Relations between sides and angles in a right-angled triangle, with applications

Minh stands $15$ m from the foot of the school flagpole. He looks up at the top of the pole along a line that makes an angle of $35^\circ$ with the horizontal, and his eyes are $1.5$ m above the ground. Without climbing up, how can we find the height of the flagpole in metres?

In the right-angled triangle formed by Minh's eye, the top of the pole and the point on the pole level with his eye, we know one side ( $15$ m) and one angle ( $35^\circ$ ). This lesson shows how to find the other sides.

Relations between sides and angles

Take triangle $ABC$ with a right angle at $A$ , and write $BC=a$ , $CA=b$ , $AB=c$ . The two sides that form the right angle, $b$ and $c$ , are called the legs.

The right-angled triangle

ABC

with sides

a

b

c

Explore

From the definition, $\sin B=\dfrac{b}{a}$ . Multiplying both sides by $a$ gives $b=a\sin B$ . For example, if $a=10$ and $\angle B=30^\circ$ , then $b=10\cdot\dfrac12=5$ . Doing the same with the other ratios gives four relations.

Why? Each equation is just a definition rewritten. For example, $\cos C=\dfrac{b}{a}$ (the side adjacent to $C$ is $b$ ), so $b=a\cos C$ ; and $\tan B=\dfrac bc$ , so $b=c\tan B$ .

When we know a leg, we use the tangent or the cotangent.

Solving a right-angled triangle

Solving a right-angled triangle means finding all the remaining sides and angles from two known facts, at least one of which is a side. We use three tools: Pythagoras' theorem, $\angle B+\angle C=90^\circ$ , and the relations above.

Applications: measuring heights and distances

When we look up at an object, the angle between the line of sight and the horizontal is called the angle of elevation. When we look down, it is called the angle of depression. We draw a right-angled triangle with one horizontal side, then use the relations.

Measuring the height of the flagpole

Example The height of the flagpole

Answer the question from the start (rounded to one decimal place).

Solution. Understand: $M$ is Minh's eye, $K$ is the point on the pole level with his eye, and $D$ is the top of the pole; $MK=15$ m, $\angle DMK=35^\circ$ , $KF=1.5$ m. Plan: in triangle $MKD$ with a right angle at $K$ , $DK$ is opposite angle $M$ and $MK$ is adjacent to it, so we use the tangent. The height of the pole is $DK$ plus the height $KF$ from Minh's eye to the ground. Solve: In triangle $MKD$ with a right angle at $K$ , we have

$DK=MK\tan35^\circ=15\tan35^\circ\approx10.50;\qquad DF=DK+KF\approx10.50+1.5=12.0.$

So the flagpole is about $12.0$ m tall.

Watch out Common mistakes

Choosing the wrong ratio. Wrong: $AC=BC\cos B$ . Why it is wrong: $AC$ is opposite angle $B$ , so it goes with sine. Right: $AC=BC\sin B$ (or $BC\cos C$ ).
Dividing instead of multiplying. Wrong: $AC=\dfrac{AB}{\tan B}$ . Why it is wrong: $\tan B=\dfrac{AC}{AB}$ , so $AC=AB\tan B$ . Right: divide only when the side you want is in the denominator, for example $BC=\dfrac{AC}{\sin B}$ .
Rounding too early. Wrong: in the kite problem, taking $\sin38^\circ\approx0.6$ and working out $50\cdot0.6=30$ (m). Why it is wrong: $50\sin38^\circ\approx30.78$ , almost $0.8$ m more, and the error grows over several steps. Right: keep the full value on the calculator and round only the final answer.

Key terms

English	Tiếng Việt
leg of a right-angled triangle	cạnh góc vuông
hypotenuse	cạnh huyền
opposite angle	góc đối
adjacent angle	góc kề
solving a right-angled triangle	giải tam giác vuông
angle of elevation	góc nâng
angle of depression	góc hạ
horizontal	phương nằm ngang

Summary

Triangle $ABC$ with a right angle at $A$ : $b=a\sin B=a\cos C$ , $c=a\sin C=a\cos B$ .
$b=c\tan B=c\cot C$ , $c=b\tan C=b\cot B$ .
Solving a right-angled triangle: Pythagoras' theorem, $\angle B+\angle C=90^\circ$ and the relations; at least one side must be known.
Real-life problems: draw a picture, find the right-angled triangle, choose the ratio, round at the last step.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Triangle $ABC$ has a right angle at $A$ , $BC=8$ cm and $\angle C=30^\circ$ . Find $AB$ and $AC$ .

