Mathematics • Year 9 • Unit 3 • Lesson 13

Elevation in the Real World

Use the angle of elevation in everyday surveyor problems: a lighthouse beacon, a Sydney landmark, a stargazing telescope, a school flagpole, and a drone above a park. Then explain when eye height matters.

Apply · Real-World Maths

1. Word problems

Each problem uses the angle of elevation. Sketch first, label sides, then choose tan (most common), sin or cos. Show your working — a single final answer with no working only earns half marks. Round heights to 2 d.p. and angles to 1 d.p. unless stated.

1.1 — Sydney Tower from Hyde Park. Sydney Tower's observation deck is 250 m above the ground. From a flat spot in Hyde Park 400 m away, what is the angle of elevation to the deck (1 d.p.)? Assume eye is at ground level for simplicity.

Stuck? opp = 250, adj = 400. Use tan⁻¹.

1.2 — School flagpole. Mia is 1.5 m tall and stands 18 m from the base of her school's flagpole. She measures the angle of elevation to the top as 42°.

(a) Find the flagpole's height ABOVE eye level (2 d.p.).
(b) Find the flagpole's TOTAL height above the ground.
(c) State one assumption you are making about the ground. 3 marks

Stuck on (a)? opp = 18 × tan 42°. For (b), add Mia's eye height.

1.3 — Lighthouse beacon. A ship's navigator standing on the deck (assume eye at sea level for this problem) sees a lighthouse beacon at an angle of elevation of 6°. The lighthouse beacon is known to be 80 m above sea level. How far (horizontally) is the ship from the lighthouse base, to the nearest metre?

Stuck? opp = 80, θ = 6°. Rearrange tan 6° = 80 / adj → adj = 80 / tan 6°.

1.4 — Stargazing telescope. Astronomy club member Jake aims his telescope at a star. From his observation site, the star appears at an angle of elevation of 67° above the eastern horizon. Jake's eye is 1.7 m above the ground; his telescope sits at the same height.

(a) Explain in one sentence why the angle of elevation depends on Jake's location AND on the time, but not on his eye height (for a star that is effectively infinitely far away).
(b) If Jake instead pointed his telescope at the top of a tower 200 m away that's exactly 470 m tall, what angle of elevation would he see (1 d.p., assume same eye height)? 3 marks

Stuck on (b)? Tower height above eye level = 470 − 1.7 = 468.3 m, distance = 200 m. tan θ = 468.3 / 200.

1.5 — Drone above a park. A drone hovers directly above point A in a flat park. Observer Sarah, 1.6 m tall, stands at point B 30 m horizontally from A. She measures the angle of elevation to the drone as 55°.

(a) Find the drone's height above the GROUND (2 d.p.).
(b) Civil drone rules in Australia limit recreational drones to 120 m above the ground. Is Sarah's drone within the limit? 3 marks

Stuck on (a)? Step 1: opp above eye = 30 × tan 55°. Step 2: add Sarah's eye height of 1.6 m.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate writes: "From 50 m away, with an angle of elevation of 40°, the building must be 50 / tan 40° ≈ 59.59 m tall." In your own words, explain (i) what mistake they have made (think about which side is "opposite" and which is "adjacent"), (ii) what the correct formula is, and (iii) what the correct height is. Refer to "opposite" and "adjacent" relative to the angle of elevation somewhere in your explanation.

Stuck? The vertical building is the OPP, the horizontal ground is the ADJ — so tan 40° = opp / adj = building / 50, giving building = 50 × tan 40° (multiply, don't divide).

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Sydney Tower from Hyde Park

tan θ = 250 / 400 = 0.625. θ = tan⁻¹(0.625) ≈ 32.0° — fairly steep, as expected when you're under a tall tower.

1.2 — School flagpole

(a) Above eye: 18 × tan 42° ≈ 18 × 0.9004 ≈ 16.21 m.
(b) Total: 16.21 + 1.5 = 17.71 m.
(c) Assumption: the ground between Mia and the base of the flagpole is FLAT (horizontal), so the "adjacent" really is a true horizontal distance.

1.3 — Lighthouse beacon

tan 6° = 80 / adj. So adj = 80 / tan 6° ≈ 80 / 0.1051 ≈ 761 m from the base of the lighthouse.
Sense check: small angle (6°) means the lighthouse is far away, which matches our 761 m answer.

1.4 — Stargazing telescope

(a) The star is so far away that moving Jake's eye up or down by 1.7 m makes essentially zero difference to the angle (the geometry of "infinity" — parallel light rays). What does change the angle: where Jake stands on Earth (changes the horizon's direction), and what time it is (the star moves due to Earth's rotation).
(b) Tower height above eye = 470 − 1.7 = 468.3 m. tan θ = 468.3 / 200 = 2.3415. θ = tan⁻¹(2.3415) ≈ 66.9°.

1.5 — Drone above a park

(a) Above eye: 30 × tan 55° ≈ 30 × 1.4281 ≈ 42.84 m. Total above ground: 42.84 + 1.6 = 44.44 m.
(b) 44.44 m is well below the 120 m limit, so Sarah's drone is within the legal recreational ceiling.

2.1 — Explain your thinking (sample response)

My classmate has put the OPP and ADJ in the wrong places. Relative to the angle of elevation at the observer's eye, the OPPOSITE side is the VERTICAL building height and the ADJACENT side is the HORIZONTAL ground distance. So the correct relationship is tan 40° = opp / adj = building / 50. To solve for the building, multiply both sides by 50: building = 50 × tan 40° (not 50 / tan 40°). This gives 50 × 0.8391 ≈ 41.95 m. The classmate's wrong answer (≈ 59.59 m) is actually the LINE OF SIGHT (hypotenuse), not the height.

Marking: 1 mark for naming the swap (opp/adj mix-up); 1 for naming the correct ratio; 1 for the correct formula (50 × tan 40°); 1 for the correct value (≈ 41.95 m).