Mathematics • Year 9 • Unit 3 • Lesson 19
Multi-Step Trig — Mixed Challenge
Combine surveyor-style two-angle problems, Pythagoras-then-trig, 3D space diagonals, and angle-with-the-base calculations. Spot a rounding mistake. Then design your own surveyor scenario from scratch.
1. Mixed problems — combine your tools
Each question needs two or more steps. Sketch BEFORE you compute. Show your working. 3 marks each
1.1 A rectangle is 7 cm by 24 cm. Find (a) the diagonal and (b) the acute angle the diagonal makes with the shorter side (1 d.p.).
1.2 From point A on flat ground, the elevation to a tower top is 30°. From point B, 80 m closer along the same line, the elevation is 45°. Find the tower's height (2 d.p.).
1.3 A box is 6 cm by 8 cm by 24 cm. (a) Find the space diagonal (clean integer or surd). (b) Find the angle this space diagonal makes with the base (1 d.p.).
1.4 A 6 m wire is anchored at the top of a flagpole and stretched to a peg in the ground. The wire makes an angle of 50° with the ground. (a) Find the height of the flagpole. (b) Find how far the peg is from the base. (2 d.p. each.)
1.5 A staircase has a horizontal run of 3.6 m and a vertical rise of 2.7 m. Find (a) the length of the handrail along the slope (2 d.p.) and (b) the angle the staircase makes with the floor (1 d.p.).
1.6 From the top of a 40 m cliff, the angle of depression to a buoy at sea is 22°. A boat on the same sight line is at depression 35°. Find (a) the horizontal distance from the cliff to the buoy and (b) the distance between the buoy and the boat (2 d.p. each).
2. Find the mistake
A student has tried the surveyor problem: from A the elevation to a tower is 22°; from B (which is 60 m closer) it is 38°. Their working is shown below. Exactly one line contains a mistake (a classic mid-step rounding error). Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — tower from A (22°) and B (38°), AB = 60 m:
Line 1: From B: h = d tan 38°. From A: h = (d + 60) tan 22°.
Line 2: Equate: d tan 38° = (d + 60) tan 22° → d (tan 38° − tan 22°) = 60 tan 22°.
Line 3: tan 38° ≈ 0.78, tan 22° ≈ 0.40 (each rounded to 2 d.p.).
Line 4: d ≈ 60 × 0.40 / (0.78 − 0.40) = 24 / 0.38 ≈ 63.16 m.
Line 5: h = 63.16 × 0.78 ≈ 49.26 m. So the tower is about 49.26 m tall.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working (keeping full precision) and the corrected final tower height (2 d.p.).
Stuck? The denominator (tan 38° − tan 22°) is a small number — rounding the two tan values amplifies the error a LOT.3. Open-ended challenge — design your own surveyor problem
This question has many valid answers. 4 marks
3.1 Design a surveyor-style problem (two observation points along a line, two elevations to the same peak) so that the answer comes out to exactly 100 m for the height of the peak.
Your design must satisfy ALL the following:
(i) The two elevation angles are different and both between 10° and 80°.
(ii) The baseline (distance between the two observation points) is a whole number of metres between 50 m and 500 m.
(iii) The peak height computed by the surveyor formula equals 100 m (within 0.5 m).
For your design:
(a) State your two elevation angles θ1 (from far A) and θ2 (from near B), and your baseline L.
(b) Set up and solve the two equations for d (distance from B to the peak).
(c) Compute the height and confirm it's 100 m (within 0.5 m).
Bonus: What happens to your design if you make the baseline 10× larger but keep both angles the same?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Rectangle 7 by 24
(a) d = √(49 + 576) = √625 = 25 cm (7-24-25 triple).
(b) Angle with the SHORTER side (7 cm): tan θ = 24 / 7 ≈ 3.4286; θ = tan−¹(3.4286) ≈ 73.7°.
1.2 — Surveyor tower (30° from A, 45° from B, AB = 80)
From B: h = d tan 45° = d (since tan 45° = 1).
