Mathematics • Year 9 • Unit 3 • Lesson 13
Elevation — Mixed Challenge
Pull together angle of elevation, inverse trig (Lesson 11), the complementary-angle rule (Lesson 12), Pythagoras, and eye-height corrections. Choose the right tool per problem, spot a plausible Year 9 mistake, and tackle a two-observer surveyor challenge.
1. Mixed problems — choose the right rule
Each question uses a different combination of elevation, complementary angles, eye-height, and Pythagoras. Decide which approach applies before you start writing. Show your working. 3 marks each
1.1 Standing 50 m from a building, the angle of elevation to the top is 38°. (a) Find the height of the building above eye level (2 d.p.). (b) If the observer's eye is 1.6 m above the ground, find the building's total height above the ground.
1.2 An observer 60 m from a tree sees its top at elevation 52°. Find the angle of elevation to the BASE of the tree from a hot-air balloon hovering directly above the observer at 50 m altitude. (Hint: it's a depression — but here just use 90° minus the relevant complementary angle if helpful, OR set up a fresh triangle from the balloon.)
1.3 A right triangle has the line of sight as the hypotenuse (length 100 m) and the angle of elevation 25°. (a) Find the height the line of sight reaches above the observer's eye. (b) Find the horizontal distance to the foot of the vertical object.
1.4 A 30 m tall pole and an observer at ground level (assume eye at ground) form an elevation of 60°. (a) Find the observer's distance from the pole's base (2 d.p.). (b) Without using inverse trig, state the angle of elevation if the observer moved to exactly the same distance away from a 60 m tall pole. Explain.
1.5 From a point on flat ground, the angle of elevation to the top of a tower is 35°. The horizontal distance is 80 m. (a) Find the height of the tower above eye level (2 d.p.). (b) Find the LINE-OF-SIGHT length (hypotenuse) from the observer's eye to the top.
1.6 A 50 m flagpole stands at the top of a 30 m hill. From a point on flat ground at the foot of the hill, the angle of elevation to the TOP of the flagpole is 40°. Find the horizontal distance from the observer to the foot of the hill (2 d.p.). [Hint: the line of sight skims the top of the flagpole and ends at the observer; opp = 30 + 50 = 80 m.]
2. Find the mistake
A Year 9 student is asked: from 40 m away, the angle of elevation to the top of a building is 60°. The observer's eye height is 1.5 m. Find the building's TOTAL height above the ground (2 d.p.). Their working is shown. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — building total height, 40 m away, elevation 60°, eye 1.5 m:
Line 1: opp / adj = tan 60° → opp = 40 × tan 60° ≈ 40 × 1.7321 ≈ 69.28 m
Line 2: This 69.28 m is the height of the building above eye level.
Line 3: To get total height above ground, SUBTRACT eye height: 69.28 − 1.5 = 67.78 m.
Line 4: Total building height ≈ 67.78 m.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct total height.
Stuck? The opp in the triangle is the height above the OBSERVER'S EYE. To get height above the GROUND, you must ADD the eye height (not subtract), because the observer's eye is above the ground.3. Open-ended challenge — the two-observer method
This question has a single answer but lets you choose the unknowns. 4 marks
3.1 A surveyor wants to find the height of a tower but can't measure the distance to the base directly. From point A on flat ground, the angle of elevation to the top of the tower is 40°. The surveyor then walks 30 m FURTHER from the tower along the same straight line to point B. From B, the angle of elevation is 25°. Find the tower's height (2 d.p.), assuming the surveyor's eye is at ground level.
Required: (i) Set up two trig equations, one for each observation point. (ii) Eliminate one unknown by equating. (iii) Solve for the tower's height. (iv) Sense-check: which angle is smaller, and is the larger distance indeed at the smaller angle?
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Building, 50 m, 38°
(a) Above eye: opp = 50 × tan 38° ≈ 50 × 0.7813 ≈ 39.07 m.