What you need and how to start

What you need: $c=a\sin C$ , $b=a\cos C$ ; $\sin30^\circ=\dfrac12$ , $\cos30^\circ=\dfrac{\sqrt3}2$ .

Steps:

$AB$ is opposite angle $C$ .
$AC$ is adjacent to angle $C$ .

Show solution

In the right-angled triangle $ABC$ , $AB=BC\sin C=8\sin30^\circ=4$ (cm) and $AC=BC\cos C=8\cos30^\circ=4\sqrt3\approx6.93$ (cm).

Answer. $AB=4$ cm, $AC=4\sqrt3\approx6.93$ cm.

Exercise 2

Triangle $ABC$ has a right angle at $A$ , $AC=10$ cm and $\angle C=42^\circ$ . Find $AB$ and $BC$ (rounded to two decimal places).

What you need and how to start

What you need: $c=b\tan C$ ; $b=a\cos C\Rightarrow a=\dfrac{b}{\cos C}$ .

Steps:

$AB=AC\tan C$ .
$BC=\dfrac{AC}{\cos C}$ .

Show solution

In the right-angled triangle $ABC$ , $AB=AC\tan C=10\tan42^\circ\approx9.00$ (cm). Since $AC=BC\cos C$ , $BC=\dfrac{AC}{\cos C}=\dfrac{10}{\cos42^\circ}\approx13.46$ (cm).

Answer. $AB\approx9.00$ cm, $BC\approx13.46$ cm.

Exercise 3

Solve triangle $ABC$ with a right angle at $A$ , given $AB=9$ cm and $BC=15$ cm (angles to the nearest degree).

What you need and how to start

What you need: Pythagoras' theorem; $\sin C=\dfrac{AB}{BC}$ .

Steps:

Find $AC$ .
Find angle $C$ , then angle $B=90^\circ-\angle C$ .

Show solution

In the right-angled triangle $ABC$ , $AC=\sqrt{BC^2-AB^2}=\sqrt{225-81}=12$ (cm) (Pythagoras' theorem). We have $\sin C=\dfrac{AB}{BC}=\dfrac9{15}=0.6$ , so $\angle C\approx37^\circ$ . Hence $\angle B\approx90^\circ-37^\circ=53^\circ$ .

Answer. $AC=12$ cm, $\angle C\approx37^\circ$ , $\angle B\approx53^\circ$ .

Exercise 4

Solve triangle $ABC$ with a right angle at $A$ , given $AC=4$ cm and $\angle B=50^\circ$ (sides to two decimal places).

What you need and how to start

What you need: $c=b\cot B$ ; $b=a\sin B$ .

Steps:

$\angle C=90^\circ-50^\circ$ .
$AB=AC\cot B=\dfrac{AC}{\tan B}$ ; $BC=\dfrac{AC}{\sin B}$ .

Show solution

In the right-angled triangle $ABC$ , $\angle C=90^\circ-50^\circ=40^\circ$ . Since $AC=AB\tan B$ , $AB=\dfrac{AC}{\tan B}=\dfrac{4}{\tan50^\circ}\approx3.36$ (cm). Since $AC=BC\sin B$ , $BC=\dfrac{AC}{\sin B}=\dfrac{4}{\sin50^\circ}\approx5.22$ (cm).

Answer. $\angle C=40^\circ$ , $AB\approx3.36$ cm, $BC\approx5.22$ cm.

Exercise 5

Triangle $ABC$ has a right angle at $A$ , $AB=2\sqrt3$ cm and $\angle B=60^\circ$ . Find the exact values of $AC$ and $BC$ .

What you need and how to start

What you need: $\tan60^\circ=\sqrt3$ , $\cos60^\circ=\dfrac12$ (see trigonometric ratios of acute angles).

Steps:

$AC=AB\tan B$ .
$BC=\dfrac{AB}{\cos B}$ .

Show solution

In the right-angled triangle $ABC$ , $AC=AB\tan B=2\sqrt3\cdot\sqrt3=6$ (cm). Since $AB=BC\cos B$ , $BC=\dfrac{AB}{\cos B}=\dfrac{2\sqrt3}{1/2}=4\sqrt3$ (cm).

Answer. $AC=6$ cm, $BC=4\sqrt3$ cm.

Exercise 6

A $6$ m ladder leans against a wall and makes an angle of $70^\circ$ with the ground. How far is its foot from the wall, and how high is its top (rounded to two decimal places)?

What you need and how to start

What you need: the ladder is the hypotenuse; the distance to the wall is adjacent to the $70^\circ$ angle.

Steps:

Distance $=6\cos70^\circ$ .
Height $=6\sin70^\circ$ .

