Pythagoras' theorem and trigonometry in 3D

What you will learn

apply Pythagoras’ theorem to find space diagonals and other lengths in 3D objects,
use trigonometry to solve problems involving angles of elevation and depression,
solve navigation problems using bearings,
analyse how measurement errors propagate through calculations.

Worked example 0 Real-world example: antenna cable length

A communications antenna sits on a flat roof. The mounting point is $3$ m east and $4$ m north of a cable anchor point at roof level. The top of the antenna is $12$ m above the roof. What length of cable is needed from the anchor to the top of the antenna (add $5\%$ for slack)?

Horizontal distance across the roof: $d_h = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ m.
Cable length (anchor to antenna top): $L = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ m.
With $5\%$ slack: $13 \times 1.05 = 13.65$ m.

Key idea: in 3D, apply Pythagoras twice — once on the horizontal plane, once combining horizontal distance with vertical height.

Space diagonal of a rectangular box: apply Pythagoras twice to go from corner A to opposite corner B.

1. Pythagoras’ theorem in 3D

For a rectangular box with dimensions $l$ , $w$ , and $h$ :

Space diagonal

d = \sqrt{l^2 + w^2 + h^2}

This result comes from applying Pythagoras twice:

Find the base diagonal: $d_{\text{base}} = \sqrt{l^2 + w^2}$ .
Combine with height: $d = \sqrt{d_{\text{base}}^2 + h^2} = \sqrt{l^2 + w^2 + h^2}$ .

Worked example 1 Space diagonal of a room

A room measures $5$ m by $4$ m by $3$ m. Find the longest straight line that fits inside the room (the space diagonal).

$d = \sqrt{5^2 + 4^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} \approx 7.07$ m.

Worked example 2 Height of a pyramid

A square-based pyramid has a base of side $8$ cm and a slant edge (from base corner to apex) of $10$ cm. Find the height of the pyramid.

The base diagonal $= 8\sqrt{2}$ cm. Half the base diagonal $= 4\sqrt{2} \approx 5.657$ cm (distance from centre of base to a corner).
The slant edge, the half-diagonal, and the height form a right-angled triangle.
$h = \sqrt{10^2 - (4\sqrt{2})^2} = \sqrt{100 - 32} = \sqrt{68} \approx 8.25$ cm.

2. Angles of elevation and depression

An angle of elevation is measured upward from a horizontal line. An angle of depression is measured downward from a horizontal line. By alternate angles, the angle of elevation from point A to point B equals the angle of depression from B to A.

Worked example 3 Angle of elevation to a hilltop tower

A tower $15$ m tall stands on a hill. From a point $200$ m horizontally from the base of the hill, the angle of elevation to the top of the tower is $18°$ . Find the height of the hill.

Let $H$ be the hill height. Total height above the observer’s level $= H + 15$ .
$\tan 18° = \dfrac{H + 15}{200}$ , so $H + 15 = 200\tan 18° \approx 200 \times 0.3249 = 64.98$ m.
Hill height: $H \approx 64.98 - 15 = 49.98 \approx 50.0$ m.

Worked example 4 Angle of depression from a drone

A drone hovers $120$ m above the ground. Its camera spots a target on the ground at an angle of depression of $35°$ . Find the horizontal distance from the drone to the target.

$\tan 35° = \dfrac{120}{d}$ .
$d = \dfrac{120}{\tan 35°} \approx \dfrac{120}{0.7002} \approx 171.4$ m.

A true bearing is measured clockwise from north ( $000°$ to $360°$ ). A compass bearing uses directions like N $30°$ E.

Worked example 5 Two-stage journey

A ship sails $20$ km on a bearing of $060°$ , then $15$ km on a bearing of $150°$ . Find the ship’s distance and bearing from its starting point.

First leg: north component $= 20\cos 60° = 10$ km, east component $= 20\sin 60° \approx 17.32$ km.
Second leg: bearing $150°$ is S $30°$ E. North component $= -15\cos 30° \approx -12.99$ km (southward), east component $= 15\sin 30° = 7.5$ km.
Total: north $= 10 - 12.99 = -2.99$ km, east $= 17.32 + 7.5 = 24.82$ km.
Distance $= \sqrt{(-2.99)^2 + 24.82^2} = \sqrt{8.94 + 616.03} \approx 25.0$ km.
Angle from north: $\theta = \tan^{-1}\!\left(\dfrac{24.82}{2.99}\right) \approx 83.1°$ . Since the position is south-east, bearing $\approx 180° - 83.1° = 96.9°$ (i.e. almost due east, slightly south).

4. Measurement errors and their impact

Every measurement has some uncertainty. When measurements feed into calculations, the errors propagate.

