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.