Mathematics • Year 9 • Unit 3 • Lesson 19
Multi-Step Trig: Real-World Surveying
Use two-step trigonometry to solve real surveying and architecture problems: an inaccessible mountain, a roof truss, a TV antenna anchor, a 3D shipping container diagonal, and a school flagpole-and-shadow check. Then explain why mid-step rounding gives the wrong answer.
1. Word problems
Each problem requires two or more steps, combining trig and/or Pythagoras. Keep full calculator precision through intermediate steps and round only the FINAL answer. Show your working — a single final answer with no working only earns half marks.
1.1 — Inaccessible mountain (surveyor method). A surveyor on flat ground measures the elevation to a mountain peak as 14° from point A. Walking 250 m closer along the same straight line, the elevation from point B is 24°.
(a) Set up the two equations (one from each observation point) for the mountain height h.
(b) Find the distance d from B to the foot of the mountain (2 d.p.).
(c) Find the height of the mountain (2 d.p.). 4 marks
1.2 — Roof truss. A symmetric roof truss has a base (span) of 9.6 m. The two roof slopes each rise at 30° to the horizontal and meet at the apex.
(a) Find the height of the apex above the base of the truss (2 d.p.).
(b) Find the length of one of the two sloping rafters (2 d.p.). 3 marks
1.3 — TV antenna anchor wires. A TV antenna is held up by two anchor wires from the same point at the top. Wire 1 is fixed to the ground 4 m from the base of the antenna and makes an angle of 70° with the ground.
(a) Find the height of the antenna (2 d.p.).
(b) Wire 2 is fixed 6 m from the base on the OTHER side. Find the angle it makes with the ground (1 d.p.) AND its length (2 d.p.). 4 marks
1.4 — 3D shipping container diagonal. A standard 20-ft shipping container is 6.06 m long, 2.44 m wide, and 2.59 m tall.
(a) Find the floor diagonal (2 d.p.).
(b) Find the space diagonal — the longest straight bar that could fit inside the container (2 d.p.).
(c) Find the angle the space diagonal makes with the floor (1 d.p.). 3 marks
1.5 — Flagpole and shadow check. A school flagpole casts a shadow 9 m long when the sun's elevation angle is 35°. A nearby tree casts a shadow 4 m long at the SAME time.
(a) Find the height of the flagpole (2 d.p.).
(b) Find the height of the tree (2 d.p.).
(c) Why is it valid to use the SAME angle (35°) for both calculations? Answer in one sentence. 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate tackles the surveyor problem in 1.1 by computing tan 24° on the calculator, writing it down rounded to 2 d.p. (0.45), then carrying THAT rounded number through the rest of the calculation. They end up with a mountain height several metres different from the correct answer. In your own words, explain (i) why rounding mid-step is mathematically risky in multi-step problems, (ii) what they should have done instead, and (iii) describe a simple test you could run to check whether early rounding has hurt your answer. Refer to "amplifies errors" somewhere in your answer.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Surveyor mountain (14° from A, 24° from B, AB = 250 m)
(a) From B: h = d tan 24°. From A: h = (d + 250) tan 14°.
(b) Equate: d tan 24° = (d + 250) tan 14° ⇒ d(tan 24° − tan 14°) = 250 tan 14°.
d = 250 × tan 14° / (tan 24° − tan 14°) ≈ 250 × 0.2493 / (0.4452 − 0.2493) ≈ 62.33 / 0.1959 ≈ 318.18 m.
(c) h = 318.18 × tan 24° ≈ 318.18 × 0.4452 ≈ 141.65 m. (Use unrounded d to keep precision.)
1.2 — Symmetric roof truss (span 9.6 m, 30°)
Split the truss into two right triangles, each with half-base 4.8 m and 30° at the wall plate.
(a) Apex height: h = 4.8 tan 30° ≈ 4.8 × 0.5774 ≈ 2.77 m.
(b) Rafter length (hypotenuse) = 4.8 / cos 30° ≈ 4.8 / 0.8660 ≈ 5.54 m.
1.3 — TV antenna with two anchor wires
(a) Wire 1: tan 70° = h / 4 ⇒ h = 4 tan 70° ≈ 4 × 2.7475 ≈ 10.99 m tall.
(b) Wire 2 (6 m from base): tan θ = h / 6 = 10.99 / 6 ≈ 1.832; θ = tan−¹(1.832) ≈ 61.4°.
Wire 2 length = √(h² + 6²) = √(120.78 + 36) = √156.78 ≈ 12.52 m.
(Or wire 2 length = 6 / cos 61.4° ≈ 12.53 m — agrees within rounding.)
1.4 — Shipping container 6.06 × 2.44 × 2.59 m
(a) Floor diagonal = √(6.06² + 2.44²) = √(36.72 + 5.95) = √42.67 ≈ 6.53 m.
(b) Space diagonal = √(6.06² + 2.44² + 2.59²) = √(36.72 + 5.95 + 6.71) = √49.38 ≈ 7.03 m. (Longest straight bar that fits.)
(c) Angle space diagonal makes with the floor: tan θ = 2.59 / 6.53 ≈ 0.3966; θ = tan−¹(0.3966) ≈ 21.6°.
1.5 — Flagpole and tree shadows (35° sun elevation)
(a) Flagpole height = 9 tan 35° ≈ 9 × 0.7002 ≈ 6.30 m.
(b) Tree height = 4 tan 35° ≈ 4 × 0.7002 ≈ 2.80 m.
(c) The sun is so far away that its rays hit any two nearby objects on the school field at essentially the same angle — so the elevation angle 35° is the same for both the flagpole and the tree.
2.1 — Why mid-step rounding is risky (sample response)
Rounding mid-step is risky because the rounded value then gets MULTIPLIED, SUBTRACTED, or DIVIDED by other numbers in later steps, and that amplifies the error — a tiny rounding mistake at one stage can become a much bigger error by the end. Writing 0.45 instead of the true 0.4452 looks like only a 0.005 difference, but after the surveyor's d-calculation that difference shows up multiplied by 250 m and divided by a small denominator (tan 24° − tan 14° ≈ 0.196), which inflates it dramatically. Instead, the classmate should have kept the full calculator value in memory (e.g. using the "Ans" key or storing in a variable) and only rounded the FINAL height to 2 d.p. A simple test: redo the calculation keeping every intermediate to as many decimals as the calculator shows, and compare to the rounded-mid-step answer — if they differ in a way that matters (more than 0.5% or 0.1 m for these distances), early rounding hurt the answer.
Marking: 1 mark for the "amplifies errors" idea; 1 for "keep full precision, round at end"; 1 for the comparison/test idea; 1 for a clear, full-sentence explanation.