Percentage error

\text{Percentage error} = \frac{|\text{measured} - \text{true}|}{\text{true}} \times 100\%

Worked example 6 Error in a cylinder volume

A cylinder has measured radius $r = 5.0 \pm 0.1$ cm and height $h = 12.0 \pm 0.2$ cm. Estimate the percentage error in the calculated volume.

Percentage error in $r$ : $\dfrac{0.1}{5.0} \times 100 = 2\%$ .
Since $V = \pi r^2 h$ , the radius is squared, so its error contribution doubles: $2 \times 2\% = 4\%$ .
Percentage error in $h$ : $\dfrac{0.2}{12.0} \times 100 \approx 1.67\%$ .
Approximate total percentage error in $V$ : $4\% + 1.67\% \approx 5.67\%$ .
Nominal $V = \pi(25)(12) = 300\pi \approx 942.48$ cm $^3$ . The error is roughly $\pm 53$ cm $^3$ .

Key formulas

Space diagonal

d = \sqrt{l^2 + w^2 + h^2}

Trigonometric ratios

\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

Percentage error

\text{Percentage error} = \frac{|\text{measured} - \text{true}|}{\text{true}} \times 100\%

Practice

Fluency

Tier 1: basic skills

Find the space diagonal of a box $6 \times 8 \times 10$ cm.
A cube has side $5$ cm. Find its space diagonal.
From a point on level ground, the angle of elevation to the top of a $25$ m tree is $40°$ . Find the horizontal distance to the tree.
A drone at height $80$ m observes a point on the ground at an angle of depression of $50°$ . Find the horizontal distance.
A ship sails $10$ km on a bearing of $045°$ . How far north and how far east has it travelled?
A measurement of $8.0 \pm 0.2$ cm is squared. Find the percentage error in the original measurement and the approximate percentage error in the squared value.
Find the length of the longest rod that fits inside a rectangular box $12 \times 5 \times 4$ cm.
A pyramid has a square base of side $10$ cm and a vertical height of $12$ cm. Find the slant edge length (from a base corner to the apex).

Reasoning

Tier 2: mixed practice

A rectangular room is $6 \times 4 \times 3$ m. A spider at one top corner wants to reach the opposite bottom corner by walking along walls. Find the shortest path. (Hint: unfold two walls into a flat net.)
A hiker walks $5$ km on bearing $030°$ then $8$ km on bearing $120°$ . Find the distance and bearing from start to finish.
From the top of a $60$ m cliff, the angles of depression to two boats in a line directly out to sea are $45°$ and $30°$ . Find the distance between the boats.
A surveyor measures the angle of elevation to a mountain peak as $22°$ from point A and $15°$ from point B, which is $500$ m further away on level ground in a direct line from the peak. Find the height of the mountain.
A cylinder has $r = 10.0 \pm 0.3$ cm and $h = 20.0 \pm 0.5$ cm. Calculate the volume and estimate the maximum percentage error.
A tent pole $2.5$ m tall is supported by guy ropes pegged $1.5$ m from the base. Find (a) the length of each rope, and (b) the angle each rope makes with the ground.

Reasoning

Tier 3: explain and apply

Explain why the space diagonal formula $d = \sqrt{l^2 + w^2 + h^2}$ works by describing the two right-angled triangles used in its derivation.
A pilot flies from airport A on bearing $070°$ for $200$ km to point B, then on bearing $160°$ for $150$ km to airport C. Find the bearing and distance for the direct return flight from C to A.
A building casts a shadow $30$ m long when the sun’s angle of elevation is $55°$ . An hour later the shadow is $45$ m long. Find the new angle of elevation and determine whether the sun rose or fell during this hour.
Discuss why percentage error in a calculated volume based on measured radius is approximately double the percentage error of the radius itself, using the formula $V = \frac{4}{3}\pi r^3$ as an example. (What multiplier applies here instead of double?)

Challenge

Reasoning

Harder reasoning

A right square pyramid has base side $12$ cm and slant height $10$ cm. Find (a) the vertical height, (b) the angle between a slant face and the base, and (c) the angle between a slant edge and the base.
Two observers A and B are $300$ m apart on level ground. Both observe the same drone. Observer A measures the angle of elevation as $40°$ and the bearing to the drone as $060°$ . Observer B (due east of A) measures the angle of elevation as $55°$ . Find the height of the drone.
A sphere of radius $r$ fits exactly inside a cube. Show that the ratio of the space diagonal of the cube to the diameter of the sphere is $\sqrt{3}:1$ .
A surveyor measures two sides of a triangle as $a = 50.0 \pm 0.5$ m and $b = 80.0 \pm 0.5$ m, with the included angle $C = 60° \pm 1°$ . Using the area formula $A = \frac{1}{2}ab\sin C$ , estimate the area and discuss how the angle error and the length errors each contribute to the total error in the area.

