Mathematics • Year 9 • Unit 3 • Lesson 13
Angles of Elevation
Build fluency with the angle of elevation — the angle measured from the horizontal UPWARD to your line of sight. Walk from a fully worked cliff problem through one guided fill-in to eight graduated practice problems on heights, distances and eye-height corrections.
1. I do — fully worked example
Read every step. The standard pattern for elevation problems: sketch, label sides, choose ratio (usually tan), solve.
Problem. You stand 80 m from the base of a cliff. The angle of elevation to the top is 35°. How tall is the cliff (2 d.p.)? Assume your eye is at ground level.
Step 1 — Sketch the right triangle.
Horizontal ground = adjacent. Vertical cliff = opposite. Line of sight = hypotenuse. Angle of elevation θ = 35° sits at your eye, between the horizontal and the line of sight.
Reason: the angle of elevation is ALWAYS from the horizontal looking UP.
Step 2 — Label the sides.
adj = 80 m (horizontal distance). opp = ? (cliff height). θ = 35°.
Reason: we want the opp (cliff) and we have the adj (ground).
Step 3 — Pick the ratio: opp + adj → tan.
tan 35° = opp / 80
Reason: TOA — opp over adj is tan.
Step 4 — Solve for opp.
opp = 80 × tan 35° ≈ 80 × 0.7002 ≈ 56.01 m
Reason: multiply both sides by 80 to isolate the unknown.
Step 5 — State answer with units.
Cliff height ≈ 56.01 m (assuming eye at ground level).
Answer: ≈ 56.01 m.
2. We do — fill in the missing steps
Same structure as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. A surveyor 70 m from a tower measures the angle of elevation to the top as 28°. Find the tower's height above eye level (2 d.p.).
Step 1 — Sketch: a right triangle with the horizontal ground as the ________ side and the tower height as the ________ side.
Step 2 — Label:
adj = ____ m, opp = ?, θ = ____ °
Step 3 — Pick the ratio: opp + adj → ________ .
Step 4 — Set up and solve:
tan 28° = opp / ____
opp = ____ × tan 28° ≈ ____ × ________ ≈ ________ m
Step 5 — State the answer: Height above eye level ≈ ________ m.
3. You do — independent practice
Show working under each problem. Problems 3.1–3.4 are foundation (height given angle and distance). Problems 3.5–3.6 are standard (angle given heights and distances). Problems 3.7–3.8 are extension (involve eye height).
Foundation — find the height
3.1 From 40 m away, the angle of elevation to the top of a tree is 30°. Find the tree's height (2 d.p., assume eye at ground). 1 mark
3.2 From 60 m away, the angle of elevation to a tower top is 45°. Find the tower's height. 1 mark
3.3 From 100 m away, the angle of elevation to a building top is 25°. Find the building's height above eye level (2 d.p.). 1 mark
3.4 A line of sight 50 m long rises at 30° to the horizontal. Find the height reached above the observer's eye. 1 mark
Standard — find the angle
3.5 A 20 m flagpole is viewed from 50 m away on flat ground. Find the angle of elevation to the top (1 d.p.). 2 marks
3.6 From 30 m away, the top of a 20 m flagpole is seen. Find the angle of elevation (1 d.p.). 2 marks
Extension — include eye height
3.7 A 1.6 m tall observer stands 12 m from a tree. The angle of elevation to the top is 50°. Find the tree's TOTAL height above the ground (2 d.p.). 2 marks
3.8 From 100 m away, the angle of elevation to the top of a flagpole is 30°. The observer's eye height is 1.5 m. Find the flagpole's TOTAL height above the ground (2 d.p.). 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (tower from 70 m at 28°)
Step 1: horizontal ground = adjacent; tower height = opposite.
Step 2: adj = 70 m, θ = 28°.
Step 3: opp + adj → tan.
Step 4: tan 28° = opp / 70; opp = 70 × tan 28° ≈ 70 × 0.5317 ≈ 37.23 m.
Step 5: ≈ 37.23 m above eye level.
3.1 — 40 m, 30°
height = 40 × tan 30° ≈ 40 × 0.5774 ≈ 23.09 m.
3.2 — 60 m, 45°
height = 60 × tan 45° = 60 × 1 = 60 m. (At 45° the height equals the distance.)
3.3 — 100 m, 25°
height = 100 × tan 25° ≈ 100 × 0.4663 ≈ 46.63 m.
3.4 — Line of sight 50 m at 30°
opp = hyp × sin θ = 50 × sin 30° = 50 × 0.5 = 25 m above the observer's eye.
3.5 — 20 m pole at 50 m
tan θ = 20 / 50 = 0.4. θ = tan⁻¹(0.4) ≈ 21.8°.
3.6 — 20 m pole at 30 m
tan θ = 20 / 30 ≈ 0.6667. θ = tan⁻¹(0.6667) ≈ 33.7°.
3.7 — Tree, 12 m, 50°, eye 1.6 m
Above eye: 12 × tan 50° ≈ 12 × 1.1918 ≈ 14.30 m.
Total above ground: 14.30 + 1.6 = 15.90 m.
3.8 — Flagpole, 100 m, 30°, eye 1.5 m
Above eye: 100 × tan 30° ≈ 100 × 0.5774 ≈ 57.74 m.
Total above ground: 57.74 + 1.5 = 59.24 m.