Mathematics • Year 9 • Unit 3 • Lesson 15
Two-Angle Trig in the Real World
Use two-angle elevation/depression scenarios to model real careers: a surveyor measuring an unreachable height, a coastguard tracking two boats, a hiker comparing two summits, a window-cleaner above and below a balcony, and a rooftop observer of two passing cars. Then explain the two-equation strategy in your own words.
1. Word problems
Each problem has TWO angles in the same setup. Sketch ONE big diagram first, identify the two right triangles, label them, and set up TWO trig equations. Show your working. Round distances to 2 d.p. unless stated.
1.1 — Surveyor measures an unreachable height. A surveyor cannot get to the base of a tower because of a fence. From point A on the same flat ground, the angle of elevation to the tower top is 35°. From point B, 25 m further from the tower along the same line, the elevation drops to 22°. Find the tower's height (2 d.p.) and the distance from B to the tower's base.
1.2 — Coastguard. A coastguard tower stands 40 m above sea level. The coastguard spots boat A at angle of depression 24° and boat B at depression 8° (both on the same straight line out from the tower base).
(a) Find each boat's distance from the tower base (2 d.p.).
(b) Find the distance between the two boats.
(c) The coastguard's launch can reach boats within 500 m before they pass into international waters. Both boats are heading further out. Which boat needs urgent attention first? 4 marks
1.3 — Two mountain peaks from a lookout. Maya stands on a lookout at altitude 800 m. To the east, peak X has an angle of elevation of 5° and is known to be 12 km away horizontally. To the west, peak Y has an angle of depression of 3° and is known to be 8 km away horizontally.
(a) Find peak X's altitude above sea level (assume Maya's altitude is exactly 800 m).
(b) Find peak Y's altitude above sea level.
(c) Which peak is higher? 3 marks
1.4 — Window cleaner. A window cleaner is on a platform 18 m above ground. From her platform she can see a window above her (on the same building face) at elevation 32°, and a window below her at depression 41°. The horizontal distance to both windows is the same (the building is vertical): treat it as d = 5 m (the platform is 5 m horizontally away from the building face).
(a) Find the height of the upper window above her platform.
(b) Find the depth of the lower window below her platform.
(c) Find the vertical distance between the two windows. 3 marks
1.5 — Two cars from a rooftop. From a rooftop 25 m above the road, an observer measures the angle of depression to car X as 38° and to car Y, further down the road in the same direction, as 14°.
(a) Find the horizontal distance to each car (2 d.p.).
(b) Find the distance between the two cars.
(c) Sense check: which car has the smaller depression angle, and is it the closer or further one? Does this match your distances? 3 marks
2. Explain your thinking
This question is about communication, not just answers. Use full sentences. 4 marks
2.1 A classmate is asked to find the height of a tower using the two-angle method (from one point the elevation is 30°, from a second point 20 m further the elevation is 20°). They try to add the two angles to get 50° and use a single triangle. In your own words, explain (i) why "adding the angles" doesn't work, (ii) what the correct method is (the two-equation method), and (iii) what shared quantity links the two triangles. Use the phrase "two right triangles" somewhere in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Surveyor, elevations 35° and 22°, 25 m apart
Let h = tower height, d_A = distance from A (closer point) to tower base.
From A: h = d_A × tan 35°. From B: h = (d_A + 25) × tan 22°.
Equate: d_A × tan 35° = (d_A + 25) × tan 22°.
d_A × (tan 35° − tan 22°) = 25 × tan 22°.
d_A = 25 × tan 22° / (tan 35° − tan 22°) ≈ 25 × 0.4040 / (0.7002 − 0.4040) ≈ 10.10 / 0.2962 ≈ 34.10 m from A to base.
h = 34.10 × tan 35° ≈ 34.10 × 0.7002 ≈ 23.88 m tower height.
Distance from B to base = 34.10 + 25 = 59.10 m.
1.2 — Coastguard tower 40 m
(a) d_A = 40 / tan 24° ≈ 40 / 0.4452 ≈ 89.85 m. d_B = 40 / tan 8° ≈ 40 / 0.1405 ≈ 284.70 m.
(b) Gap = 284.70 − 89.85 ≈ 194.85 m.
(c) Boat B is at 284.70 m, closer to the 500 m international-waters boundary than boat A (89.85 m). If both are heading out, B needs urgent attention first.
1.3 — Two mountain peaks
(a) Peak X above Maya = 12000 × tan 5° ≈ 12000 × 0.0875 ≈ 1049.66 m. So sea-level altitude of X ≈ 800 + 1049.66 ≈ 1849.66 m.
(b) Peak Y below Maya = 8000 × tan 3° ≈ 8000 × 0.0524 ≈ 419.27 m. So sea-level altitude of Y ≈ 800 − 419.27 ≈ 380.73 m.
(c) Peak X (≈ 1850 m) is much higher than peak Y (≈ 381 m).
1.4 — Window cleaner, d = 5 m
(a) Upper window above platform = 5 × tan 32° ≈ 5 × 0.6249 ≈ 3.12 m.
(b) Lower window below platform = 5 × tan 41° ≈ 5 × 0.8693 ≈ 4.35 m.
(c) Vertical distance between windows = 3.12 + 4.35 ≈ 7.47 m.
1.5 — Two cars from a 25 m rooftop
(a) d_X = 25 / tan 38° ≈ 25 / 0.7813 ≈ 32.00 m. d_Y = 25 / tan 14° ≈ 25 / 0.2493 ≈ 100.27 m.
(b) Gap = 100.27 − 32.00 ≈ 68.27 m.
(c) Car Y has the smaller depression (14°) and is the further car — matches the rule "smaller depression angle = further away" ✓.
2.1 — Explain your thinking (sample response)
Adding the two angles to get 50° doesn't work because the 30° and 20° belong to two right triangles, NOT to one triangle. Each angle of elevation sits at a different observation point and defines its OWN right triangle, so combining them arithmetically has no geometric meaning. The correct method is the two-equation method: write h = d × tan 30° for the first triangle, and h = (d + 20) × tan 20° for the second triangle (where h is the tower height and d is the distance from the first observation point to the base). The shared quantity that LINKS the two triangles is the tower's height h — it's the OPPOSITE side in both triangles, even though each triangle has a different adjacent side. Equating the two expressions for h lets us solve for d, and then for h itself.
Marking: 1 mark for explaining why adding angles fails; 1 for naming the two-equation method; 1 for identifying the shared quantity (height h); 1 for the phrase "two right triangles" with a clear, full-sentence explanation.