From A: h = (d + 80) tan 30°.
Equate: d = (d + 80)(1/√3) ⇒ d√3 = d + 80 ⇒ d(√3 − 1) = 80 ⇒ d = 80 / (√3 − 1) ≈ 80 / 0.7321 ≈ 109.28 m.
h = d = 109.28 m.
1.3 — Box 6 by 8 by 24
(a) Space diagonal = √(36 + 64 + 576) = √676 = 26 cm (clean integer).
(b) Base diagonal = √(36 + 64) = √100 = 10. Angle with base: tan θ = 24 / 10 = 2.4; θ = tan−¹(2.4) ≈ 67.4°.
1.4 — 6 m wire at 50°
(a) Flagpole height = 6 sin 50° ≈ 6 × 0.7660 ≈ 4.60 m.
(b) Peg distance from base = 6 cos 50° ≈ 6 × 0.6428 ≈ 3.86 m.
1.5 — Staircase (run 3.6 m, rise 2.7 m)
(a) Handrail = √(3.6² + 2.7²) = √(12.96 + 7.29) = √20.25 = 4.50 m (3-4-5 scaled by 0.9).
(b) Angle: tan θ = 2.7 / 3.6 = 0.75; θ = tan−¹(0.75) ≈ 36.9°.
1.6 — 40 m cliff, buoy at 22°, boat at 35°
From the cliff top (40 m above sea level), horizontal distance d for a given depression θ satisfies tan θ = 40 / d, so d = 40 / tan θ.
(a) Horizontal distance to BUOY: 40 / tan 22° ≈ 40 / 0.4040 ≈ 99.00 m.
(b) Horizontal distance to BOAT: 40 / tan 35° ≈ 40 / 0.7002 ≈ 57.13 m. Distance between buoy and boat = 99.00 − 57.13 ≈ 41.87 m (with the boat closer to the cliff than the buoy).
2 — Find the mistake (surveyor tower, mid-step rounding)
(a) The mistake is on Line 3.
(b) Rounding tan 38° to 0.78 and tan 22° to 0.40 introduces small errors that get AMPLIFIED when you subtract them in the denominator (0.78 − 0.40 = 0.38, but the true value is closer to 0.7813 − 0.4040 = 0.3773 — nearly 1% off). That rolls through the calculation and changes the final height by several metres.
(c) Corrected working (full precision):
d = 60 × tan 22° / (tan 38° − tan 22°) = 60 × 0.40403 / (0.78129 − 0.40403) = 24.2418 / 0.37726 ≈ 64.26 m.
h = 64.26 × tan 38° ≈ 64.26 × 0.78129 ≈ 50.20 m. (Compare with the student's 49.26 — about 1 m off.)
Sanity check: keeping full precision throughout gives a more reliable answer; the lesson rule "round only at the end" really matters here.
3 — Design your own surveyor problem (sample solution)
Design: θ1 = 30° (from far A), θ2 = 45° (from near B), L = baseline = 74 m.
From B: h = d tan 45° = d.
From A: h = (d + 74) tan 30° = (d + 74)/√3.
Set equal: d = (d + 74)/√3 ⇒ d√3 − d = 74 ⇒ d(√3 − 1) = 74 ⇒ d = 74 / 0.7321 ≈ 101.08 m.
h = d ≈ 101.08 m — within 1.1 m of the target. To hit exactly 100 m, use L = (100)(√3 − 1) ≈ 73.21 m, but L must be a whole number, so 73 m gives h ≈ 99.71 m (within 0.5 m ✓).
Bonus: If you keep both angles the same and multiply L by 10, the linearity of the formula means d scales by 10 and h scales by 10 — so a baseline 10× longer gives a peak 10× taller (assuming the same angles). The formula is purely scale-invariant in L for fixed angles.
Marking: 1 mark for two valid distinct angles in 10-80° range; 1 mark for whole-number baseline 50-500 m; 1 mark for setting up and solving the surveyor equations; 1 mark for confirming h is within 0.5 m of 100 m (and noting the linearity in the bonus).