Mathematics • Year 10 • Unit 3 • Lesson 19
Applications — Master Challenge
Pull every Lesson 19 idea together: triangulation from two angles, multi-leg navigation with bearings, two-point elevation, depression problems, back-bearings, and a self-designed surveying scenario. Diagram first, formula second, sanity check always.
1. Mixed problems — choose the right idea
For each: draw a diagram (with North if bearings are involved), pick the formula, solve. Round lengths to 1 dp / nearest metre and angles/bearings to the nearest degree unless told otherwise. 2-3 marks each
1.1 Two surveyors 80 m apart sight a chimney. The first sees it at angle 50° from their baseline; the second at angle 65°. Find the distance from the first surveyor to the chimney. 2 marks
1.2 A bushwalker walks 5 km on bearing 045°, then 8 km on bearing 135°. Find (a) the direct distance back to the start, (b) the bearing back to the start. 3 marks
1.3 A drone hovers 60 m above ground. The angle of depression to a target on the ground is 24°. Find the horizontal distance from the drone to the target. 2 marks
1.4 A flagpole on top of a building is observed from a point 40 m from the building. The angle of elevation to the bottom of the flagpole (top of the building) is 35°. The angle of elevation to the top of the flagpole is 42°. Find the length of the flagpole. 3 marks
1.5 A ship sails 20 km East then 15 km South. Find the bearing of the start from the ship's current position. 2 marks
1.6 A plane flies 100 km on bearing 060°, then 80 km on bearing 150°. The bearing change is 90° (interior angle 90°). Find the area of the triangle traced out by start, turn and finish, in km². 2 marks
2. Find the mistake
A Year 10 student tries to find the bearing back to start for a ship that sails 30 km on bearing 080°, then 40 km on bearing 150°. Their working is shown. Exactly one step contains a mistake. Spot it, explain why, then re-do correctly. 3 marks
Student's working:
Line 1: Bearing change = 150° − 80° = 70°.
Line 2: So the interior angle at the turn = 70°.
Line 3: d² = 30² + 40² − 2(30)(40) cos 70° = 2500 − 2400 × 0.342 = 2500 − 821 = 1679.
Line 4: d = √1679 ≈ 41.0 km.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Re-do the working from that line and give the corrected direct distance.
Stuck? Picture the ship continuing in its original direction (080°) and then needing to swing onto 150°. The 70° bearing change is the external turning angle — the interior angle of the triangle is the supplement, 180° − 70°.3. Open-ended challenge — design a surveying problem
Be creative but show every number. 4 marks
3.1 Design a real Australian surveying scenario with a one-sentence context (e.g. a council surveying a triangular park, a hiker reporting back to base, a yacht race) that requires you to use all three of the Lesson 19 building blocks: (a) a bearing/back-bearing, (b) an angle of elevation or depression, and (c) the sine rule or cosine rule to find a missing length or angle.
For your scenario:
(i) State the context and all given numbers clearly.
(ii) Draw a labelled diagram (with North if relevant).
(iii) Solve all three parts and report the final answers with sensible units and rounding.
Bonus: Quote at least one of your final answers to the appropriate number of significant figures and justify the choice in one sentence.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Two surveyors 80 m apart, angles 50° and 65°
Third angle = 180° − 50° − 65° = 65°. Distance from first surveyor is opposite the 65° angle at the second surveyor's end:
d / sin 65° = 80 / sin 65°, so d = 80 m. (Isosceles triangle — the two base angles are equal.)
1.2 — Bushwalker (5 km on 045°, 8 km on 135°)
Bearing change = 135° − 45° = 90°. Interior angle at turn = 90°.
(a) Right triangle: d = √(5² + 8²) = √89 ≈ 9.4 km.
(b) Angle off first leg back to start: tan α = 8/5 → α ≈ 58.0°. Bearing from start to finish = 045° + 58.0° = 103.0°. Bearing back to start = 103.0° + 180° = 283° (3 s.f.).
1.3 — Drone (60 m up, depression 24°)
tan 24° = 60 / d → d = 60 / 0.4452 ≈ 134.8 m.
1.4 — Flagpole on building (40 m away, elevations 35° and 42°)
Height to top of building = 40 × tan 35° = 40 × 0.7002 ≈ 28.0 m.
Height to top of flagpole = 40 × tan 42° = 40 × 0.9004 ≈ 36.0 m.
Flagpole length = 36.0 − 28.0 = 8.0 m.
1.5 — Ship 20 km East then 15 km South
Ship is now 20 km E and 15 km S of the start, i.e. on bearing 180° − tan⁻¹(20/15) from the start. Easier: bearing of start from ship is in the NW quadrant. From the ship, start is 20 km West and 15 km North. Angle West of North = tan⁻¹(20/15) = tan⁻¹(1.333) ≈ 53.1°.
Bearing = 360° − 53.1° = 306.9° ≈ 307°.
1.6 — Plane (100 km on 060°, 80 km on 150°)
Bearing change = 90° → interior angle 90° → right triangle. Area = ½ × 100 × 80 = 4 000 km².
2 — Find the mistake
(a) The mistake is on Line 2.
(b) A 70° change of bearing is the external turning angle (how far the ship swings round). The interior angle of the triangle at the turn is its supplement, 180° − 70° = 110°. The student used 70° in the cosine rule instead of 110°.
(c) Corrected (using interior angle 110°):
d² = 30² + 40² − 2(30)(40) cos 110° = 2 500 − 2 400 × (−0.342) = 2 500 + 820.8 = 3 320.8.
d = √3 320.8 ≈ 57.6 km. (Compare with the student's wrong 41.0 km — the obtuse interior angle makes the triangle's third side longer, not shorter.)
3 — Open-ended (sample scenario)
Scenario: A surveyor at start point S walks 200 m on bearing 060° to point A, then 150 m on bearing 130° to point B at the base of a tower. From A she measured the angle of elevation to the top of the tower (at B) as 7°. Find (a) the direct distance SB, (b) the back-bearing from B to S, (c) the height of the tower (measured from A's eye level).
(a) Cosine rule (SAS): Interior angle at A = 180° − (130° − 60°) = 110°. SB² = 200² + 150² − 2(200)(150) cos 110° = 40 000 + 22 500 − 60 000 × (−0.342) = 62 500 + 20 520 = 83 020. SB = √83 020 ≈ 288 m.
(b) Bearing/back-bearing: Find angle at S using sine rule: sin θ / 150 = sin 110° / 288. sin θ = 150 × 0.9397 / 288 ≈ 0.490. θ ≈ 29.3°. Bearing S→B = 060° + 29.3° = 089.3°. Back-bearing B→S = 089.3° + 180° = 269.3° ≈ 269°.
(c) Elevation: The horizontal distance from A to the tower's base B is 150 m. Tower height (above A's eye) = 150 × tan 7° = 150 × 0.1228 ≈ 18.4 m.
Bonus: The input distances were given to 3 s.f. (200, 150) and the angle to 2 s.f., so quoting the tower height as "18 m" (2 s.f.) is most defensible. Reporting "18.42 m" would imply false precision.
Marking: 1 mark for a coherent scenario with clean numbers, 1 mark for a correct cosine-rule application, 1 mark for a correctly computed back-bearing, 1 mark for a sensible elevation calculation and rounding justification.