Year 10 Mathematics | Victorian Curriculum 2.0
Pythagoras' theorem and trigonometry in 3D
Topic 11 | Measurement | Practice

What you will learn

Worked example 0 Real-world example: antenna cable length

A communications antenna sits on a flat roof. The mounting point is 33 m east and 44 m north of a cable anchor point at roof level. The top of the antenna is 1212 m above the roof. What length of cable is needed from the anchor to the top of the antenna (add 5%5\% for slack)?

  1. Horizontal distance across the roof: dh=32+42=25=5d_h = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 m.
  2. Cable length (anchor to antenna top): L=52+122=169=13L = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 m.
  3. With 5%5\% slack: 13×1.05=13.6513 \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.

d_baselwhspace diagonal dABd = sqrt(l^2 + w^2 + h^2)
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 ll, ww, and hh:

Space diagonal
d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2}

This result comes from applying Pythagoras twice:

  1. Find the base diagonal: dbase=l2+w2d_{\text{base}} = \sqrt{l^2 + w^2}.
  2. Combine with height: d=dbase2+h2=l2+w2+h2d = \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 55 m by 44 m by 33 m. Find the longest straight line that fits inside the room (the space diagonal).

  1. d=52+42+32=25+16+9=50≈7.07d = \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 88 cm and a slant edge (from base corner to apex) of 1010 cm. Find the height of the pyramid.

  1. The base diagonal =82= 8\sqrt{2} cm. Half the base diagonal =42≈5.657= 4\sqrt{2} \approx 5.657 cm (distance from centre of base to a corner).
  2. The slant edge, the half-diagonal, and the height form a right-angled triangle.
  3. h=102−(42)2=100−32=68≈8.25h = \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 1515 m tall stands on a hill. From a point 200200 m horizontally from the base of the hill, the angle of elevation to the top of the tower is 18°18°. Find the height of the hill.

  1. Let HH be the hill height. Total height above the observer’s level =H+15= H + 15.
  2. tan⁡18°=H+15200\tan 18° = \dfrac{H + 15}{200}, so H+15=200tan⁡18°≈200×0.3249=64.98H + 15 = 200\tan 18° \approx 200 \times 0.3249 = 64.98 m.
  3. Hill height: H≈64.98−15=49.98≈50.0H \approx 64.98 - 15 = 49.98 \approx 50.0 m.
Worked example 4 Angle of depression from a drone

A drone hovers 120120 m above the ground. Its camera spots a target on the ground at an angle of depression of 35°35°. Find the horizontal distance from the drone to the target.

  1. tan⁡35°=120d\tan 35° = \dfrac{120}{d}.
  2. d=120tan⁡35°≈1200.7002≈171.4d = \dfrac{120}{\tan 35°} \approx \dfrac{120}{0.7002} \approx 171.4 m.

3. Bearings and navigation

A true bearing is measured clockwise from north (000°000° to 360°360°). A compass bearing uses directions like N30°30°E.

Worked example 5 Two-stage journey

A ship sails 2020 km on a bearing of 060°060°, then 1515 km on a bearing of 150°150°. Find the ship’s distance and bearing from its starting point.

  1. First leg: north component =20cos⁡60°=10= 20\cos 60° = 10 km, east component =20sin⁡60°≈17.32= 20\sin 60° \approx 17.32 km.
  2. Second leg: bearing 150°150° is S30°30°E. North component =−15cos⁡30°≈−12.99= -15\cos 30° \approx -12.99 km (southward), east component =15sin⁡30°=7.5= 15\sin 30° = 7.5 km.
  3. Total: north =10−12.99=−2.99= 10 - 12.99 = -2.99 km, east =17.32+7.5=24.82= 17.32 + 7.5 = 24.82 km.
  4. Distance =(−2.99)2+24.822=8.94+616.03≈25.0= \sqrt{(-2.99)^2 + 24.82^2} = \sqrt{8.94 + 616.03} \approx 25.0 km.
  5. Angle from north: θ=tan⁡−1 ⁣(24.822.99)≈83.1°\theta = \tan^{-1}\!\left(\dfrac{24.82}{2.99}\right) \approx 83.1°. Since the position is south-east, bearing ≈180°−83.1°=96.9°\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
Percentage error=∣measured−true∣true×100%\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±0.1r = 5.0 \pm 0.1 cm and height h=12.0±0.2h = 12.0 \pm 0.2 cm. Estimate the percentage error in the calculated volume.

