Mathematics • Year 9 • Unit 3 • Lesson 20
Trigonometry Synthesis and Review
Build fluency selecting the right tool for any Year 9 trig problem: Pythagoras for sides, SOH-CAH-TOA for sides from angles, inverse trig for angles from sides, elevation/depression for up/down, bearings for navigation, multi-step when none of the above is enough on its own.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A 10 m ladder leans against a wall and reaches 8 m up. Find (a) the distance from the wall to the foot of the ladder and (b) the angle the ladder makes with the ground, to 1 d.p.
Step 1 — List given and wanted.
Given: hypotenuse (ladder) = 10 m; one leg (wall-height) = 8 m. Wanted: other leg (ground-distance) and the angle the ladder makes with the ground.
Reason: writing this down BEFORE picking a tool stops you from accidentally solving a different question.
Step 2 — Pick the tool for (a): two sides → third side → Pythagoras.
a² + b² = c²: ground² + 8² = 10²
ground² = 100 − 64 = 36 ⇒ ground = 6 m.
Reason: when given two sides of a right triangle and asked for the third, Pythagoras is the right tool (6-8-10 is a triple).
Step 3 — Pick the tool for (b): two sides → angle → inverse trig.
From the ground angle: opp = 8 (wall-height); adj = 6 (ground); hyp = 10 (ladder).
Use any of three pairings (sin: opp/hyp; cos: adj/hyp; tan: opp/adj). Pick sin since both 8 and 10 are given exactly: sin θ = 8 / 10 = 0.8.
θ = sin−¹(0.8) ≈ 53.130°
Reason: opp + hyp pair → sin (SOH). With two sides plus the right angle known, any of the three pairings works; pick the most exact.
Step 4 — Round only at the end.
θ ≈ 53.1° (1 d.p.).
Reason: 1 d.p. for angles, 2 d.p. for distances is the standard convention.
Answer: (a) 6 m from the wall; (b) angle ≈ 53.1° with the ground.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. From a point 30 m from the foot of a tower, the elevation to the top is 42°. Find (a) the height of the tower and (b) the straight-line distance from the observer's eye-level point to the top of the tower, to 2 d.p.
Step 1 — Given/wanted: given the horizontal distance __________ m and the elevation angle __________ °. Wanted: tower HEIGHT (opposite the angle) and SIGHT LINE (hypotenuse).
Step 2 — Tool for height (angle + adj → opp → tan):
tan __________ ° = h / __________ ⇒ h = __________ × tan __________ ° ≈ __________ m
Step 3 — Tool for sight line (angle + adj → hyp → cos):
cos __________ ° = __________ / sight-line ⇒ sight-line = __________ / cos __________ ° ≈ __________ m
Step 4 — Round and answer: Height ≈ __________ m; sight line ≈ __________ m (2 d.p.).
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single rule from one earlier lesson). The middle two are standard (combine two ideas). The last two are extension (mixed, two or more tools).
Foundation — single rule
3.1 A right triangle has legs 8 and 15. Find the hypotenuse. 1 mark
3.2 In a right triangle, hyp = 20, θ = 40°. Find the side opposite θ (2 d.p.). 1 mark
3.3 In a right triangle, opp = 9, adj = 12. Find the acute angle θ (1 d.p.). 1 mark
3.4 A boat sails 12 km on true bearing 060°. Find the eastward component (2 d.p.). 1 mark
Standard — combine two ideas
3.5 From a 15 m cliff, the angle of depression to a swimmer in the water is 28°. Find the horizontal distance from the cliff base to the swimmer (2 d.p.). 2 marks
3.6 Convert the compass bearing N40°W to a true bearing, then find its reverse bearing. 2 marks
Extension — synthesis (mixed tools)
3.7 A 5 m ladder leans against a wall and reaches 4 m up. Find (a) how far the foot is from the wall, (b) the angle the ladder makes with the ground (1 d.p.), and (c) the angle the ladder makes with the wall (1 d.p.). 3 marks
3.8 A walker travels 4 km on bearing 050°, then 3 km on bearing 140°. Find (a) the straight-line distance from start (2 d.p.) and (b) the true bearing of the end-point from the start (2 d.p.). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (30 m, elevation 42°)
Step 1: given horizontal distance 30 m and elevation angle 42°.
Step 2: tan 42° = h / 30 ⇒ h = 30 × tan 42° ≈ 30 × 0.9004 ≈ 27.01 m.
Step 3: cos 42° = 30 / sight-line ⇒ sight-line = 30 / cos 42° ≈ 30 / 0.7431 ≈ 40.37 m.
Step 4: height ≈ 27.01 m; sight line ≈ 40.37 m.
3.1 — Legs 8 and 15
Hyp = √(64 + 225) = √289 = 17. (8-15-17 triple.)
3.2 — Hyp 20, θ = 40°, find opp
opp = hyp × sin θ = 20 sin 40° ≈ 20 × 0.6428 ≈ 12.86.
3.3 — opp = 9, adj = 12
tan θ = 9 / 12 = 0.75; θ = tan−¹(0.75) ≈ 36.9°. (Or: this is the angle of a 3-4-5 triangle scaled by 3, with the 9 opposite and 12 adjacent — the smaller acute angle.)
3.4 — 12 km on 060° (E-comp)
E = 12 sin 60° ≈ 12 × 0.866 ≈ 10.39 km east.
3.5 — Cliff 15 m, depression 28°
Depression from top of cliff = elevation from swimmer's level. tan 28° = 15 / d (since the height 15 m is OPPOSITE the angle, the horizontal distance d is ADJACENT).
d = 15 / tan 28° ≈ 15 / 0.5317 ≈ 28.21 m from the cliff base.
3.6 — N40°W → true; reverse
N40°W in NW quadrant: true = 360 − 40 = 320°.
Reverse = 320 − 180 = 140°.
3.7 — 5 m ladder reaching 4 m up
(a) Foot-to-wall distance: ground² + 4² = 5² ⇒ ground² = 25 − 16 = 9 ⇒ ground = 3 m. (3-4-5 triple.)
(b) Angle with ground: sin θ = 4 / 5 = 0.8; θ = sin−¹(0.8) ≈ 53.1°.
(c) Angle with wall: 90° − 53.1° = 36.9° (angles in a right triangle sum to 180°; the two non-right angles sum to 90°).
3.8 — 4 km on 050° then 3 km on 140°
Leg 1: N1 = 4 cos 50° ≈ 2.57; E1 = 4 sin 50° ≈ 3.06.
Leg 2: N2 = 3 cos 140° ≈ −2.30; E2 = 3 sin 140° ≈ 1.93.
Net N ≈ 0.27 km; Net E ≈ 4.99 km.
(a) Distance = √(0.27² + 4.99²) ≈ √(0.073 + 24.90) ≈ 5.00 km.
(b) End-point in NE quadrant (both N and E positive). Angle from N: tan−¹(4.99 / 0.27) ≈ 86.9°. True bearing ≈ 086.90° (nearly due east — almost on the equator of the compass).