Show solution

The ladder, the wall and the ground form a right-angled triangle whose hypotenuse is the $6$ m ladder. The distance from the foot of the ladder to the wall is $6\cos70^\circ\approx2.05$ (m). The height of the top of the ladder is $6\sin70^\circ\approx5.64$ (m).

Answer. About $2.05$ m and $5.64$ m.

Exercise 7

From the top of a lighthouse $40$ m above the sea, the keeper sees a boat at an angle of depression of $12^\circ$ . How far is the boat from the foot of the lighthouse (to the nearest metre)?

What you need and how to start

What you need: the angle of depression equals the angle between the line of sight and the sea (alternate angles).

Steps:

In the right-angled triangle, the $40$ m side is opposite the $12^\circ$ angle at the boat.
Distance $=\dfrac{40}{\tan12^\circ}$ .

Show solution

Let $d$ (m) be the distance from the boat to the foot of the lighthouse. The angle between the line of sight and the sea is $12^\circ$ (alternate angles). In the right-angled triangle formed by the lighthouse, the sea and the line of sight, $40=d\tan12^\circ$ , so $d=\dfrac{40}{\tan12^\circ}\approx188$ (m).

Answer. About $188$ m.

Exercise 8

When the sun's rays make an angle of $52^\circ$ with the ground, a tree casts a shadow $9$ m long on flat ground. Find the height of the tree (rounded to one decimal place).

What you need and how to start

What you need: the height is opposite the $52^\circ$ angle and the shadow is adjacent to it.

Steps:

Draw the right-angled triangle: tree, shadow, sun's ray.
Height $=9\tan52^\circ$ .

Show solution

The tree, its shadow and the sun's ray form a right-angled triangle. The height of the tree is $9\tan52^\circ\approx11.5$ (m).

Answer. About $11.5$ m.

Exercise 9

Hà is flying a kite. The string is straight and $50$ m long, and makes an angle of $38^\circ$ with the horizontal; Hà holds the string $1.2$ m above the ground.

a) How high is the kite above the ground (rounded to one decimal place)?

b) What is the horizontal distance from Hà's hand to the point directly below the kite (rounded to one decimal place)?

c) Hà lets out more string until it is $60$ m long, keeping the same angle. How many metres higher does the kite go (rounded to one decimal place)?

What you need and how to start

What you need: the string is the hypotenuse; the height is opposite the $38^\circ$ angle.

Steps:

Height above the hand: $50\sin38^\circ$ ; add $1.2$ .
Horizontal distance: $50\cos38^\circ$ .
The rise: $(60-50)\sin38^\circ$ .

Show solution

a) The height of the kite above Hà's hand is $50\sin38^\circ\approx30.78$ (m). Hence its height above the ground is about $30.78+1.2\approx32.0$ (m).

b) The horizontal distance is $50\cos38^\circ\approx39.4$ (m).

c) The kite rises by $60\sin38^\circ-50\sin38^\circ=10\sin38^\circ\approx6.2$ (m).

Answer. a) $32.0$ m; b) $39.4$ m; c) $6.2$ m.

Exercise 10 Challenge

Two friends stand on the same straight line through the foot of a building, on the same side, $20$ m apart. The nearer one sees the top of the building at an angle of elevation of $45^\circ$ , the farther one at $30^\circ$ . Find the height of the building (ignore eye height; give an exact form, then round to two decimal places).

What you need and how to start

What you need: the distance to the foot of the building is $h\cot\alpha$ ; rationalising the denominator.

Steps:

Let $h$ be the height. The two distances are $h\cot45^\circ=h$ and $h\cot30^\circ=h\sqrt3$ .
Form the equation $h\sqrt3-h=20$ and solve it.

Show solution

Let $h$ (m) be the height of the building ( $h>0$ ). The distances from the two friends to the foot of the building are $h\cot45^\circ=h$ and $h\cot30^\circ=h\sqrt3$ . From the question, $h\sqrt3-h=20$ . Hence $h=\dfrac{20}{\sqrt3-1}=\dfrac{20\left(\sqrt3+1\right)}{2}=10\left(\sqrt3+1\right)\approx27.32$ (m) (which satisfies the condition).

Answer. $10\left(\sqrt3+1\right)\approx27.32$ m.

a) $BC=12$ cm.	True	False
b) $AC=6\sqrt3$ cm.	True	False
c) $\angle B=30^\circ$ .	True	False
d) $AC=BC\sin C$ .	True	False

Relations between sides and angles

Explore

Solving a right-angled triangle

Applications: measuring heights and distances

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer