Mathematics Standard • Year 11 • Module 2 • Lesson 16

Angles of Elevation and Depression — Skill Drill

Build fluency with the elevation/depression setup: draw the horizontal reference line, place the right angle correctly, then apply SOHCAHTOA.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 The angle of elevation is measured ____________ from the horizontal. The angle of depression is measured ____________ from the horizontal.

Q1.2 When the horizontal lines at A (below) and B (above) are parallel, the angle of depression from B to A equals the angle of ____________ from A to B (reason: ____________ angles).

Q1.3 In an elevation/depression triangle, the right angle is placed at the foot of the ____________ (vertical) line, not at the observer.

Stuck? Revisit lesson § Looking Up and § Looking Down. Horizontal reference line is non-negotiable.

2. Worked example — basic angle of elevation

Follow each line of working. Every step has a reason on the right.

Problem. From a point 45 m from the base of a vertical building, the angle of elevation to the top is 38°. Find the height of the building, to 2 d.p.

Step 1 — Diagram and ratio.

A = 45 m (horizontal), O = h (height), angle 38°. Use tan θ = O ÷ A.

Reason: no hypotenuse involved → tan; horizontal A, vertical O.

Step 2 — Rearrange.

h = 45 × tan 38°

Reason: unknown in numerator; multiply both sides by 45.

Step 3 — Evaluate.

h = 45 × 0.7813... ≈ 35.16 m

Reason: tan 38° ≈ 0.7813; round to 2 d.p.

Conclusion. The building height is 35.16 m.

3. Faded example — angle of depression (use equal angles)

From the top of a cliff 80 m high, a boat at sea is observed with an angle of depression of 24°. Find the horizontal distance from the base of the cliff to the boat, to 2 d.p. Fill in each blank. 4 marks

Step 1 — Redraw using equal angles: the angle of elevation at the boat = ____________ ° (alternate angles).

Step 2 — Identify O, A, ratio:

O = ____________ m (cliff), A = x (horizontal), θ = ____________ °. Use tan θ = O ÷ A.

Step 3 — Rearrange for x (unknown in denominator):

x = ____________ ÷ tan ____________ °

Step 4 — Evaluate:

x = 80 ÷ ____________ ≈ ____________ m (to 2 d.p.)

Conclusion. The horizontal distance is ____________ m.

Stuck? Revisit lesson § Worked Example 2 — Angle of Depression. Equal-alternate-angles is the key trick.

4. Graduated practice — Elevation, depression, and finding angles

For every question: draw a labelled diagram (horizontal reference line, angle, right angle at the foot of the vertical) before computing.

Foundation — basic elevation and depression (4 questions)

Q	Problem	Answer (to 2 d.p. unless noted)
4.1 1	From 60 m from the base of a vertical tree, angle of elevation = 42°. Find height.
4.2 1	From the top of a 120 m cliff, angle of depression to a boat = 31°. Find horizontal distance to the boat.
4.3 1	From a lighthouse 65 m above sea level, a ship has angle of depression 18°. Find the straight-line distance (line of sight) from lighthouse to ship.
4.4 1	Observer at ground level, building top is 35 m above observer's eye level and 50 m horizontally away. Find angle of elevation to the nearest minute.

Standard — typical HSC difficulty (6 questions)

Show ratio, substitution and rounding. Where degrees and minutes are asked, convert at the end.

4.5 A 15 m flagpole stands on top of a 25 m building. An observer is 80 m from the base. Find the angle of elevation to the top of the flagpole, to the nearest minute. 2 marks

4.6 From the top of a 90 m tower, a car on the road below is 140 m from the tower base. Find the angle of depression, to the nearest minute. 2 marks

4.7 An observer is 100 m from a tower. Angle of elevation to the top of the tower is 48°. A flag on top adds 5 m of height. Find the new angle of elevation to the top of the flag, to the nearest minute. 3 marks

4.8 From 30 m from the base of a vertical wall, angle of elevation to the top = 56°. Find the height. 2 marks

4.9 From the top of a 50 m building, angle of depression to a car on the street = 32°. Find the horizontal distance to the car. 2 marks

4.10 A boat is 220 m horizontally from a 34 m cliff. Find the angle of elevation from the boat to the top of the cliff, to the nearest minute. 2 marks

Extension — two-observation problems (2 questions)

4.11 From point P, the angle of elevation to the top of a tower is 45°. From point Q, 20 m further from the tower than P (on the same side, both on level ground), the angle of elevation is 30°. Find the height of the tower, to 2 d.p. 3 marks

4.12 Two observers on opposite sides of a vertical cliff: Observer A is 80 m from the base on level ground, with angle of elevation 55°. Observer B on the other side has angle of elevation 40°. Find the cliff height (to 2 d.p.) and the total horizontal distance between the two observers (to 2 d.p.). 3 marks

Stuck on 4.11? Let h = height, d = distance from P. From P: h = d × tan 45° = d. From Q: h = (d + 20) × tan 30°. Set equal and solve for d.

