Finding the Hypotenuse and Other Sides by Dividing

When the side you want is underneath in the ratio, multiply up and then divide. This finds the hypotenuse from either shorter side, a side from the other shorter side, and every side of a triangle.

Before you start

Choosing the ratio (lesson 10.2): name the side you know and the side you want from the given angle. Opposite and hypotenuse: $\sin$ . Adjacent and hypotenuse: $\cos$ . Opposite and adjacent: $\tan$ .
Solving an equation with the unknown underneath: if $\dfrac{12}{y} = 3$ , then $12 = 3y$ and $y = \dfrac{12}{3} = 4$ .

Why We Divide

In lesson 10.2 the side we wanted was on top of the ratio, so we multiplied. When the side we want is underneath, multiply both sides by it first, and then divide.

Example 1: The hypotenuse from the opposite side

In $\triangle PQR$ , $\angle R = 90^{\circ}$ , $\angle P = 37^{\circ}$ and $QR = 5$ cm. Find $PQ$ .

Solution

From $\angle P$ , $QR = 5$ cm is opposite and $PQ$ is the hypotenuse: sine.

$\sin 37^{\circ} = \dfrac{5}{PQ}$

Multiply both sides by $PQ$ : $PQ \sin 37^{\circ} = 5$

Divide both sides by $\sin 37^{\circ}$ : $PQ = \dfrac{5}{\sin 37^{\circ}}$ (key 5 ÷ sin 37 =)

$= 8.308\ldots = 8.31$ cm (3 s.f.)

Check: the hypotenuse, $8.31$ cm, is longer than the side given, $5$ cm.

$PQ = 8.31$ cm (to 3 s.f.)

Finding a side underneath in the ratio

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

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

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

You need not learn these: write the ratio with the numbers in, multiply by the unknown, then divide by the sine, cosine or tangent.

Example 2: The hypotenuse from the adjacent side

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle B = 40^{\circ}$ and $BC = 9$ cm. Find $AB$ .

Solution

From $\angle B$ , $BC = 9$ cm is adjacent and $AB$ is the hypotenuse: cosine.

$\cos 40^{\circ} = \dfrac{9}{AB}$ , so $AB \cos 40^{\circ} = 9$

$AB = \dfrac{9}{\cos 40^{\circ}} = 11.748\ldots = 11.7$ cm (3 s.f.)

$AB = 11.7$ cm (to 3 s.f.)

Example 3: The adjacent side from the opposite side

In $\triangle XYZ$ , $\angle Z = 90^{\circ}$ , $\angle X = 35^{\circ}$ and $YZ = 12$ cm. Find $XZ$ .

Solution

From $\angle X$ , $YZ = 12$ cm is opposite and $XZ$ is adjacent: tangent.

$\tan 35^{\circ} = \dfrac{12}{XZ}$ , so $XZ \tan 35^{\circ} = 12$

$XZ = \dfrac{12}{\tan 35^{\circ}} = 17.137\ldots = 17.1$ cm (3 s.f.)

$XZ = 17.1$ cm (to 3 s.f.)

The Other Way: Use the Other Angle

When the two shorter sides are involved, there is another way. The other acute angle is $90^{\circ} - \theta$ , and from it the opposite and adjacent sides change places, so the side you want comes on top and you can multiply.

This does not work for the hypotenuse: it is underneath in sine and cosine from either angle, so it is always found by dividing.

Example 4: Example 3 by the other angle

In $\triangle XYZ$ of Example 3 ( $\angle Z = 90^{\circ}$ , $\angle X = 35^{\circ}$ , $YZ = 12$ cm), find $XZ$ by using $\angle Y$ .

Solution

$\angle Y = 90^{\circ} - 35^{\circ} = 55^{\circ}$

From $\angle Y$ , $XZ$ is opposite and $YZ = 12$ cm is adjacent: $\tan 55^{\circ} = \dfrac{XZ}{12}$

$XZ = 12 \tan 55^{\circ} = 17.137\ldots = 17.1$ cm (3 s.f.), as before.

$XZ = 17.1$ cm, the same by either angle.

Is the Answer Reasonable?

A quick check

The hypotenuse is always longer than each of the other two sides.

The sine and the cosine of an acute angle are less than $1$ , so dividing by them gives a larger number. A "hypotenuse" that comes out shorter than a given side means you multiplied instead of dividing.

