Pythagoras' theorem and trigonometry applications

What you will learn

apply Pythagoras’ theorem to solve problems in 2D and 3D,
use the trigonometric ratios (sin, cos, tan) in right-angled triangles,
solve problems involving angles of elevation and depression,
work with compass bearings and true bearings,
use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ on the Cartesian plane.

Worked example 0 Real-world example: how tall is the building?

A surveyor stands 40 m from the base of a building. The angle of elevation to the top is $32°$ . The surveyor’s eye level is 1.6 m above the ground. How tall is the building?

Draw a right-angled triangle: the horizontal distance is 40 m, the angle of elevation is $32°$ , and the opposite side is the height above eye level.
Use $\tan 32° = \dfrac{\text{opposite}}{40}$ , so opposite $= 40 \times \tan 32° \approx 40 \times 0.6249 = 24.996$ m.
Add the surveyor’s eye height: total height $\approx 25.0 + 1.6 = 26.6$ m.

Key idea: angles of elevation let us find heights we cannot measure directly, as long as we know the horizontal distance and can measure the angle.

1. Pythagoras’ theorem in 2D

In any right-angled triangle with hypotenuse $c$ and shorter sides $a$ and $b$ :

Pythagoras' theorem

c^2 = a^2 + b^2

To find a shorter side, rearrange: $a^2 = c^2 - b^2$ .

Worked example 1 Ladder against a wall

A 5 m ladder leans against a wall with its base 2 m from the wall. How high up the wall does the ladder reach?

The ladder is the hypotenuse: $c = 5$ .
The base distance is $b = 2$ .
$a^2 = 5^2 - 2^2 = 25 - 4 = 21$ .
$a = \sqrt{21} \approx 4.58$ m.

The ladder reaches approximately 4.58 m up the wall.

Worked example 2 Pythagoras in 3D — diagonal of a box

A rectangular box has dimensions 3 m $\times$ 4 m $\times$ 12 m. Find the length of the space diagonal (corner to opposite corner).

First find the diagonal of the base: $d_{\text{base}} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ m.
Now use the base diagonal and the height to find the space diagonal: $d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ m.

Shortcut: for a box with sides $l$ , $w$ , $h$ , the space diagonal is $d = \sqrt{l^2 + w^2 + h^2}$ .

Space diagonal of a rectangular box: apply Pythagoras twice.

2. Trigonometric ratios

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

Trigonometric ratios (SOH CAH TOA)

Sine

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

Cosine

\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

Tangent

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

Choose the ratio that uses the two sides you know (or the side you know and the side you want).

Worked example 3 Finding a side using trigonometry

In a right-angled triangle, the hypotenuse is 15 cm and one angle is $38°$ . Find the side opposite this angle.

We have the hypotenuse and want the opposite side, so use sin.
$\sin 38° = \dfrac{\text{opp}}{15}$ .
$\text{opp} = 15 \times \sin 38° \approx 15 \times 0.6157 = 9.24$ cm.

Worked example 4 Finding an angle

A ramp rises 1.2 m over a horizontal distance of 5 m. Find the angle of incline.

We have the opposite (1.2 m) and the adjacent (5 m), so use tan.
$\tan\theta = \dfrac{1.2}{5} = 0.24$ .
$\theta = \tan^{-1}(0.24) \approx 13.5°$ .

3. Angles of elevation, depression, and bearings

Worked example 5 Angle of depression

A lighthouse keeper 30 m above sea level spots a boat at an angle of depression of $12°$ . How far is the boat from the base of the lighthouse?

The angle of depression from the keeper equals the angle of elevation from the boat (alternate angles).
$\tan 12° = \dfrac{30}{d}$ .
$d = \dfrac{30}{\tan 12°} \approx \dfrac{30}{0.2126} \approx 141.1$ m.

Worked example 6 Bearings

A ship sails on a bearing of $065°$ for 12 km. How far east and how far north has it travelled?

The bearing $065°$ is $65°$ clockwise from north.
Northward (adjacent to the $65°$ angle): $12 \cos 65° \approx 12 \times 0.4226 = 5.07$ km.
Eastward (opposite the $65°$ angle): $12 \sin 65° \approx 12 \times 0.9063 = 10.88$ km.

4. The distance formula

Pythagoras’ theorem on the Cartesian plane gives the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ :

Distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Worked example 7 Distance between two points

Find the distance between $A(3, -1)$ and $B(-2, 7)$ .

$d = \sqrt{(-2 - 3)^2 + (7 - (-1))^2} = \sqrt{(-5)^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89}$ .
$d \approx 9.43$ units.

Worked example 8 Is a triangle right-angled?

Points $P(1, 2)$ , $Q(4, 6)$ , and $R(8, 3)$ form a triangle. Determine whether it is right-angled.