(b) Total above ground: 39.07 + 1.6 = 40.67 m.
1.2 — Balloon above observer, tree at 60 m, elevation 52°
From the balloon (50 m altitude, directly above the observer), the LINE TO THE BASE OF THE TREE is a depression looking down: vertical drop = 50 m, horizontal distance = 60 m. So depression angle = tan⁻¹(50/60) ≈ 39.8° (which equals the angle of elevation from the base of the tree looking up to the balloon, by alternate angles).
Note: the lesson on depression hasn't been covered yet (that's Lesson 14), but the calculation here uses the same right-triangle reasoning.
1.3 — Hyp = 100 m, elevation = 25°
(a) opp = hyp × sin θ = 100 × sin 25° ≈ 100 × 0.4226 ≈ 42.26 m above eye.
(b) adj = hyp × cos θ = 100 × cos 25° ≈ 100 × 0.9063 ≈ 90.63 m horizontal distance.
1.4 — 30 m pole at elevation 60°
(a) tan 60° = 30 / adj → adj = 30 / tan 60° = 30 / √3 ≈ 17.32 m.
(b) At the same 17.32 m distance, a 60 m pole would have opp = 60, so tan θ = 60 / 17.32 = 2 × tan 60° = 2√3 ≈ 3.464. θ = tan⁻¹(3.464) ≈ 73.9°. (Doubling the height roughly doubles the steepness of the angle from 60° up to about 74° — but NOT to 120° because tan is non-linear; this is a common misconception.)
1.5 — Tower, 80 m, elevation 35°
(a) opp = 80 × tan 35° ≈ 80 × 0.7002 ≈ 56.02 m above eye.
(b) hyp = opp / sin 35° = 56.02 / 0.5736 ≈ 97.66 m (or equivalently 80 / cos 35° ≈ 97.66 m).
1.6 — Flagpole on a hill, 50 + 30 m
Total height of the flagpole top above the observer = 30 + 50 = 80 m.
tan 40° = 80 / adj → adj = 80 / tan 40° ≈ 80 / 0.8391 ≈ 95.34 m horizontal distance to the foot of the hill.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The observer's eye is ABOVE the ground, so the computed opp (69.28 m) is the height ABOVE THE EYE, not above the ground. To get total height above the ground we ADD the eye height (not subtract), since the building extends an extra 1.5 m down from the line of sight to the actual ground.
(c) Corrected working:
opp = 40 × tan 60° ≈ 40 × 1.7321 ≈ 69.28 m (height above eye level).
Total above ground = 69.28 + 1.5 = 70.78 m.
The subtraction would only be correct if the observer were INSIDE the building looking up from a basement — not the usual scenario.
3 — Two-observer method (sample solution)
(i) Let h = tower height, d = distance from A to the base. Equation 1 (from A): h = d × tan 40°. Equation 2 (from B, which is 30 m further away): h = (d + 30) × tan 25°.
(ii) Equate: d × tan 40° = (d + 30) × tan 25°.
Expand: d × tan 40° = d × tan 25° + 30 × tan 25°.
Collect d: d × (tan 40° − tan 25°) = 30 × tan 25°.
Solve: d = 30 × tan 25° / (tan 40° − tan 25°) ≈ 30 × 0.4663 / (0.8391 − 0.4663) ≈ 13.99 / 0.3728 ≈ 37.52 m.
(iii) Substitute back: h = 37.52 × tan 40° ≈ 37.52 × 0.8391 ≈ 31.48 m. Tower is about 31.48 m tall.
(iv) Sense check: 25° is the smaller angle (from B), and B is the further point (d + 30 = 67.52 m); indeed smaller angle = further away ✓. The height of 31.48 m is a sensible Year 9 building/tower scale.
Marking: 1 mark for setting up two equations; 1 for the elimination step; 1 for the correct h ≈ 31.48 m; 1 for the sense-check or correct sign convention.