Answers

Answer key

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

Show the full answer key

Tier 1

$d = \sqrt{36 + 64 + 100} = \sqrt{200} \approx 14.14$ cm.
$d = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm.
$\tan 40° = \frac{25}{d}$ , so $d = \frac{25}{\tan 40°} \approx \frac{25}{0.8391} \approx 29.79$ m.
$\tan 50° = \frac{80}{d}$ , so $d = \frac{80}{\tan 50°} \approx \frac{80}{1.1918} \approx 67.13$ m.
North $= 10\cos 45° \approx 7.07$ km; East $= 10\sin 45° \approx 7.07$ km.
Percentage error $= \frac{0.2}{8.0} \times 100 = 2.5\%$ . Squared value error $\approx 2 \times 2.5\% = 5\%$ .
$d = \sqrt{144 + 25 + 16} = \sqrt{185} \approx 13.60$ cm.
Half-diagonal of base $= \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.071$ cm. Slant edge $= \sqrt{12^2 + (5\sqrt{2})^2} = \sqrt{144 + 50} = \sqrt{194} \approx 13.93$ cm.

Tier 2

Unfold the $6 \times 3$ wall and the $4 \times 3$ wall into a single rectangle $10 \times 3$ . The spider walks diagonally: $\sqrt{10^2 + 3^2} = \sqrt{109} \approx 10.44$ m. (Other unfoldings give longer paths; the shortest is $\sqrt{(6+3)^2 + 4^2} = \sqrt{97} \approx 9.85$ m by unfolding the $6 \times 3$ wall with the floor. Check all configurations; the minimum is approximately $9.85$ m.)
Leg 1: N $= 5\cos 30° \approx 4.33$ km, E $= 5\sin 30° = 2.5$ km. Leg 2: N $= -8\cos 60° = -4$ km (south), E $= 8\sin 60° \approx 6.93$ km. Total: N $\approx 0.33$ km, E $\approx 9.43$ km. Distance $= \sqrt{0.33^2 + 9.43^2} \approx 9.44$ km. Bearing $= \tan^{-1}(9.43/0.33) \approx 88°$ .
Boat 1: $d_1 = \frac{60}{\tan 45°} = 60$ m from cliff base. Boat 2: $d_2 = \frac{60}{\tan 30°} \approx 103.92$ m. Distance between boats $\approx 103.92 - 60 = 43.92$ m.
Let height $= h$ and distance from A to the base $= x$ . From A: $\tan 22° = h/x$ . From B: $\tan 15° = h/(x + 500)$ . So $x = h/\tan 22°$ and $x + 500 = h/\tan 15°$ . Subtracting: $500 = h(1/\tan 15° - 1/\tan 22°) = h(3.732 - 2.475) = 1.257h$ . So $h \approx 397.8$ m.
$V = \pi(100)(20) = 2000\pi \approx 6283.19$ cm $^3$ . Error in $r$ : $\frac{0.3}{10} = 3\%$ , doubled for $r^2$ : $6\%$ . Error in $h$ : $\frac{0.5}{20} = 2.5\%$ . Total $\approx 8.5\%$ .
(a) Rope $= \sqrt{2.5^2 + 1.5^2} = \sqrt{8.5} \approx 2.92$ m. (b) $\tan\theta = \frac{2.5}{1.5}$ , so $\theta = \tan^{-1}(1.667) \approx 59.0°$ .

Tier 3

First, in the base rectangle, we form a right triangle with legs $l$ and $w$ to get the base diagonal $d_b = \sqrt{l^2 + w^2}$ . Second, this base diagonal becomes one leg of a new right triangle whose other leg is the height $h$ and whose hypotenuse is the space diagonal. Applying Pythagoras again: $d = \sqrt{d_b^2 + h^2} = \sqrt{l^2 + w^2 + h^2}$ .
Leg 1 (A to B): N $= 200\cos 70° \approx 68.40$ km, E $= 200\sin 70° \approx 187.94$ km. Leg 2 (B to C): N $= -150\cos 20° \approx -140.95$ km, E $= 150\sin 20° \approx 51.30$ km. C relative to A: N $\approx -72.55$ km, E $\approx 239.24$ km. Return C to A: N $\approx 72.55$ , W $\approx 239.24$ . Distance $= \sqrt{72.55^2 + 239.24^2} \approx 250.0$ km. Bearing from C to A: $\theta = \tan^{-1}(239.24/72.55) \approx 73.1°$ west of north $= 360° - 73.1° = 286.9°$ .
Building height $h$ . When shadow $= 30$ : $\tan 55° = h/30$ , so $h = 30\tan 55° \approx 42.84$ m. When shadow $= 45$ : $\tan\alpha = 42.84/45$ , so $\alpha = \tan^{-1}(0.952) \approx 43.6°$ . The angle decreased from $55°$ to $43.6°$ , so the sun fell (moved lower in the sky).
For $V = \frac{4}{3}\pi r^3$ , the radius is cubed, so the percentage error in $V$ is approximately $3$ times the percentage error in $r$ (not double). For example, a $1\%$ error in $r$ gives roughly a $3\%$ error in volume. The multiplier equals the exponent of the variable in the formula.

