratio links two sides. So the question to ask is: which two sides does this problem involve, the one I know and the one I want? The ratio that uses exactly those two is the one to choose.

Choosing the ratio

1. Find the right angle, and mark the hypotenuse opposite it.

2. From the given angle, name the side you know and the side you want: hypotenuse, opposite or adjacent.

3. Choose the ratio that uses those two sides.

The two sides involved	The ratio to use
opposite and hypotenuse	$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$
adjacent and hypotenuse	$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$
opposite and adjacent	$\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$

The Unknown on Top

In this lesson the side you want is on the top of the ratio. Write the ratio with the numbers in, then multiply both sides by the bottom.

Finding a side on top of the ratio

opposite $=$ hypotenuse $\times \sin \theta$

adjacent $=$ hypotenuse $\times \cos \theta$

opposite $=$ adjacent $\times \tan \theta$

Key the whole product in one go and round only the answer: lengths to 3 significant figures unless the question says otherwise.

Example 1: Opposite from the hypotenuse

In $\triangle ABC$ , $\angle ACB = 90^{\circ}$ , $\angle BAC = 72^{\circ}$ and $AB = 12$ cm. Find $BC$ .

Solution

The right angle is at $C$ , so $AB = 12$ cm is the hypotenuse.

From the $72^{\circ}$ angle at $A$ : $BC$ does not touch $A$ , so the side we want is the opposite. We know the hypotenuse. Opposite and hypotenuse: use sine.

$\sin 72^{\circ} = \dfrac{BC}{12}$

$BC = 12 \sin 72^{\circ}$ (key 12 sin 72 =)

$= 11.412\ldots = 11.4$ cm (3 s.f.)

$BC = 11.4$ cm (to 3 s.f.)

Example 2: Adjacent from the hypotenuse

In $\triangle XYZ$ , $\angle Z = 90^{\circ}$ , $\angle X = 55^{\circ}$ and $XY = 12$ m. Find $XZ$ .

Solution

The right angle is at $Z$ , so $XY = 12$ m is the hypotenuse.

From $\angle X$ : $XZ$ touches $X$ and is not the hypotenuse, so it is adjacent. Adjacent and hypotenuse: use cosine.

$\cos 55^{\circ} = \dfrac{XZ}{12}$

$XZ = 12 \cos 55^{\circ} = 6.882\ldots = 6.88$ m (3 s.f.)

$XZ = 6.88$ m (to 3 s.f.)

Example 3: Opposite from the adjacent

In $\triangle PQR$ , $\angle R = 90^{\circ}$ , $\angle P = 42^{\circ}$ and $PR = 15$ cm. Find $QR$ .

Solution

From $\angle P$ : $QR$ does not touch $P$ , so it is opposite; $PR = 15$ cm is adjacent. The hypotenuse is not involved. Opposite and adjacent: use tangent.

$\tan 42^{\circ} = \dfrac{QR}{15}$

$QR = 15 \tan 42^{\circ} = 13.506\ldots = 13.5$ cm (3 s.f.)

$QR = 13.5$ cm (to 3 s.f.)

The Angle at Either Corner

The given angle may be at either acute corner, and the triangle may be drawn any way round. Always name the sides from the angle you are using.

Example 4: Two sides from one angle

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle B = 35^{\circ}$ and $AB = 10$ cm. Find $AC$ and $BC$ .

Solution

The hypotenuse is $AB = 10$ cm. The angle is at $B$ , so name the sides from $B$ .

$AC$ does not touch $B$ : it is opposite. Opposite and hypotenuse: sine.

$AC = 10 \sin 35^{\circ} = 5.735\ldots = 5.74$ cm (3 s.f.)

$BC$ touches $B$ : it is adjacent. Adjacent and hypotenuse: cosine.

$BC = 10 \cos 35^{\circ} = 8.191\ldots = 8.19$ cm (3 s.f.)

$AC = 5.74$ cm and $BC = 8.19$ cm (to 3 s.f.)

When one acute angle is known, so is the other: they add up to $90^{\circ}$ . Either angle can then be used, and they give the same answer, because the side opposite one acute angle is adjacent to the other. Once two sides are known, Pythagoras' theorem gives the third side another way, and is a good check (Example 5(c)).

Example 5: Using the other angle

In $\triangle LMN$ , $\angle M = 90^{\circ}$ , $\angle L = 61^{\circ}$ and $LN = 8.5$ cm. (a) Find $MN$ . (b) Find $\angle N$ , and use it to find $LM$ . (c) Check $LM$ by Pythagoras' theorem.

Solution

(a) From $\angle L$ , $MN$ is opposite: $MN = 8.5 \sin 61^{\circ} = 7.434\ldots = 7.43$ cm (3 s.f.)

