Mathematics Standard • Year 11 • Module 2 • Lesson 15
Right-Angled Trig: Finding Angles — Problem Set
Apply inverse trig to safety regulations, aviation, surveying, sport and engineering scenarios — interpret the angle in context, not just compute it.
Problem 1 — Telephone-pole support wire (safety check)
A 15 m vertical telephone pole has a support wire from its top to an anchor on the ground. Safety regulations require the angle between the wire and the ground to be between 55° and 70°.
Set up: What are we solving for?
(i) If the wire is anchored 9 m from the base of the pole, find the angle the wire makes with the ground, in degrees and minutes. 2 marks
(ii) Does this satisfy the regulation? Justify in one sentence. 1 mark
(iii) Find the minimum anchor distance (in metres, to 1 d.p.) at which the angle would still satisfy the regulation. 2 marks
Stuck on (iii)? The regulation upper bound is 70°. tan 70° = 15 ÷ d_min ⇒ d_min = 15 ÷ tan 70°.Problem 2 — Pilot's angle of depression to a runway
A pilot is flying at an altitude of 1500 m. She spots a runway 4200 m horizontally from directly below the aircraft.
Set up: What are we solving for?
(i) Sketch the labelled right-angled triangle (right angle at the point directly below the aircraft; altitude 1500 m vertical; horizontal distance 4200 m; angle of depression at the aircraft). 1 mark
(ii) Find the angle of depression from the aircraft to the runway, in degrees and minutes. 2 marks
(iii) The pilot wants to begin a steady descent so that the angle of depression to the runway threshold stays the same all the way down. By what angle (to the nearest minute) would the descent path deviate from horizontal? Explain in one sentence why this answer equals the angle in (ii). 2 marks
Stuck on (iii)? The descent path follows the same line of sight; the descent angle equals the angle of depression.Problem 3 — Skate park drop-in ramp
A skate park is fitting a drop-in ramp. The vertical drop is 2.4 m and the curved (run) distance along the ground is 3.6 m. The local council wants the angle of inclination between 30° and 45°.
Set up: What are we solving for?
(i) Treating the ramp as the hypotenuse of a right-angled triangle with vertical leg 2.4 m and horizontal leg 3.6 m, sketch and label the triangle. 1 mark
(ii) Find the angle of inclination of the ramp, in degrees and minutes. 2 marks
(iii) Does this design satisfy the council's 30°-to-45° requirement? Justify in one sentence using your value from (ii). 2 marks
Stuck? With O = 2.4 m and A = 3.6 m: θ = tan⁻¹(2.4 ÷ 3.6).Problem 4 — Bushwalker's bearing
A bushwalker starts at a campsite and walks 6.2 km due north, then 4.5 km due east, ending at a lookout. She wants to know the straight-line distance back to camp and the bearing from camp to the lookout.
Set up: What are we solving for?
(i) Find the straight-line distance from camp to lookout, to 2 d.p. (km). (Use Pythagoras.) 2 marks
(ii) Find the angle the camp→lookout line makes with due north, in degrees and minutes. 2 marks
(iii) Express the bearing from camp to lookout in the form N___°___'E. 1 mark
Stuck? The angle east of north equals tan⁻¹(east ÷ north) = tan⁻¹(4.5 ÷ 6.2).Problem 5 — Surveyor sighting a tower (two-step)
A surveyor uses a clinometer to measure the angle of elevation to the top of a vertical tower from two points on level ground. From point A, 120 m from the tower base, she records an angle of elevation. From point B, directly between A and the tower and 80 m from the tower base, the angle she records is different.
She measures the tower height (independently) as 96 m using a laser range-finder.
Set up: What are we solving for?
(i) Find the angle of elevation from A, in degrees and minutes. 2 marks
(ii) Find the angle of elevation from B, in degrees and minutes. 2 marks
(iii) State, in one sentence, why the angle at B is bigger than the angle at A, even though the same tower is being measured. 1 mark
Stuck? B is closer, so the same height "fills" a larger fraction of the field of view → larger angle.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Pole support wire
Set up. Use tan⁻¹(15/9) for the angle, then check against 55°-70°; for the minimum-distance case, the angle is 70°.
(i) tan θ = 15/9 = 1.6667; θ = tan⁻¹(1.6667) ≈ 59.04° = 59°2'.
(ii) 55° < 59°2' < 70° → Yes, the angle satisfies the regulation.
(iii) At the upper-bound angle 70°: tan 70° = 15 ÷ d_min ⇒ d_min = 15 ÷ tan 70° ≈ 15 ÷ 2.7475 ≈ 5.5 m (any larger anchor distance lowers the angle below 70°, so 5.5 m is the minimum that still meets the regulation).
Problem 2 — Pilot at 1500 m, runway 4200 m horizontal
Set up. Right angle at point directly below aircraft; depression at aircraft.
(i) Diagram: vertical leg 1500 m down to point below aircraft; horizontal leg 4200 m to runway; hypotenuse line-of-sight; depression angle at aircraft between horizontal reference and line-of-sight.
(ii) tan θ = 1500 / 4200 ≈ 0.3571; θ = tan⁻¹(0.3571) ≈ 19.65° = 19°39'.
(iii) The descent path angle is the same as the angle of depression — it is exactly 19°39' below horizontal, because the descent line follows the same line-of-sight that defined the angle of depression in the first place.
Problem 3 — Skate park ramp
Set up. Use tan⁻¹(2.4 / 3.6); compare to 30°-45° band.
(i) Diagram: right angle at the base; vertical 2.4 m; horizontal 3.6 m; angle of inclination at the base of the ramp (between horizontal and ramp surface).
(ii) θ = tan⁻¹(2.4 / 3.6) = tan⁻¹(0.6667) ≈ 33.69° = 33°41'.
(iii) 30° < 33°41' < 45° → Yes, the ramp satisfies the council's inclination range.
Problem 4 — Bushwalker bearing
Set up. Pythagoras for distance; tan⁻¹(east/north) for bearing.
(i) d = √(6.2² + 4.5²) = √(38.44 + 20.25) = √58.69 ≈ 7.66 km.
(ii) tan⁻¹(4.5 / 6.2) = tan⁻¹(0.7258) ≈ 35.97° = 35°58'.
(iii) Bearing from camp to lookout = N35°58'E.
Problem 5 — Surveyor sights tower
Set up. Same tower height 96 m, two ground distances 120 m and 80 m; use tan⁻¹(96 / ground distance).
(i) From A: tan⁻¹(96/120) = tan⁻¹(0.8) ≈ 38.66° = 38°40'.
(ii) From B: tan⁻¹(96/80) = tan⁻¹(1.2) ≈ 50.19° = 50°12'.
(iii) B is closer to the tower, so the same height (96 m) sits over a smaller horizontal distance — the height is a larger fraction of the base, so tan⁻¹ returns a larger angle.