5. Self-check the easy 3

Tick the first three once you've checked your method works.

For 4.1 I drew the horizontal reference line from the observer (not the tree), placed the right angle at the base of the tree, and used tan.

For 4.2 I redrew using equal alternate angles — the 31° appears at the boat — then used tan with the unknown in the denominator.

For 4.4 I used tan⁻¹(35 ÷ 50) and converted the decimal degrees to minutes at the end.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Direction of measurement

Elevation: upward from horizontal. Depression: downward from horizontal.

Q1.2 — Equal angles

Angle of depression from B = angle of elevation from A. Reason: alternate interior angles (or alternate angles) between parallel horizontal lines cut by the line of sight.

Q1.3 — Right-angle location

At the foot of the vertical (height) line — not at the observer.

Q3 — Faded example (cliff 80 m, depression 24°)

Step 1: Equal-angles → elevation at boat = 24°.
Step 2: O = 80, θ = 24°. Use tan θ = O ÷ A.
Step 3: x = 80 ÷ tan 24°.
Step 4: x = 80 ÷ 0.4452 ≈ 179.69 m.
Conclusion: horizontal distance ≈ 179.69 m.

Q4.1 — Tree from 60 m, elevation 42°

h = 60 × tan 42° ≈ 60 × 0.9004 ≈ 54.05 m.

Q4.2 — Cliff 120 m, depression 31°

Elevation at boat = 31°. d = 120 ÷ tan 31° ≈ 120 ÷ 0.6009 ≈ 199.71 m.

Q4.3 — Lighthouse 65 m, depression 18° — line of sight

Line of sight is the hypotenuse: sin 18° = 65 ÷ H ⇒ H = 65 ÷ sin 18° ≈ 65 ÷ 0.3090 ≈ 210.34 m.

Q4.4 — Tan⁻¹(35/50)

tan⁻¹(0.7) ≈ 34.99° = 34°59'.

Q4.5 — Flagpole on building, observer 80 m away

Total height = 25 + 15 = 40 m. tan θ = 40 ÷ 80 = 0.5; θ = tan⁻¹(0.5) ≈ 26.565° = 26°34'.

Q4.6 — Tower 90 m, car 140 m from base

tan θ = 90 ÷ 140 ≈ 0.6429; θ = tan⁻¹(0.6429) ≈ 32.74° = 32°44'.

Q4.7 — Tower + flag (5 m), observer 100 m

Tower height = 100 × tan 48° ≈ 100 × 1.1106 ≈ 111.06 m. With flag: total = 116.06 m. New angle = tan⁻¹(116.06/100) = tan⁻¹(1.1606) ≈ 49.26° = 49°16'.

Q4.8 — Wall from 30 m, elevation 56°

h = 30 × tan 56° ≈ 30 × 1.4826 ≈ 44.47 m.

Q4.9 — Building 50 m, depression 32°

Elevation at car = 32°. d = 50 ÷ tan 32° ≈ 50 ÷ 0.6249 ≈ 80.02 m.

Q4.10 — Cliff 34 m, boat 220 m horizontal

tan θ = 34 ÷ 220 ≈ 0.1545; θ = tan⁻¹(0.1545) ≈ 8.78° = 8°47'.

Q4.11 — Two-observation (45° from P, 30° from Q, gap 20 m)

From P: h = d × tan 45° = d. From Q: h = (d + 20) × tan 30°. Set equal: d = (d + 20) × tan 30° ⇒ d − d × tan 30° = 20 tan 30° ⇒ d(1 − tan 30°) = 20 tan 30° ⇒ d = (20 × tan 30°) ÷ (1 − tan 30°) ≈ (20 × 0.5774) ÷ (1 − 0.5774) ≈ 11.547 ÷ 0.4226 ≈ 27.32 m. So h ≈ 27.32 m.

Q4.12 — Cliff with observers on both sides

From A: h = 80 × tan 55° ≈ 80 × 1.4281 ≈ 114.25 m. From B: d_B = h ÷ tan 40° ≈ 114.25 ÷ 0.8391 ≈ 136.16 m. Total distance between observers = 80 + 136.16 = 216.16 m.