  1. Percentage error in rr: 0.15.0×100=2%\dfrac{0.1}{5.0} \times 100 = 2\%.
  2. Since V=πr2hV = \pi r^2 h, the radius is squared, so its error contribution doubles: 2×2%=4%2 \times 2\% = 4\%.
  3. Percentage error in hh: 0.212.0×100≈1.67%\dfrac{0.2}{12.0} \times 100 \approx 1.67\%.
  4. Approximate total percentage error in VV: 4%+1.67%≈5.67%4\% + 1.67\% \approx 5.67\%.
  5. Nominal V=π(25)(12)=300π≈942.48V = \pi(25)(12) = 300\pi \approx 942.48 cm3^3. The error is roughly ±53\pm 53 cm3^3.

Key formulas

Space diagonal
d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2}
Trigonometric ratios
sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj\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
Percentage error=∣measured−true∣true×100%\text{Percentage error} = \frac{|\text{measured} - \text{true}|}{\text{true}} \times 100\%

Practice

Fluency

Tier 1: basic skills

    1. Find the space diagonal of a box 6×8×106 \times 8 \times 10 cm.
    2. A cube has side 55 cm. Find its space diagonal.
    3. From a point on level ground, the angle of elevation to the top of a 2525 m tree is 40°40°. Find the horizontal distance to the tree.
    4. A drone at height 8080 m observes a point on the ground at an angle of depression of 50°50°. Find the horizontal distance.
    5. A ship sails 1010 km on a bearing of 045°045°. How far north and how far east has it travelled?
    6. A measurement of 8.0±0.28.0 \pm 0.2 cm is squared. Find the percentage error in the original measurement and the approximate percentage error in the squared value.
    7. Find the length of the longest rod that fits inside a rectangular box 12×5×412 \times 5 \times 4 cm.
    8. A pyramid has a square base of side 1010 cm and a vertical height of 1212 cm. Find the slant edge length (from a base corner to the apex).
Reasoning

Tier 2: mixed practice

    1. A rectangular room is 6×4×36 \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.)
    2. A hiker walks 55 km on bearing 030°030° then 88 km on bearing 120°120°. Find the distance and bearing from start to finish.
    3. From the top of a 6060 m cliff, the angles of depression to two boats in a line directly out to sea are 45°45° and 30°30°. Find the distance between the boats.
    4. A surveyor measures the angle of elevation to a mountain peak as 22°22° from point A and 15°15° from point B, which is 500500 m further away on level ground in a direct line from the peak. Find the height of the mountain.
    5. A cylinder has r=10.0±0.3r = 10.0 \pm 0.3 cm and h=20.0±0.5h = 20.0 \pm 0.5 cm. Calculate the volume and estimate the maximum percentage error.
    6. A tent pole 2.52.5 m tall is supported by guy ropes pegged 1.51.5 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

    1. Explain why the space diagonal formula d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2} works by describing the two right-angled triangles used in its derivation.
    2. A pilot flies from airport A on bearing 070°070° for 200200 km to point B, then on bearing 160°160° for 150150 km to airport C. Find the bearing and distance for the direct return flight from C to A.
    3. A building casts a shadow 3030 m long when the sun’s angle of elevation is 55°55°. An hour later the shadow is 4545 m long. Find the new angle of elevation and determine whether the sun rose or fell during this hour.
    4. 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=43πr3V = \frac{4}{3}\pi r^3 as an example. (What multiplier applies here instead of double?)

Challenge

Reasoning

Harder reasoning

    1. A right square pyramid has base side 1212 cm and slant height 1010 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.
    2. Two observers A and B are 300300 m apart on level ground. Both observe the same drone. Observer A measures the angle of elevation as 40°40° and the bearing to the drone as 060°060°. Observer B (due east of A) measures the angle of elevation as 55°55°. Find the height of the drone.
    3. A sphere of radius rr fits exactly inside a cube. Show that the ratio of the space diagonal of the cube to the diameter of the sphere is 3:1\sqrt{3}:1.
    4. A surveyor measures two sides of a triangle as a=50.0±0.5a = 50.0 \pm 0.5 m and b=80.0±0.5b = 80.0 \pm 0.5 m, with the included angle C=60°±1°C = 60° \pm 1°. Using the area formula A=12absin⁡CA = \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.
Year 10 Mathematics study companion | Practice