Mathematics • Year 9 • Unit 3 • Lesson 9
Tangent Ratio — Finding a Side
Use tan θ = opp/adj to find the opposite (opp = adj × tan θ) or the adjacent (adj = opp ÷ tan θ). The leg-only ratio — no hypotenuse needed! Build from a fully-worked building-height example through guided practice to eight independent problems.
1. I do — fully worked example
Standing 25 m from a building, you look up at the top at an angle of 40°. How tall is the building (ignoring eye height)?
Problem. adj = 25 m (ground distance to building), θ = 40° (angle of elevation), opp = ? (building height). Round to 2 d.p. DEG mode (check: sin 30° = 0.5).
Step 1 — Label sides relative to θ.
adj = 25 m (horizontal, next to θ — the ground), opp = ? (vertical building height, across from θ), hyp = NOT mentioned and NOT wanted.
Reason: opp + adj involved (no hyp) → use tangent (TOA). This is the signature setup for tan.
Step 2 — Set up the tangent equation.
tan 40° = opp / 25
Reason: tan θ = opp/adj.
Step 3 — Rearrange to isolate opp.
opp = 25 × tan 40°
Reason: opp is on top → multiply both sides by the bottom (25) to free it.
Step 4 — Compute.
opp ≈ 25 × 0.8391 ≈ 20.98 m
Reason: keep calculator precision, round the final answer.
Step 5 — Add units, sanity check.
opp ≈ 20.98 m tall. Reasonable? Yes — a building seen from 25 m at 40° is about 21 m tall (roughly a 7-storey building).
Answer: Building is approximately 20.98 m tall (above eye level).
2. We do — fill in the missing steps
A tower is 60 m tall. From an observation point at the same level as its base, the top is seen at 35° above horizontal. How far away is the observation point? 4 marks
Step 1 — Label. opp = ____ m (the tower height), adj = ? (the ground distance), θ = ____ ° . Two sides: opp and adj (no hyp), so use __________ .
Step 2 — Set up:
tan ____ ° = ____ / adj
Step 3 — Rearrange to put adj on its own:
adj = ____ / tan ____ °
Step 4 — Compute:
adj ≈ 60 / ________ ≈ ________ m
Step 5 — Sanity check. For a 35° angle of elevation, should the ground distance be bigger or smaller than the tower height? __________ (Hint: 35° < 45°, so tan 35° < 1, so adj > opp.) Is your answer? __________
3. You do — independent practice
Round to 2 d.p. unless told. Foundation (3.1–3.4) is one rearrangement. Standard (3.5–3.6) is contextual. Extension (3.7–3.8) is multi-step.
Foundation — opp = adj × tan θ, or adj = opp ÷ tan θ
3.1 adj = 20, θ = 35°. Find opp. 1 mark
3.2 adj = 12, θ = 45°. Find opp. (Use tan 45° = 1 exactly.) 1 mark
3.3 opp = 9, θ = 25°. Find adj. 1 mark
3.4 opp = 7, θ = 60°. Find adj. (Use tan 60° ≈ 1.7321.) 1 mark
Standard — practical contexts
3.5 Standing 30 m from a flagpole, you look up at the top at 22°. How tall is the flagpole (above your eye level)? 2 marks
3.6 A 45 m tall radio tower is viewed at an angle of elevation of 18° from a point on level ground. How far is the observer from the base of the tower? 2 marks
Extension — push your thinking
3.7 From a point 80 m from a cliff base, the top of the cliff is seen at 35°. Find the cliff height. Then, if you move 20 m closer (now 60 m from the base), find the NEW angle of elevation. (Hint: cliff height stays the same; use tan θ = opp/adj to find the new θ with tan⁻¹.) 3 marks
3.8 A road climbs at 8° below the horizontal as you look back from the top. After driving 1200 m down the road (the slope distance is irrelevant — only the horizontal distance from the foot of the road matters). Suppose the bottom of the road is 1100 m horizontally from the top. Find the vertical drop (use opp = adj × tan 8°) and state whether 1200 m of road is consistent with an 8° slope (use Pythagoras to estimate). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (tower 60 m, θ = 35°)
Step 1: opp = 60, θ = 35°, ratio is tan.
Step 2: tan 35° = 60 / adj.
Step 3: adj = 60 / tan 35°.
Step 4: adj ≈ 60 / 0.7002 ≈ 85.69 m.
Step 5: For 35° < 45°, adj > opp expected. 85.69 > 60 ✓.
3.1 — adj = 20, θ = 35°
opp = 20 tan 35° ≈ 20 × 0.7002 ≈ 14.00.
3.2 — adj = 12, θ = 45°
opp = 12 tan 45° = 12 × 1 = 12 exactly. (At 45°, opp = adj — isosceles right triangle.)
3.3 — opp = 9, θ = 25°
adj = 9 / tan 25° ≈ 9 / 0.4663 ≈ 19.30.
3.4 — opp = 7, θ = 60°
adj = 7 / tan 60° ≈ 7 / 1.7321 ≈ 4.04.
3.5 — Flagpole at 30 m, 22°
Flagpole height (above eye level) = 30 tan 22° ≈ 30 × 0.4040 ≈ 12.12 m.
3.6 — 45 m radio tower at 18°
Distance from base: adj = 45 / tan 18° ≈ 45 / 0.3249 ≈ 138.50 m.
3.7 — Cliff height, then new angle
Height = 80 tan 35° ≈ 80 × 0.7002 ≈ 56.01 m.
From 60 m: tan θ_new = 56.01 / 60 ≈ 0.9335. θ_new = tan⁻¹(0.9335) ≈ 43°.
Moving closer makes the angle steeper — that's why you "look up more" as you approach a tall cliff.
3.8 — 8° slope, 1100 m horizontal
Vertical drop = 1100 tan 8° ≈ 1100 × 0.1405 ≈ 154.59 m.
Slope ≈ √(1100² + 154.59²) ≈ √(1,210,000 + 23,898) ≈ √1,233,898 ≈ 1110.81 m.
1200 m of road is NOT consistent — the actual slope distance for a true 8° road over 1100 m horizontal is about 1111 m. 1200 m would imply either a steeper angle or a longer horizontal.