Mathematics Standard • Year 11 • Module 2 • Lesson 16
Angles of Elevation and Depression — Problem Set
Apply elevation and depression to ships, towers, drones and two-observation problems — diagram first, equal-angles second, then SOHCAHTOA.
Problem 1 — Ship approaching a lighthouse
A ship is sailing toward a lighthouse. The lighthouse is 45 m tall, standing at the edge of a cliff at sea level. From the ship, the angle of elevation to the top of the lighthouse is initially 12°. After the ship travels closer, the angle increases to 28°.
Set up: What are we solving for?
(i) Sketch and label both right-angled triangles (one for each ship position) sharing the same 45 m vertical lighthouse. 1 mark
(ii) Find the initial horizontal distance from the ship to the base of the lighthouse, to the nearest metre. 2 marks
(iii) Find the new horizontal distance after the angle becomes 28°, and the distance the ship has travelled towards the lighthouse, both to the nearest metre. 2 marks
Stuck? At each ship position: tan(elevation) = 45 ÷ (horizontal distance). The travel = first distance − second distance.Problem 2 — Drone observation over a park
A drone hovers 120 m vertically above the ground. From the drone, the angle of depression to a picnic shelter on the ground is 26°. Later, the drone tilts its camera and observes a second picnic shelter, also on the ground, at an angle of depression of 18°. The two shelters are on the same straight horizontal line as the point directly below the drone.
Set up: What are we solving for?
(i) Sketch and label the diagram: drone at top, vertical 120 m down to a point on the ground, two shelters at different horizontal distances along the ground. Mark both depression angles at the drone using the horizontal reference line. 1 mark
(ii) Find the horizontal distance from the point directly below the drone to each shelter, to the nearest metre. 2 marks
(iii) Find the distance between the two shelters along the ground (assume both are on the same side of the point below the drone), to the nearest metre. 2 marks
Stuck? Larger depression angle = closer shelter. Distance apart = farther minus closer.Problem 3 — Antenna on top of a building
A 12 m antenna sits on top of a 40 m building. An observer stands 75 m horizontally from the base.
Set up: What are we solving for?
(i) Find the angle of elevation to the top of the antenna, to the nearest minute. 2 marks
(ii) Find the angle of elevation to the top of the building (without the antenna), to the nearest minute. 1 mark
(iii) Find the angle subtended by the antenna at the observer (i.e. the difference between the two angles in (i) and (ii)), to the nearest minute. 2 marks
Stuck? Subtract the two angles (in degrees and minutes — be careful with borrowing).Problem 4 — Surveyor with two observation points
A surveyor stands on level ground at point P and measures the angle of elevation to the top of a vertical tower as 55°. She then walks 40 m directly away from the tower (still on level ground, on the same side) to point Q, and measures the angle of elevation as 35°.
Set up: What are we solving for?
(i) Let d = horizontal distance from P to the base of the tower, and h = height of the tower. Write one equation for h using the angle at P, and one using the angle at Q (in terms of d). 1 mark
(ii) By setting the two equations equal, find d to 2 d.p. 3 marks
(iii) Find the height of the tower h to 2 d.p. 1 mark
Stuck? d × tan 55° = (d + 40) × tan 35°. Expand and collect d on one side.Problem 5 — Lighthouse with two boats
A vertical lighthouse is 50 m tall and stands at the water's edge (the base is at sea level). The keeper, at the top, observes two boats out at sea on the same bearing. The closer boat has angle of depression 36° and the farther boat has angle of depression 21°.
Set up: What are we solving for?
(i) Sketch and label the diagram (one lighthouse, two boats, two depression angles at the top, horizontal reference line). 1 mark
(ii) Find the horizontal distance from the lighthouse base to each boat, to 2 d.p. 2 marks
(iii) Find the distance between the two boats (assume they are on the same straight line from the lighthouse), to 2 d.p. 1 mark
(iv) The closer boat is travelling away from the lighthouse at 9 km/h. How long, in minutes (to the nearest minute), until it is as far away as the farther boat? 2 marks
Stuck on (iv)? Convert distance (m) to km, then use T = D ÷ S. Convert hours to minutes at the end.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ship approaching lighthouse
Set up. Equal-angles or direct tan: tan(elev) = 45 ÷ horizontal distance.
(i) Two triangles, same vertical 45 m, two different horizontal legs, two elevation angles (12° and 28°) at the ship in each.
(ii) d₁ = 45 ÷ tan 12° ≈ 45 ÷ 0.2126 ≈ 211.73 ≈ 212 m.
(iii) d₂ = 45 ÷ tan 28° ≈ 45 ÷ 0.5317 ≈ 84.64 ≈ 85 m. Distance travelled = 212 − 85 = 127 m.
Problem 2 — Drone observing two shelters
Set up. Drone-down vertical 120 m; depression at drone equals elevation at shelter.
(i) Drone at top; vertical 120 m to a ground point; two shelters along ground; horizontal reference line at drone for depression angles 26° and 18°.
(ii) Shelter 1: d₁ = 120 ÷ tan 26° ≈ 120 ÷ 0.4877 ≈ 246.04 ≈ 246 m. Shelter 2: d₂ = 120 ÷ tan 18° ≈ 120 ÷ 0.3249 ≈ 369.31 ≈ 369 m.
(iii) Distance apart = 369 − 246 = 123 m.
Problem 3 — Antenna on building
Set up. Two elevations from the same observer, subtract.
(i) Top of antenna total height = 40 + 12 = 52 m. tan θ_top = 52 ÷ 75 ≈ 0.6933; θ_top = tan⁻¹(0.6933) ≈ 34.72° = 34°43'.
(ii) tan θ_bld = 40 ÷ 75 ≈ 0.5333; θ_bld = tan⁻¹(0.5333) ≈ 28.07° = 28°4'.
(iii) Antenna subtended angle = 34°43' − 28°4' = 6°39'.
Problem 4 — Two-observation tower
Set up. Same height h; two equations in d.
(i) From P: h = d × tan 55°. From Q: h = (d + 40) × tan 35°.
(ii) d × tan 55° = (d + 40) × tan 35° ⇒ d × (1.4281 − 0.7002) = 40 × 0.7002 ⇒ d × 0.7279 = 28.008 ⇒ d ≈ 38.48 m.
(iii) h = 38.48 × tan 55° ≈ 38.48 × 1.4281 ≈ 54.95 m.
Problem 5 — Lighthouse with two boats
Set up. Same vertical 50 m; equal-angles gives elevation at each boat.
(i) Lighthouse vertical 50 m; two boats on the same side at sea; depression 36° (closer) and 21° (farther) at the top; horizontal reference at the top.
(ii) Closer: d₁ = 50 ÷ tan 36° ≈ 50 ÷ 0.7265 ≈ 68.82 m. Farther: d₂ = 50 ÷ tan 21° ≈ 50 ÷ 0.3839 ≈ 130.24 m.
(iii) Distance apart = 130.24 − 68.82 = 61.42 m.
(iv) Distance to travel = 61.42 m = 0.06142 km. T = 0.06142 ÷ 9 ≈ 0.006824 h × 60 ≈ 0.41 min ≈ 0 min (about 25 seconds; less than a minute, so to the nearest minute the closer boat reaches the position in under 1 min).