(b) $\angle N = 90^{\circ} - 61^{\circ} = 29^{\circ}\text{.}$ From $\angle N$ , $LM$ is opposite:

$LM = 8.5 \sin 29^{\circ} = 4.120\ldots = 4.12$ cm (3 s.f.)

From $\angle L$ it is adjacent, and $8.5 \cos 61^{\circ}$ gives the same $4.120\ldots$ , because $\sin 29^{\circ} = \cos 61^{\circ}$ .

(c) $LM = \sqrt{8.5^2 - 7.434\ldots^2} = \sqrt{16.981\ldots} = 4.120\ldots$ cm, the same. Use the full value of $MN$ here, not $7.43$ .

(a) $MN = 7.43$ cm (b) $\angle N = 29^{\circ}$ , $LM = 4.12$ cm (to 3 s.f.)

Sides in a Setting

In a word problem, draw the right-angled triangle first. A wall standing on level ground makes a right angle with it, so a ladder against a wall is the hypotenuse.

Example 6: A ladder against a wall

A ladder $3.5$ m long leans against a vertical wall. It makes an angle of $72^{\circ}$ with the level ground. How high up the wall does it reach?

Solution

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

From the $72^{\circ}$ angle at $F$ , the height $WT$ is opposite. Opposite and hypotenuse: sine.

$h = 3.5 \sin 72^{\circ} = 3.328\ldots = 3.33$ (3 s.f.)

The ladder reaches $3.33$ m up the wall.

Common Mistakes

Choosing the ratio by habit. Name the side you know and the side you want first: $AC$ in $\triangle ABC$ with $\angle C = 90^{\circ}$ is adjacent to $\angle A$ but opposite $\angle B$ .
Rounding the ratio before multiplying. $40 \tan 63^{\circ} = 78.504\ldots = 78.5\text{,}$ but $40 \times 1.96 = 78.4$ . Key the whole product in one go.
Writing $\sin 72^{\circ} = \frac{12}{BC}$ , with the sides upside down. The side opposite the angle is always on top of sine.
Leaving the calculator in radian mode: check that $\sin 30^{\circ}$ gives $0.5$ .
Leaving out the unit, or giving more figures than the question asks for.

Practice Questions

Give lengths correct to 3 significant figures unless the question says otherwise. Try each question before opening its solution.

Question 1

Which ratio would you use? (a) Given the angle $40^{\circ}$ and the hypotenuse; find the opposite side. (b) Given $60^{\circ}$ and the adjacent side; find the opposite side. (c) Given $25^{\circ}$ and the opposite side; find the hypotenuse. (d) Given $70^{\circ}$ and the hypotenuse; find the adjacent side.

ShowHide solution

(a) Opposite and hypotenuse: sine.

(b) Opposite and adjacent: tangent.

(d) Adjacent and hypotenuse: cosine.

(a) $\sin$ (b) $\tan$ (c) $\sin$ (d) $\cos$

Question 2

In $\triangle DEF$ , $\angle E = 90^{\circ}$ , $\angle D = 28^{\circ}$ and $DF = 15$ cm. Find $EF$ .

ShowHide solution

From $\angle D$ , $EF$ is opposite and $DF$ is the hypotenuse: sine.

$EF = 15 \sin 28^{\circ} = 7.042\ldots$

$EF = 7.04$ cm

Question 3

In $\triangle KLM$ , $\angle L = 90^{\circ}$ , $\angle K = 64^{\circ}$ and $KM = 9.5$ cm. Find $KL$ .

ShowHide solution

From $\angle K$ , $KL$ is adjacent and $KM$ is the hypotenuse: cosine.

$KL = 9.5 \cos 64^{\circ} = 4.164\ldots$

$KL = 4.16$ cm

Question 4

In $\triangle STU$ , $\angle T = 90^{\circ}$ , $\angle S = 33^{\circ}$ and $ST = 20$ m. Find $TU$ .

ShowHide solution

From $\angle S$ , $TU$ is opposite and $ST$ is adjacent: tangent.

$TU = 20 \tan 33^{\circ} = 12.988\ldots = 13.0$ m (3 s.f.)

$TU = 13.0$ m

Question 5

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle B = 47^{\circ}$ and $AB = 18$ m. Find $BC$ , correct to 1 decimal place.

ShowHide solution

From $\angle B$ , $BC$ is adjacent and $AB$ is the hypotenuse: cosine.

$BC = 18 \cos 47^{\circ} = 12.275\ldots = 12.3$ m (1 d.p.)

