Mathematics • Year 9 • Unit 3 • Lesson 8
Cosine Ratio — Mixed Challenge
Push your cosine skills further: combine cos with sin, use Pythagoras as a verification tool, spot a mistake in someone's working, and design a safe ladder-placement range. DEG mode, 2 d.p. unless stated.
1. Mixed problems — pick the right cosine setup
Decide whether to MULTIPLY or DIVIDE by cos θ before you start. 3 marks each
1.1 A right triangle has hyp = 18 and θ = 55°. Find adj and then find opp (use sin θ since you know hyp).
1.2 A right triangle has adj = 11 and θ = 38°. Find hyp.
1.3 A 5 m ladder leans at 70° to the ground. Find the foot distance from the wall (adj) and the height up the wall (opp), then verify with Pythagoras (opp² + adj² ≈ hyp²).
1.4 A car drives 250 m up a road that rises at 6° to the horizontal. Find the horizontal distance covered (the ground footprint). Then state how much higher (vertically) the car has climbed (use sin 6°).
1.5 A right triangle has adj = 14 and you want to find the hypotenuse. The angle θ is unknown but you know cos θ = 0.7. Find hyp without using your calculator's cos⁻¹ button.
1.6 A 5 m ladder is pulled away from a wall. At first the angle with the ground is 75°, so the foot is close in. After the pull, the angle is 60°. Find the foot distance from the wall in BOTH positions (2 d.p.) and state how far the foot moved.
2. Find the mistake
A student tries to find the length of a guy wire given the anchor distance and the angle. Exactly one line contains a Year 9 trig mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Problem: A guy wire is anchored 8 m from the base of a tower at 65° to the ground. Find the length of the wire.
Student's working:
Line 1: adj = 8, θ = 65°, hyp = ?
Line 2: cos 65° = adj / hyp = 8 / hyp
Line 3: hyp = 8 × cos 65°
Line 4: hyp ≈ 8 × 0.4226 ≈ 3.38 m
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong, and what should have happened.
(c) Write out the corrected working from that line onward, including the corrected final answer.
Stuck? Sanity check: 3.38 m of wire to reach a tower base 8 m away? The wire would have to be at least 8 m to even reach the anchor along the ground. Hyp must always be ≥ adj.3. Open-ended challenge — safe ladder zone
This question has more than one valid answer. 4 marks
3.1 A 4 m ladder will be leaned against a wall. Workplace safety code says the angle the ladder makes with the ground must be between 65° (gentlest allowed — won't slip) and 80° (steepest allowed — won't fall back).
Find the range of safe foot-distances from the wall:
(i) Find the foot distance at θ = 65° (the gentlest safe angle).
(ii) Find the foot distance at θ = 80° (the steepest safe angle).
(iii) State the foot must be in the range "from ___ m to ___ m from the wall".
(iv) A worker places the foot 1.5 m from the wall. Is this in the safe range? Justify.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — hyp = 18, θ = 55°
adj = 18 cos 55° ≈ 18 × 0.5736 ≈ 10.32.
opp = 18 sin 55° ≈ 18 × 0.8192 ≈ 14.74.
Check: opp² + adj² ≈ 14.74² + 10.32² ≈ 217.27 + 106.50 ≈ 323.77 ≈ 18² = 324 ✓.
1.2 — adj = 11, θ = 38°
hyp = 11 / cos 38° ≈ 11 / 0.7880 ≈ 13.96.
1.3 — 5 m ladder at 70°
adj = 5 cos 70° ≈ 5 × 0.3420 ≈ 1.71 m.
opp = 5 sin 70° ≈ 5 × 0.9397 ≈ 4.70 m.
Pythagoras check: 1.71² + 4.70² ≈ 2.92 + 22.09 ≈ 25.01 ≈ 5² ✓.
1.4 — Car 250 m up 6° road
Horizontal = 250 cos 6° ≈ 250 × 0.9945 ≈ 248.63 m.
Vertical climb = 250 sin 6° ≈ 250 × 0.1045 ≈ 26.14 m.
For a shallow 6° road, the horizontal is almost the same as the slope distance.
1.5 — adj = 14, cos θ = 0.7
cos θ = adj/hyp, so 0.7 = 14/hyp, giving hyp = 14 / 0.7 = 20 exactly.
No need to know θ — the ratio is enough.
1.6 — Ladder pulled from 75° to 60°
Initial foot = 5 cos 75° ≈ 5 × 0.2588 ≈ 1.29 m.
Final foot = 5 cos 60° = 5 × 0.5 = 2.50 m.
Foot moved 2.50 − 1.29 = 1.21 m further from the wall.
2 — Find the mistake
(a) The mistake is on Line 3.
(b) From cos 65° = 8 / hyp, the student multiplied both sides by cos 65° instead of by hyp. To isolate hyp, first multiply both sides by hyp (giving cos 65° × hyp = 8) and then divide by cos 65°. The correct rearrangement is hyp = 8 / cos 65°, not 8 × cos 65°.
(c) Corrected from Line 3:
hyp = 8 / cos 65°
≈ 8 / 0.4226
≈ 18.93 m.
Sense-check: hyp (18.93 m) > adj (8 m), and the hyp is always the longest side. The student's answer of 3.38 m was shorter than the adj of 8 m — physically impossible.
3 — Safe ladder zone (sample solution)
Foot distance from wall = 4 cos θ (since hyp = 4 m and adj = foot distance).
(i) θ = 65°: foot = 4 cos 65° ≈ 4 × 0.4226 ≈ 1.69 m.
(ii) θ = 80°: foot = 4 cos 80° ≈ 4 × 0.1736 ≈ 0.69 m.
(iii) Foot must be in the range from approximately 0.69 m to 1.69 m from the wall.
(iv) 1.5 m is within the range 0.69 m to 1.69 m, so yes, this placement is safe. (It corresponds to θ = cos⁻¹(1.5/4) ≈ cos⁻¹(0.375) ≈ 68° — comfortably between 65° and 80°.)
Marking: 1 mark for each of (i), (ii), (iii); 1 mark for the (iv) justification showing 0.69 ≤ 1.5 ≤ 1.69.