Mathematics • Year 10 • Unit 3 • Lesson 19
Geometric Applications — Skill Drill
Build fluency with the three application building blocks from Lesson 19: triangulation (sine rule from two known angles and one side), bearings (clockwise from North 000° to 360°, with back-bearing = ±180°), and angles of elevation/depression (h = d × tan θ). Diagram first, then formula.
1. I do — fully worked example
Read every step. The diagram is the key — once it is right, the formula falls out.
Problem. Points A and B are 200 m apart on flat ground. From A, the angle to a tree T is 35°. From B, the angle to T is 55°. Find the distance from A to T.
Step 1 — Draw and label the triangle.
△ABT with AB = 200 m, ∠BAT = 35°, ∠ABT = 55°. We want AT.
Reason: two angles and the side between them = ASA pattern.
Step 2 — Find the third angle.
∠ATB = 180° − 35° − 55° = 90°
Step 3 — Apply the sine rule.
AT / sin 55° = 200 / sin 90°
AT = 200 × sin 55° / 1 = 200 × 0.819
Step 4 — Calculate and verify.
AT ≈ 163.8 m
Check: BT = 200 × sin 35° ≈ 114.7 m. Pythagoras: √(163.8² + 114.7²) ≈ 200 ✓
Answer: AT ≈ 164 m (3 s.f.).
2. We do — fill in the missing steps
Angle of elevation. Fill in each blank. 5 marks
Problem. From a point 80 m horizontally from the base of a flagpole at Centennial Park, the angle of elevation to the top of the flagpole is 25°. Find the height of the flagpole.
Step 1 — Draw and label. A right-angled triangle with horizontal distance ______ m along the ground, height ______ at the top, and angle of elevation ______° at the ground.
Step 2 — Identify the trig ratio. Relative to 25°, the height is the ______ side and 80 m is the ______ side. The ratio that pairs these is ______ θ = ______ / ______.
Step 3 — Substitute:
tan 25° = h / 80
h = 80 × tan 25° = 80 × ______ ≈ ______
Step 4 — Final answer with units:
Height ≈ ______ m
Step 5 — Sanity check. The angle is small (only 25°), so the height should be (less than / more than) the horizontal distance of 80 m: ______. Is your answer consistent? ______
3. You do — independent practice
Show your working. Round lengths to 1 dp, angles to 0.1°, bearings to the nearest degree.
Foundation — single-step applications
3.1 From a point 100 m from the base of a building, the angle of elevation to the top is 30°. Find the building's height. 2 marks
3.2 A bearing of B from A is 070°. What is the back-bearing (bearing of A from B)? 1 mark
3.3 Two observers 50 m apart measure angles of 40° and 60° from each end of their baseline to a distant point. Find the distance from the first observer to that point. 2 marks
3.4 A flagpole casts a 12 m shadow. The angle of elevation from the tip of the shadow back to the top of the flagpole is 35°. Find the height of the flagpole. 2 marks
Standard — bearings and triangulation
3.5 A hiker walks 6 km on bearing 030°, then 5 km on bearing 120°. The change in bearing (120° − 30°) is 90°, so the interior angle at the turn is 180° − 90° = 90° (a right angle). Find the straight-line distance back to the start. 3 marks
3.6 From the deck of a Manly ferry, the angle of depression to a buoy is 18°. The deck is 12 m above water level. How far is the buoy from the ferry along the water? 3 marks
Extension — push your thinking
3.7 From two points 100 m apart on a straight road, the angles of elevation to the top of a tower are 25° (from the closer point) and 40° (from the farther point — sorry, swap that: 40° from the closer, 25° from the farther). Set up two equations using tan and solve for the tower height. 3 marks
3.8 A yacht sails 12 km on bearing 040°, then 15 km on bearing 100°. Find (a) the straight-line distance from start, and (b) the bearing of the start from the finish. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (Centennial Park flagpole)
Horizontal = 80 m, height = h, elevation = 25°.
Height is the opposite side, 80 m is the adjacent side. Ratio: tan θ = opp/adj.
tan 25° = h / 80 → h = 80 × 0.4663 ≈ 37.3 m.
Sanity: small angle → height much less than 80 m. ✓
3.1 — Building, 100 m away, elevation 30°
h = 100 × tan 30° = 100 × 0.5774 ≈ 57.7 m.
3.2 — Back-bearing of 070°
070° + 180° = 250°.
3.3 — Two observers, 50 m apart, angles 40° and 60°
Third angle = 180° − 40° − 60° = 80°. Distance from first observer is opposite the 60° angle.
d / sin 60° = 50 / sin 80°. d = 50 × 0.866 / 0.985 ≈ 44.0 m.
3.4 — Flagpole, 12 m shadow, elevation 35°
tan 35° = h / 12 → h = 12 × 0.7002 ≈ 8.4 m.
3.5 — Hiker 6 km on 030°, 5 km on 120° (right angle at turn)
Right triangle (interior angle 90°). d = √(6² + 5²) = √61 ≈ 7.8 km.
3.6 — Manly ferry, 12 m deck, depression 18°
tan 18° = 12 / d → d = 12 / tan 18° = 12 / 0.3249 ≈ 36.9 m. (Depression angle from deck = elevation angle from buoy looking up.)
3.7 — Tower, two points 100 m apart, elevations 40° and 25°
Let x = horizontal distance from closer point to tower base, h = tower height.
tan 40° = h / x → h = x × 0.8391.
tan 25° = h / (x + 100) → h = (x + 100) × 0.4663.
Equate: 0.8391 x = 0.4663 (x + 100) = 0.4663 x + 46.63.
0.3728 x = 46.63 → x ≈ 125.1 m. Then h = 125.1 × 0.8391 ≈ 105 m.
3.8 — Yacht 12 km on 040°, 15 km on 100°
Bearing change = 100° − 40° = 60°. Interior angle at turn = 180° − 60° = 120°.
(a) d² = 12² + 15² − 2(12)(15) cos 120° = 144 + 225 − 360 × (−0.5) = 369 + 180 = 549. d = √549 ≈ 23.4 km.
(b) Sine rule for angle at the start (between path-1 direction and the closing line back from finish): sin θ / 15 = sin 120° / 23.4. sin θ = 15 × 0.866 / 23.4 ≈ 0.555. θ ≈ 33.7°. Bearing from start to finish = 040° + 33.7° = 073.7°, so bearing of start from finish (back-bearing) = 073.7° + 180° = 253.7° ≈ 254°.