$PQ = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = 5$ .
$QR = \sqrt{(8-4)^2 + (3-6)^2} = \sqrt{16 + 9} = 5$ .
$PR = \sqrt{(8-1)^2 + (3-2)^2} = \sqrt{49 + 1} = \sqrt{50}$ .
Check: $PQ^2 + QR^2 = 25 + 25 = 50 = PR^2$ . Yes, the triangle is right-angled at $Q$ .

Practice

Fluency

Tier 1: basic skills

A right-angled triangle has sides 6 cm and 8 cm. Find the hypotenuse.
The hypotenuse of a right-angled triangle is 13 m and one side is 5 m. Find the other side.
A rectangular room is 4 m by 3 m. Find the length of the diagonal of the floor.
Find $\sin 40°$ , $\cos 40°$ , and $\tan 40°$ using a calculator (2 d.p.).
In a right-angled triangle with hypotenuse 20 cm and an angle of $50°$ , find the opposite side.
Find the angle whose tangent is $\dfrac{3}{4}$ .
Find the distance between the points $(1, 3)$ and $(4, 7)$ .
Find the distance between $(-2, 5)$ and $(3, -7)$ .
A 6 m ladder reaches 5.5 m up a wall. Find the angle the ladder makes with the ground.
A box has dimensions 2 m $\times$ 3 m $\times$ 6 m. Find the space diagonal.

Reasoning

Tier 2: mixed practice

A kite is flying at a height of 40 m. The string is 65 m long. What angle does the string make with the ground?
From the top of a 50 m cliff, the angle of depression to a boat is $18°$ . How far is the boat from the base of the cliff?
A hiker walks 8 km on a bearing of $040°$ and then 6 km due east. How far north and how far east is the hiker from the starting point?
Points $A(0, 0)$ , $B(6, 0)$ , and $C(3, 5)$ form a triangle. Find the perimeter.
A tent pole 2.4 m tall is supported by a rope pegged 1.8 m from the base. Find the angle the rope makes with the ground.
A rectangular prism has a base of 5 cm $\times$ 12 cm and a space diagonal of 15 cm. Find the height of the prism.
A ship sails 20 km on a bearing of $120°$ . How far south and how far east is it from its starting point?
Two buildings are 30 m apart. From the top of the shorter building (20 m tall), the angle of elevation to the top of the taller building is $35°$ . Find the height of the taller building.

Reasoning

Tier 3: explain and apply

Explain why the distance formula is a direct application of Pythagoras’ theorem. Use a diagram to support your answer.
A helicopter is directly above a point on the ground. Observer A is 500 m due north of that point and measures the angle of elevation as $28°$ . Observer B is 700 m due east and measures the angle of elevation as $20°$ . Are both observers looking at the same helicopter height? Show your working.
A surveyor needs to find the width of a river. She stands at point $A$ on one bank and sights a tree at point $B$ directly across the river. She then walks 50 m along the bank to point $C$ and measures $\angle ACB = 62°$ . Find the width of the river.
Prove that for any three points forming a triangle on the coordinate plane, the triangle inequality holds: the sum of any two side lengths exceeds the third.
A rescue helicopter is at coordinates $(3, 7)$ and must reach a boat at $(-5, 1)$ . If each grid unit represents 2 km, find the actual distance the helicopter must fly.

Challenge

Reasoning

Harder reasoning

A cone has a slant height of 13 cm and a base radius of 5 cm. Find the height of the cone and then calculate the angle between the slant surface and the base.
Two lighthouses are 10 km apart on a straight coastline running east—west. A ship at sea measures the bearing to the western lighthouse as $320°$ and the bearing to the eastern lighthouse as $050°$ . Find the ship’s perpendicular distance from the coastline.
A cube has side length $s$ . Show that the space diagonal has length $s\sqrt{3}$ and find the angle the space diagonal makes with the base of the cube.
Three mobile phone towers are at positions $A(0, 0)$ , $B(8, 0)$ , and $C(3, 6)$ on a coordinate grid (units in km). A phone receives signal from a tower only if it is within 7 km. Determine whether a phone at position $P(5, 4)$ can receive signal from all three towers.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
$a = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ m.
$d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m.
$\sin 40° \approx 0.64$ , $\cos 40° \approx 0.77$ , $\tan 40° \approx 0.84$ .
$\text{opp} = 20 \sin 50° \approx 20 \times 0.7660 = 15.32$ cm.
$\theta = \tan^{-1}\!\left(\dfrac{3}{4}\right) = \tan^{-1}(0.75) \approx 36.87° \approx 36.9°$ .
$d = \sqrt{(4-1)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.
$d = \sqrt{(3-(-2))^2 + (-7-5)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.
$\sin\theta = \dfrac{5.5}{6}$ , so $\theta = \sin^{-1}(0.9167) \approx 66.4°$ .
$d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ m.

