Mathematics • Year 9 • Unit 3 • Lesson 15
Combined Elevation and Depression
Build fluency with two-angle problems: scenarios where you must split a situation into TWO right triangles (often sharing a side) and set up one trig equation per triangle. Walk from a fully worked "two boats from a cliff" problem through one guided fill-in to eight graduated practice problems.
1. I do — fully worked example
Read every step. The standard pattern for two-angle problems: sketch ONE big diagram showing both angles, identify the TWO triangles, label them, write one trig equation per triangle, combine.
Problem. From a 50 m cliff, boat A is at depression 30° and boat B is at depression 18° (further out, same straight line). Find the distance between them (2 d.p.).
Step 1 — Sketch ONE big diagram.
Draw the cliff top with the horizontal at the observer's eye. Mark TWO depression angles going down: 30° to nearby boat A, 18° to farther boat B. Vertical drop is 50 m (same for both).
Reason: both triangles share the 50 m vertical drop as their OPPOSITE side.
Step 2 — Identify the two right triangles.
Triangle 1: cliff top, boat A, point directly below the cliff. opp = 50, θ₁ = 30°, adj = d_A.
Triangle 2: cliff top, boat B, same point below cliff. opp = 50, θ₂ = 18°, adj = d_B.
Reason: the cliff height (50 m) is the SHARED side.
Step 3 — Write one trig equation per triangle.
tan 30° = 50 / d_A → d_A = 50 / tan 30°
tan 18° = 50 / d_B → d_B = 50 / tan 18°
Reason: opp + adj → tan in each triangle.
Step 4 — Compute each distance.
d_A = 50 / tan 30° ≈ 50 / 0.5774 ≈ 86.60 m (boat A, closer)
d_B = 50 / tan 18° ≈ 50 / 0.3249 ≈ 153.88 m (boat B, further)
Reason: smaller depression angle = boat is further away.
Step 5 — Subtract for the gap.
Gap = d_B − d_A ≈ 153.88 − 86.60 ≈ 67.28 m
Answer: the boats are ≈ 67.28 m apart.
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. From the top of a 30 m cliff, the angle of depression to a swimmer is 25° and the angle of depression to a boat further out is 12°. Find the distance between the swimmer and the boat (2 d.p.).
Step 1 — Sketch: ONE diagram showing both depression angles from the cliff top. Cliff height = ________ m (shared opp).
Step 2 — Two triangles:
Swimmer triangle: opp = ________, θ = ________ °, adj = d_S.
Boat triangle: opp = ________, θ = ________ °, adj = d_B.
Step 3 — Equations:
d_S = ________ / tan ________ ° ≈ ________ m
d_B = ________ / tan ________ ° ≈ ________ m
Step 4 — Gap:
Gap = ________ − ________ ≈ ________ m
Step 5 — State: the swimmer and boat are ≈ ________ m apart.
3. You do — independent practice
Show working under each problem. Problems 3.1–3.4 are foundation (find one distance given an angle). Problems 3.5–3.6 are standard (two-angle subtract-for-the-gap). Problems 3.7–3.8 are extension (find an unknown height using two equations).
Foundation — single-angle reminders
3.1 From an 80 m cliff, the depression to a single boat is 35°. Find its horizontal distance from the cliff base (2 d.p.). 1 mark
3.2 A tower is 20 m away from an observer at ground level. The angle of elevation to the top is 60°. Find the tower's height (2 d.p.). 1 mark
3.3 From eye height 1.7 m, a wall is 8 m away. The top of the wall is at elevation 25°. Find the wall's height above the GROUND (2 d.p.). 1 mark
3.4 A bird sits 5 m above eye level and 10 m horizontally from the observer. Find the angle of elevation to the bird (1 d.p.). 1 mark
Standard — two depression angles, find the gap
3.5 From a 100 m cliff: boat A at depression 45°, boat B at depression 30° (both on the same straight line out from the cliff base). Find the distance between A and B (2 d.p.). 2 marks
3.6 From a lookout 75 m above a road: a car ahead at depression 22°, a truck further along at depression 14°. Find the distance between the car and the truck (2 d.p.). 2 marks
Extension — two-equation method
3.7 From point A on flat ground, a tower top is seen at elevation 25°. Walking 30 m CLOSER to the tower (point B), the elevation rises to 40°. Find the tower's height (2 d.p.). [Hint: let h = height, d = distance from B. Then h = d × tan 40° = (d + 30) × tan 25°.] 3 marks
3.8 From a balcony 10 m above the ground, the observer sees a bird above at elevation 20° and a dog below at depression 35°. Assume both are at the SAME horizontal distance d = 8 m. (a) Find the bird's height above the observer's eye. (b) Find the dog's vertical distance below the observer's eye. (c) Find the total vertical distance between the bird and the dog. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (30 m cliff, depressions 25° and 12°)
Step 1: cliff height = 30 m (shared opp).
Step 2: swimmer triangle opp = 30, θ = 25°; boat triangle opp = 30, θ = 12°.
Step 3: d_S = 30 / tan 25° ≈ 64.34 m; d_B = 30 / tan 12° ≈ 141.14 m.
Step 4: gap = 141.14 − 64.34 ≈ 76.80 m.
Step 5: swimmer and boat are ≈ 76.80 m apart.
3.1 — 80 m cliff, depression 35°
adj = 80 / tan 35° ≈ 80 / 0.7002 ≈ 114.25 m.
3.2 — Tower at 20 m, elevation 60°
height = 20 × tan 60° ≈ 20 × 1.7321 ≈ 34.64 m.
3.3 — Wall, 8 m away, elevation 25°, eye 1.7 m
Above eye: 8 × tan 25° ≈ 8 × 0.4663 ≈ 3.73 m. Total above ground: 3.73 + 1.7 ≈ 5.43 m.
3.4 — Bird 5 m above eye, 10 m away
tan θ = 5 / 10 = 0.5. θ = tan⁻¹(0.5) ≈ 26.6°.
3.5 — 100 m cliff, depressions 45° and 30°
d_A = 100 / tan 45° = 100 m. d_B = 100 / tan 30° ≈ 100 / 0.5774 ≈ 173.21 m.
Gap = 173.21 − 100 ≈ 73.21 m between the boats.
3.6 — 75 m lookout, depressions 22° and 14°
d_car = 75 / tan 22° ≈ 75 / 0.4040 ≈ 185.66 m. d_truck = 75 / tan 14° ≈ 75 / 0.2493 ≈ 300.85 m.
Gap = 300.85 − 185.66 ≈ 115.19 m between the car and the truck.
3.7 — Two-observer tower (closer this time)
Let d = distance from B (closer point) to the tower base. Then h = d × tan 40° (from B) and h = (d + 30) × tan 25° (from A, further away).
Equate: d × tan 40° = (d + 30) × tan 25°.
d × (tan 40° − tan 25°) = 30 × tan 25°.
d = 30 × tan 25° / (tan 40° − tan 25°) ≈ 30 × 0.4663 / (0.8391 − 0.4663) ≈ 13.99 / 0.3728 ≈ 37.52 m.
h = 37.52 × tan 40° ≈ 37.52 × 0.8391 ≈ 31.48 m.
3.8 — Bird above + dog below, both at d = 8 m
(a) Bird above eye = 8 × tan 20° ≈ 8 × 0.3640 ≈ 2.91 m.
(b) Dog below eye = 8 × tan 35° ≈ 8 × 0.7002 ≈ 5.60 m.
(c) Total vertical gap = 2.91 + 5.60 ≈ 8.52 m between bird and dog.