$BC = 12.3$ m

Question 6

In $\triangle XYZ$ , $\angle Z = 90^{\circ}$ , $\angle X = 58^{\circ}$ and $XY = 4.9$ m. Find $YZ$ . Then find $\angle Y$ and use it to find $XZ$ .

ShowHide solution

From $\angle X$ , $YZ$ is opposite: $YZ = 4.9 \sin 58^{\circ} = 4.155\ldots = 4.16$ m (3 s.f.)

$\angle Y = 90^{\circ} - 58^{\circ} = 32^{\circ}\text{.}$ From $\angle Y$ , $XZ$ is opposite:

$XZ = 4.9 \sin 32^{\circ} = 2.596\ldots = 2.60$ m (3 s.f.)

$YZ = 4.16$ m, $\angle Y = 32^{\circ}$ , $XZ = 2.60$ m

Question 7

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle A = 40^{\circ}$ and $AC = 7$ cm. Find $BC$ , then use Pythagoras' theorem to find $AB$ .

ShowHide solution

From $\angle A$ , $BC$ is opposite and $AC$ adjacent: $BC = 7 \tan 40^{\circ} = 5.873\ldots = 5.87$ cm (3 s.f.)

$AB = \sqrt{7^2 + 5.873\ldots^2} = \sqrt{83.500\ldots} = 9.137\ldots = 9.14$ cm (3 s.f.), with the full value of $BC$ .

$BC = 5.87$ cm, $AB = 9.14$ cm

Question 8

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle A = 52^{\circ}$ and $AB = 20$ cm. Hui Min writes " $AC = 20 \sin 52^{\circ} = 15.8$ cm". Explain her mistake and find $AC$ .

ShowHide solution

$AC$ touches $A$ , so it is the side adjacent to the $52^{\circ}$ angle, not the opposite side. Adjacent and hypotenuse: cosine.

$AC = 20 \cos 52^{\circ} = 12.313\ldots = 12.3$ cm (3 s.f.). (Her $15.8$ cm is $BC$ .)

She used sine for the adjacent side. $AC = 12.3$ cm.

Question 9

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $\angle P = 56^{\circ}$ and $PQ = 30$ cm. Jia Hui writes " $\tan 56^{\circ} = 1.48$ , so $QR = 30 \times 1.48 = 44.4$ cm". Is her answer correct to 3 significant figures?

ShowHide solution

From $\angle P$ , $QR$ is opposite and $PQ$ adjacent, so tangent is the right ratio.

But $\tan 56^{\circ} = 1.482\ldots$ , and rounding it to $1.48$ lost the third figure of the answer.

$QR = 30 \tan 56^{\circ} = 44.476\ldots = 44.5$ cm (3 s.f.)

No. Keyed in one go, $QR = 44.5$ cm.

Question 10

A rope $3$ m long is pulled straight from the top $T$ of a vertical tent pole $PT$ to a peg $G$ on level ground. The rope makes an angle of $50^{\circ}$ with the ground. Find (a) the distance from the peg to the foot of the pole, (b) the height of the pole.

ShowHide solution

The pole is vertical, so $\angle P = 90^{\circ}$ and the rope $GT$ is the hypotenuse.

(a) From the $50^{\circ}$ angle at $G$ , $PG$ is adjacent: $d = 3 \cos 50^{\circ} = 1.928\ldots = 1.93$ (3 s.f.)

(b) $PT$ is opposite: $h = 3 \sin 50^{\circ} = 2.298\ldots = 2.30$ (3 s.f.)

(a) $1.93$ m (b) $2.30$ m

Question 11

On a nature walk at Bukit Timah Hill, Arjun walks $120$ m along a straight path that slopes up at $8^{\circ}$ to the horizontal. How many metres higher is he at the end of the path than at the start?

ShowHide solution

Let $B$ be the point level with the start $A$ and directly below the end $T$ . Then $\angle B = 90^{\circ}$ and the path $AT$ is the hypotenuse.

The rise $BT$ is opposite the $8^{\circ}$ angle: $h = 120 \sin 8^{\circ} = 16.700\ldots = 16.7$ (3 s.f.)

He is $16.7$ m higher.

Key Points

Name the side you know and the side you want from the given angle, then choose the ratio that uses those two.
Opposite and hypotenuse: sine. Adjacent and hypotenuse: cosine. Opposite and adjacent: tangent.
With the unknown on top: opposite $=$ hypotenuse $\times \sin \theta$ , adjacent $=$ hypotenuse $\times \cos \theta$ , opposite $=$ adjacent $\times \tan \theta$ .
The other acute angle is $90^{\circ} - \theta$ ; either angle gives the same answer when its sides are named from it.
Key the whole product in one go and round only the answer, usually to 3 significant figures.