Example 5: Spot the error

For $\triangle PQR$ of Example 1 ( $\angle R = 90^{\circ}$ , $\angle P = 37^{\circ}$ , $QR = 5$ cm), Daniel writes $PQ = 5 \sin 37^{\circ} = 3.01$ cm. Without working the answer out, how can you tell he is wrong?

Solution

$PQ$ is the hypotenuse, so it must be longer than $QR = 5$ cm. His $3.01$ cm is shorter.

From $\sin 37^{\circ} = \dfrac{5}{PQ}$ the unknown is underneath, so he should have divided: $PQ = 5 \div \sin 37^{\circ} = 8.31$ cm.

A hypotenuse of $3.01$ cm is shorter than the side of $5$ cm; he multiplied instead of dividing.

Every Side of a Triangle

One side and one acute angle fix the whole triangle. Find the other angle by subtracting from $90^{\circ}$ , then each missing side from the side you were given (not from a side you rounded), and check with Pythagoras' theorem.

Example 6: Solving a triangle

In $\triangle DEF$ , $\angle E = 90^{\circ}$ , $\angle D = 52^{\circ}$ and $EF = 6.5$ cm. Find $\angle F$ , $DF$ and $DE$ .

Solution

$\angle F = 90^{\circ} - 52^{\circ} = 38^{\circ}$

From $\angle D$ , $EF = 6.5$ cm is opposite.

$DF$ is the hypotenuse: $\sin 52^{\circ} = \dfrac{6.5}{DF}$ , so $DF = \dfrac{6.5}{\sin 52^{\circ}} = 8.248\ldots = 8.25$ cm (3 s.f.)

$DE$ is adjacent: $\tan 52^{\circ} = \dfrac{6.5}{DE}$ , so $DE = \dfrac{6.5}{\tan 52^{\circ}} = 5.078\ldots = 5.08$ cm (3 s.f.)

Check: $5.078\ldots^2 + 6.5^2 = 25.789\ldots + 42.25 = 68.039\ldots\text{,}$ and $8.248\ldots^2 = 68.039\ldots$

$\angle F = 38^{\circ}$ , $DF = 8.25$ cm, $DE = 5.08$ cm (to 3 s.f.)

Two Triangles Sharing a Side

In a figure made of two right-angled triangles, work in one triangle at a time. A side the two triangles share links them; keep every length you find at its full calculator value, and round only the final answer.

Example 7: A side on a line

In the figure, $\angle ADC = 90^{\circ}$ and $B$ is a point on $DC$ . $AD = 6$ cm, $\angle ABD = 65^{\circ}$ and $\angle ACD = 40^{\circ}$ . Find $BC$ .

Solution

In $\triangle ABD$ : from $\angle ABD$ , $AD$ is opposite and $DB$ adjacent: $\tan 65^{\circ} = \dfrac{6}{DB}$ , so $DB = \dfrac{6}{\tan 65^{\circ}} = 2.797\ldots$ cm

In $\triangle ACD$ : from $\angle ACD$ , $AD$ is opposite and $DC$ adjacent: $\tan 40^{\circ} = \dfrac{6}{DC}$ , so $DC = \dfrac{6}{\tan 40^{\circ}} = 7.150\ldots$ cm

$BC = DC - DB = 7.150\ldots - 2.797\ldots = 4.352\ldots = 4.35$ cm (3 s.f.)

$BC = 4.35$ cm (to 3 s.f.)

Common Mistakes

Multiplying when the unknown is underneath: $5 \sin 37^{\circ} = 3.01$ cm cannot be the hypotenuse, which must be longer than the $5$ cm side.
Dividing the wrong way round: $\sin 37^{\circ} \div 5 = 0.120\ldots$ is not a length here. From $PQ \sin 37^{\circ} = 5$ , divide $5$ by $\sin 37^{\circ}$ .
Rounding a shared side before using it again: keep its full calculator value when you carry it into the second triangle (try this in Question 9).
Finding the hypotenuse by "the other angle": it is underneath from both angles, so always divide.

Practice Questions

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

Question 1

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle A = 67^{\circ}$ and $BC = 8.7$ cm. Find $AB$ .

ShowHide solution

From $\angle A$ , $BC$ is opposite and $AB$ the hypotenuse: $\sin 67^{\circ} = \dfrac{8.7}{AB}$

$AB = \dfrac{8.7}{\sin 67^{\circ}} = 9.451\ldots = 9.45$ cm (3 s.f.)

