Mathematics • Year 10 • Unit 3 • Lesson 4
Angles of Elevation & Depression — Skill Drill
Build fluency with the two key definitions from Lesson 4: angle of elevation (upward from horizontal to line of sight) and angle of depression (downward from horizontal to line of sight). Always start with a labelled diagram, mark the horizontal, then identify the right-angled triangle. Use the alternate-angles property where useful.
1. I do — fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. Maya stands 40 m from the base of a tower in Parramatta. The angle of elevation from her eyeline to the top of the tower is 28°. Assuming her eyes are at ground level, find the height of the tower correct to 1 decimal place.
Step 1 — Draw and label.
Horizontal ground 40 m, vertical tower h, angle of elevation = 28°.
Reason: the diagram is the most important step — examiners give marks for it.
Step 2 — Identify the right-angled triangle.
Angle = 28° (at observer), A = 40 (horizontal), O = h (vertical = tower).
Reason: the right angle is between the horizontal ground and the vertical tower. The elevation is at the observer.
Step 3 — Pick the ratio.
O + A → TAN. tan 28° = h / 40.
Reason: the hypotenuse is not involved, so tangent is the cleanest choice.
Step 4 — Rearrange and calculate.
h = 40 × tan 28° = 40 × 0.5317 = 21.27 m
Reason: unknown on top means multiply (DEG mode!).
Step 5 — Round and write the unit.
h ≈ 21.3 m (to 1 d.p.)
Answer: the tower is approximately 21.3 m tall.
2. We do — fill in the missing steps
Same five-step structure as Section 1. Fill in each blank. 5 marks
Problem. A lifeguard on top of a 12 m observation tower at Manly looks down at a swimmer in distress. The angle of depression from the lifeguard's eyes to the swimmer is 18°. How far is the swimmer from the base of the tower? Round to 1 d.p.
Step 1 — Draw the picture:
Vertical tower ______ m, horizontal water surface, depression angle ______° measured from the horizontal at the lifeguard's eyes downward.
Step 2 — Use the alternate-angles trick:
The angle of elevation from the swimmer up to the lifeguard equals the angle of depression — so the angle at the base of the triangle (at the swimmer) is also ______°.
Step 3 — Pick the ratio (relative to the angle at the swimmer):
O = ______ , A = ______ , ratio = ______ ______° = ______ / ______
Step 4 — Rearrange and calculate:
A = 12 / tan 18° = 12 / ______________ = ______________
Step 5 — Round and unit:
A ≈ ______________ m (to 1 d.p.)
3. You do — independent practice
Show working. Round side lengths to 1 d.p. and angles to the nearest degree unless told otherwise.
Foundation — direct elevation / depression set-ups
3.1 A 15 m flagpole casts a shadow on the ground. The angle of elevation of the sun is 35°. Find the length of the shadow. 1 mark
3.2 From a point 50 m from the base of a vertical cliff, the angle of elevation to the top is 42°. Find the height of the cliff. 1 mark
3.3 A drone hovers 60 m vertically above its operator. The drone observes a person on the ground. The angle of depression from the drone to that person is 25°. Find the horizontal distance from the operator (directly below the drone) to the person. 1 mark
3.4 A boat is 200 m horizontally from the base of a lighthouse, and the top of the lighthouse is 32 m above sea level. Find the angle of depression from the top of the lighthouse to the boat. 1 mark
Standard — two-step
3.5 Anya is 1.6 m tall. Standing 8 m from a tree, the angle of elevation from her eyes to the top of the tree is 50°. Find the total height of the tree. (Hint: the trig gives you the height above her eyes — add her height.) 2 marks
3.6 From a lookout 80 m above a lake, the angle of depression to a boat is 12°. Use the alternate-angles trick to identify the angle inside the triangle at the boat, then find the horizontal distance from directly below the lookout to the boat. 2 marks
Extension — push your thinking
3.7 A surveyor measures the angle of elevation to the top of a building as 38° from a certain point. She walks 20 m closer and the angle of elevation now becomes 50°. Find the height of the building. (Hint: set up two equations using tan and the unknown distance from the closer point to the base.) 3 marks
3.8 A helicopter is 200 m above the ground. The pilot looks down and sees a fire at angle of depression 40°, and a fire-truck behind the fire (further from the helicopter) at angle of depression 25°. Find the horizontal distance between the fire and the fire-truck. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (Manly lifeguard)
Step 1: 12 m tower, 18° depression.
Step 2: Angle of elevation at the swimmer equals 18° (alternate angles between parallel horizontals).
Step 3: O = 12 (tower), A = distance (swimmer to base). tan 18° = 12 / A.
Step 4: A = 12 / tan 18° = 12 / 0.3249 = 36.9293.
Step 5: A ≈ 36.9 m (to 1 d.p.).
3.1 — Flagpole shadow
tan 35° = 15 / shadow → shadow = 15 / tan 35° = 15 / 0.7002 = 21.43 ≈ 21.4 m.
3.2 — Cliff height
tan 42° = h / 50 → h = 50 × tan 42° = 50 × 0.9004 = 45.02 ≈ 45.0 m.
3.3 — Drone to person
tan 25° = 60 / d → d = 60 / tan 25° = 60 / 0.4663 = 128.66 ≈ 128.7 m.
3.4 — Lighthouse depression angle
tan θ = 32 / 200 = 0.16, so θ = tan⁻¹(0.16) = 9.09° ≈ 9°.
3.5 — Anya + tree
Height above her eyes: tan 50° = h / 8 → h = 8 × tan 50° = 8 × 1.1918 = 9.53 m.
Total tree height = 9.53 + 1.6 = 11.1 m (to 1 d.p.).
Always add the observer's eye height when the question asks for the actual height of the object.
3.6 — Lookout depression
Alternate angles: the angle at the boat is also 12°.
tan 12° = 80 / A → A = 80 / tan 12° = 80 / 0.2126 = 376.28 ≈ 376.3 m.
3.7 — Surveyor walking closer
Let d = distance from closer point to base; let h = height.
From closer point: tan 50° = h / d. → h = d × tan 50°.
From further point: tan 38° = h / (d + 20). → h = (d + 20) × tan 38°.
Set equal: d × tan 50° = (d + 20) × tan 38°.
d × 1.1918 = (d + 20) × 0.7813
1.1918 d = 0.7813 d + 15.626
0.4105 d = 15.626 → d = 38.07 m.
Height h = 38.07 × tan 50° = 38.07 × 1.1918 = 45.37 ≈ 45.4 m.
3.8 — Helicopter, fire and truck
Horizontal distance to fire: tan 40° = 200 / d_fire → d_fire = 200 / tan 40° = 200 / 0.8391 = 238.4 m.
Horizontal distance to truck: tan 25° = 200 / d_truck → d_truck = 200 / tan 25° = 200 / 0.4663 = 428.9 m.
Distance between fire and truck = 428.9 − 238.4 = 190.5 m.