Tier 2

$\sin\theta = \dfrac{40}{65} = \dfrac{8}{13}$ , so $\theta = \sin^{-1}(0.6154) \approx 38.0°$ .
$\tan 18° = \dfrac{50}{d}$ , so $d = \dfrac{50}{\tan 18°} \approx \dfrac{50}{0.3249} \approx 153.9$ m.
North: $8 \cos 40° \approx 8 \times 0.7660 = 6.13$ km. East from first leg: $8 \sin 40° \approx 8 \times 0.6428 = 5.14$ km. Total east: $5.14 + 6 = 11.14$ km. Total north: $6.13$ km.
$AB = 6$ , $BC = \sqrt{(6-3)^2 + (0-5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ , $AC = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ . Perimeter $\approx 6 + 5.83 + 5.83 = 17.66$ units.
$\tan\theta = \dfrac{2.4}{1.8} = \dfrac{4}{3}$ , so $\theta = \tan^{-1}(1.333) \approx 53.1°$ .
Space diagonal: $d^2 = l^2 + w^2 + h^2$ . So $15^2 = 5^2 + 12^2 + h^2$ , giving $225 = 25 + 144 + h^2$ , thus $h^2 = 56$ and $h = \sqrt{56} \approx 7.48$ cm.
South: $20 \cos 60° = 20 \times 0.5 = 10$ km (bearing $120°$ is $60°$ from south). East: $20 \sin 60° = 20 \times 0.8660 \approx 17.32$ km.
Let extra height above the shorter building be $x$ . $\tan 35° = \dfrac{x}{30}$ , so $x = 30 \tan 35° \approx 30 \times 0.7002 = 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

Tier 3

On the Cartesian plane, the horizontal distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $|x_2 - x_1|$ and the vertical distance is $|y_2 - y_1|$ . These form the two shorter sides of a right-angled triangle. Applying Pythagoras’ theorem: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ , giving $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
Observer A: height $= 500 \tan 28° \approx 500 \times 0.5317 = 265.9$ m. Observer B: height $= 700 \tan 20° \approx 700 \times 0.3640 = 254.8$ m. The heights differ ( $265.9 \neq 254.8$ ), so either the measurements are imprecise or the helicopter is not directly above the assumed point.
$\tan 62° = \dfrac{w}{50}$ , where $w$ is the river width. $w = 50 \tan 62° \approx 50 \times 1.8807 = 94.0$ m.
This is the standard triangle inequality. For any triangle with vertices on the coordinate plane, the shortest path between two points is the straight line (the side). Going via a third point is longer, so $AB + BC > AC$ (and cyclic permutations). A rigorous proof uses the Cauchy—Schwarz inequality or the properties of the Euclidean metric.
$d = \sqrt{(-5-3)^2 + (1-7)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ grid units. Actual distance $= 10 \times 2 = 20$ km.

Challenge

Height: $h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. Angle between slant and base: $\cos\alpha = \dfrac{5}{13}$ , so $\alpha = \cos^{-1}(0.3846) \approx 67.4°$ .
The western lighthouse is at $W$ and the eastern at $E$ , 10 km apart. Bearing $320°$ to $W$ means the angle from north is $320°$ , so the line from the ship to $W$ makes $40°$ west of north. Bearing $050°$ to $E$ means $50°$ east of north. The ship is south of the coastline. At the ship, the angle $\angle WSE = 360° - 320° + 50° = 90°$ . Using the right triangle: let perpendicular distance $= h$ . $\tan 40° = \dfrac{d_W}{h}$ and $\tan 50° = \dfrac{d_E}{h}$ where $d_W + d_E = 10$ . So $h(\tan 40° + \tan 50°) = 10$ , giving $h = \dfrac{10}{\tan 40° + \tan 50°} = \dfrac{10}{0.8391 + 1.1918} \approx \dfrac{10}{2.0309} \approx 4.92$ km.
Space diagonal: $d = \sqrt{s^2 + s^2 + s^2} = \sqrt{3s^2} = s\sqrt{3}$ . The base diagonal is $s\sqrt{2}$ . The angle $\alpha$ between the space diagonal and the base satisfies $\tan\alpha = \dfrac{s}{s\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ , so $\alpha = \tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \approx 35.3°$ .
$PA = \sqrt{5^2 + 4^2} = \sqrt{41} \approx 6.40$ km (within 7 km — yes). $PB = \sqrt{(5-8)^2 + 4^2} = \sqrt{9 + 16} = 5$ km (within 7 km — yes). $PC = \sqrt{(5-3)^2 + (4-6)^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83$ km (within 7 km — yes). The phone can receive signal from all three towers.

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