$AB = 9.45$ cm

Question 2

In $\triangle KLM$ , $\angle L = 90^{\circ}$ , $\angle K = 28^{\circ}$ and $KL = 14$ cm. Find $KM$ .

ShowHide solution

From $\angle K$ , $KL$ is adjacent and $KM$ the hypotenuse: $\cos 28^{\circ} = \dfrac{14}{KM}$

$KM = \dfrac{14}{\cos 28^{\circ}} = 15.855\ldots = 15.9$ cm (3 s.f.)

$KM = 15.9$ cm

Question 3

In $\triangle PQR$ , $\angle Q = 90^{\circ}$ , $\angle P = 62^{\circ}$ and $QR = 11$ m. Find $PQ$ .

ShowHide solution

From $\angle P$ , $QR$ is opposite and $PQ$ adjacent: $\tan 62^{\circ} = \dfrac{11}{PQ}$

$PQ = \dfrac{11}{\tan 62^{\circ}} = 5.848\ldots = 5.85$ m (3 s.f.)

Or: $\angle R = 28^{\circ}$ , and $PQ = 11 \tan 28^{\circ}$ gives the same.

$PQ = 5.85$ m

Question 4

In $\triangle STU$ , $\angle T = 90^{\circ}$ , $\angle S = 71^{\circ}$ and $ST = 4.2$ cm. Find $SU$ , marked $x$ cm in the figure.

ShowHide solution

From $\angle S$ , $ST$ is adjacent and $SU$ the hypotenuse: $\cos 71^{\circ} = \dfrac{4.2}{x}$

$x = \dfrac{4.2}{\cos 71^{\circ}} = 12.900\ldots = 12.9$ (3 s.f.)

$SU = 12.9$ cm

Question 5

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle A = 23^{\circ}$ and $BC = 5$ cm. (a) Find $AB$ . (b) Find $AC$ . (c) Find $AC$ again by using $\angle B$ .

ShowHide solution

(a) $\sin 23^{\circ} = \dfrac{5}{AB}$ , so $AB = \dfrac{5}{\sin 23^{\circ}} = 12.796\ldots = 12.8$ cm (3 s.f.)

(b) $\tan 23^{\circ} = \dfrac{5}{AC}$ , so $AC = \dfrac{5}{\tan 23^{\circ}} = 11.779\ldots = 11.8$ cm (3 s.f.)

(c) $\angle B = 67^{\circ}$ . From $\angle B$ , $AC$ is opposite and $BC$ adjacent: $AC = 5 \tan 67^{\circ} = 11.779\ldots\text{,}$ the same.

(a) $12.8$ cm (b) $11.8$ cm (c) $11.8$ cm

Question 6

In $\triangle ABC$ , $\angle C = 90^{\circ}$ , $\angle A = 30^{\circ}$ and $BC = 7$ cm. Given that $\sin 30^{\circ} = 0.5$ exactly, find $AB$ without a calculator.

ShowHide solution

$BC$ is opposite $\angle A$ and $AB$ is the hypotenuse: $\sin 30^{\circ} = \dfrac{7}{AB}$

$AB = \dfrac{7}{0.5} = 14$

In a right-angled triangle with a $30^{\circ}$ angle, the side opposite that angle is half the hypotenuse.

$AB = 14$ cm

Question 7

In $\triangle XYZ$ , $\angle Y = 90^{\circ}$ , $\angle X = 48^{\circ}$ and $XY = 6$ cm. Ravi finds the hypotenuse $XZ$ as $6 \cos 48^{\circ} = 4.01$ cm. Explain how you can tell this is wrong, and find $XZ$ .

ShowHide solution

The hypotenuse must be longer than $XY = 6$ cm, and $4.01$ cm is shorter.

$\cos 48^{\circ} = \dfrac{6}{XZ}$ has the unknown underneath, so divide: $XZ = \dfrac{6}{\cos 48^{\circ}} = 8.966\ldots = 8.97$ cm (3 s.f.)

His hypotenuse is shorter than a side. $XZ = 8.97$ cm.

Question 8

In $\triangle ABC$ , $\angle ACB = 90^{\circ}$ , $\angle CAB = 56^{\circ}$ and $AB = 8.9$ m. $H$ is the point on $AB$ such that $CH$ is perpendicular to $AB$ . Find $AC$ , $CH$ and $AH$ .