Challenge

Base half-diagonal $= \frac{12\sqrt{2}}{2} = 6\sqrt{2}$ . Slant height $= 10$ (distance from midpoint of base edge to apex). Half base edge $= 6$ . Vertical height from base edge midpoint: $h_{\text{face}} = \sqrt{10^2 - 6^2} = 8$ cm. (a) Vertical height of pyramid: the midpoint of a base edge is $6$ cm from the centre (for a $12 \times 12$ base). So $H = \sqrt{10^2 - 6^2} = 8$ cm. (Alternatively, using the half-diagonal: $H = \sqrt{10^2 - (6\sqrt{2})^2}$ … but slant height goes from base edge midpoint, not corner. Let’s recalculate. Slant height $l = 10$ is from midpoint of a base edge to apex. Half base $= 6$ . Distance from centre to midpoint of edge $= 6$ . $H = \sqrt{l^2 - 6^2} = \sqrt{100 - 36} = 8$ cm.) (b) Angle between slant face and base: $\tan\alpha = H/6 = 8/6$ , so $\alpha = \tan^{-1}(4/3) \approx 53.1°$ . (c) Slant edge (corner to apex): distance from centre to corner $= 6\sqrt{2}$ . Slant edge $= \sqrt{8^2 + (6\sqrt{2})^2} = \sqrt{64 + 72} = \sqrt{136} \approx 11.66$ cm. Angle with base: $\tan\beta = 8/(6\sqrt{2}) = 8/8.485$ , so $\beta \approx 43.3°$ .
Let A be at the origin, B at $(300, 0)$ . Drone bearing $060°$ from A means the drone’s horizontal position is along direction $060°$ from A. Let horizontal distance from A to point below drone $= d_A$ . Then $d_A \tan 40°...$ — actually the drone is at some point $(x, y, h)$ . From A: bearing $060°$ means $x = d_A \sin 60°$ , $y = d_A \cos 60°$ and $\tan 40° = h/d_A$ . From B at $(300, 0)$ : $\tan 55° = h/d_B$ where $d_B$ is horizontal distance from B to drone. Using $d_A = h/\tan 40°$ and $d_B = h/\tan 55°$ , and the geometry: $x = d_A \sin 60° = h\sin 60°/\tan 40°$ , $y = d_A \cos 60° = h\cos 60°/\tan 40°$ . Also $d_B^2 = (x - 300)^2 + y^2 = h^2/\tan^2 55°$ . Substituting and solving numerically gives $h \approx 218$ m. (Accept reasonable numerical solutions with clear working.)
Cube side $= 2r$ (the sphere diameter equals the cube side). Space diagonal of cube $= 2r\sqrt{3}$ . Diameter of sphere $= 2r$ . Ratio $= \frac{2r\sqrt{3}}{2r} = \sqrt{3}:1$ .
Nominal area $= \frac{1}{2}(50)(80)\sin 60° = 2000 \times 0.8660 \approx 1732$ m $^2$ . Length errors: $\frac{0.5}{50} = 1\%$ and $\frac{0.5}{80} = 0.625\%$ . Since each length appears to the first power, their contributions are $1\%$ and $0.625\%$ . Angle error: $1°$ in $60°$ . The sensitivity factor is $\frac{d}{dC}(\sin C) \cdot \frac{C}{\sin C}$ . At $C = 60°$ : $\frac{\cos 60°}{\sin 60°} \times 1° \times \frac{\pi}{180} \approx 0.577 \times 0.01745 \approx 1.007\%$ . Total error $\approx 1 + 0.625 + 1.007 \approx 2.6\%$ , i.e. $\pm 45$ m $^2$ . The angle error contributes roughly as much as the length errors combined.

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Pythagoras' theorem and trigonometry in 3D

What you will learn

1. Pythagoras’ theorem in 3D

2. Angles of elevation and depression

3. Bearings and navigation

4. Measurement errors and their impact

Key formulas

Practice

Challenge

Answer key

Tier 1

Tier 2

Tier 3

Challenge