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.