ShowHide solution

In $\triangle ABC$ : from $\angle A$ , $AC$ is adjacent and $AB$ the hypotenuse: $AC = 8.9 \cos 56^{\circ} = 4.976\ldots = 4.98$ m (3 s.f.)

In $\triangle ACH$ ( $\angle AHC = 90^{\circ}$ ), $AC$ is the hypotenuse. From $\angle A$ :

$CH = AC \sin 56^{\circ} = 4.976\ldots \times \sin 56^{\circ} = 4.125\ldots = 4.13$ m (3 s.f.)

$AH = AC \cos 56^{\circ} = 4.976\ldots \times \cos 56^{\circ} = 2.783\ldots = 2.78$ m (3 s.f.)

$AC = 4.98$ m, $CH = 4.13$ m, $AH = 2.78$ m

Question 9

A vertical mast $PQ$ stands on level ground, with $Q$ its foot. A wire from the top $P$ to the point $A$ on the ground, $4$ m from $Q$ , makes an angle of $60^{\circ}$ with the ground. A second wire, from $P$ to the point $B$ on the other side, makes an angle of $42^{\circ}$ with the ground. Find (a) the height $PQ$ , (b) the distance $QB$ .

ShowHide solution

(a) In $\triangle APQ$ , from the $60^{\circ}$ angle at $A$ , $PQ$ is opposite and $AQ = 4$ m adjacent: $PQ = 4 \tan 60^{\circ} = 6.928\ldots = 6.93$ m (3 s.f.)

(b) In $\triangle BPQ$ , from the $42^{\circ}$ angle at $B$ , $PQ$ is opposite and $QB$ adjacent: $\tan 42^{\circ} = \dfrac{6.928\ldots}{QB}$

$QB = \dfrac{6.928\ldots}{\tan 42^{\circ}} = 7.694\ldots = 7.69$ m (3 s.f.). (With $PQ$ rounded to $6.93$ the answer comes out as $7.70$ m, which is wrong.)

(a) $PQ = 6.93$ m (b) $QB = 7.69$ m

Question 10

A straight zip line on Sentosa runs from a platform down to a landing point $15$ m lower. It makes an angle of $12^{\circ}$ with the horizontal. Find (a) the length of the zip line, (b) the horizontal distance it covers.

ShowHide solution

Let $V$ be the point level with the landing point $E$ and directly below the platform $T$ . Then $\angle V = 90^{\circ}$ , $VT = 15$ m is opposite the $12^{\circ}$ angle at $E$ , and the zip line $ET$ is the hypotenuse.

(a) $\sin 12^{\circ} = \dfrac{15}{l}$ , so $l = \dfrac{15}{\sin 12^{\circ}} = 72.146\ldots = 72.1$ (3 s.f.)

(b) $\tan 12^{\circ} = \dfrac{15}{d}$ , so $d = \dfrac{15}{\tan 12^{\circ}} = 70.569\ldots = 70.6$ (3 s.f.)

(a) $72.1$ m (b) $70.6$ m

Question 11

A rope is pulled straight from the top of a vertical flagpole $6$ m tall to a peg on the level ground. The rope makes an angle of $58^{\circ}$ with the ground. Find (a) the length of the rope, (b) the distance from the peg to the foot of the flagpole.

ShowHide solution

The pole is vertical, so $\angle F = 90^{\circ}$ . From the $58^{\circ}$ angle at $G$ , the pole $FT = 6$ m is opposite.

(a) $\sin 58^{\circ} = \dfrac{6}{r}$ , so $r = \dfrac{6}{\sin 58^{\circ}} = 7.075\ldots = 7.08$ (3 s.f.)

(b) $\tan 58^{\circ} = \dfrac{6}{d}$ , so $d = \dfrac{6}{\tan 58^{\circ}} = 3.749\ldots = 3.75$ (3 s.f.)

(a) $7.08$ m (b) $3.75$ m

Key Points

When the side you want is underneath in the ratio, multiply both sides by it, then divide by the sine, cosine or tangent.
hypotenuse $=$ opposite $\div \sin \theta$ ; hypotenuse $=$ adjacent $\div \cos \theta$ ; adjacent $=$ opposite $\div \tan \theta$ .
For the two shorter sides, the other angle, $90^{\circ} - \theta$ , puts the unknown on top; the hypotenuse is always found by dividing.
The hypotenuse must come out longer than each of the other sides.
One side and one acute angle give every side; in two triangles, carry a shared side over at its full value.