Mathematics • Year 9 • Unit 3 • Lesson 8
Cosine Ratio — Finding a Side
Use cos θ = adj/hyp to find the adjacent (adj = hyp × cos θ) or the hypotenuse (hyp = adj ÷ cos θ). Build from a fully-worked guy-wire example through guided fill-ins to eight independent problems. DEG mode throughout.
1. I do — fully worked example
A 10 m guy wire supports a tower; the wire makes a 65° angle with the ground. How far from the tower's base is the wire anchored?
Problem. hyp = 10 m (the wire), θ = 65°, adj = ? (the ground distance from base to anchor). Round to 2 d.p. DEG mode (sin 30° should give 0.5).
Step 1 — Label sides relative to θ.
hyp = 10 (the slanted wire), adj = ? (the horizontal ground — next to θ), opp (the vertical to top of tower — not needed).
Reason: adj + hyp involved → use cosine (CAH).
Step 2 — Set up the cosine equation.
cos 65° = adj / 10
Reason: cos θ = adj/hyp.
Step 3 — Rearrange to isolate adj.
adj = 10 × cos 65°
Reason: adj is on top → multiply both sides by the bottom (10).
Step 4 — Compute.
adj ≈ 10 × 0.4226 ≈ 4.23 m
Reason: full calculator precision, round at the end.
Step 5 — Add units, sanity check.
adj ≈ 4.23 m from the base. Reasonable? Yes — at a steep 65° angle, most of the wire goes UP rather than out, so the ground distance is small.
Answer: adj ≈ 4.23 m.
2. We do — fill in the missing steps
A right triangle has adj = 15 and θ = 40°. Find the hypotenuse to 2 d.p. Fill in each blank. 4 marks
Step 1 — Label. adj = ____ , hyp = ? , θ = ____ ° . The two sides involved are adj and hyp, so the ratio is __________ .
Step 2 — Set up:
cos ____ ° = ____ / hyp
Step 3 — Rearrange to put hyp on its own:
hyp = ____ / cos ____ °
Step 4 — Compute (DEG mode):
hyp ≈ 15 / ________ ≈ ________
Step 5 — Sanity check. Is your hyp BIGGER than adj? __________ Should it be? __________
3. You do — independent practice
Round to 2 d.p. unless told otherwise. Foundation (3.1–3.4) is one rearrangement. Standard (3.5–3.6) is contextual. Extension (3.7–3.8) is multi-step.
Foundation — adj = hyp × cos θ, or hyp = adj ÷ cos θ
3.1 hyp = 14, θ = 30°. Find adj. (Use cos 30° ≈ 0.8660.) 1 mark
3.2 hyp = 20, θ = 60°. Find adj. (Use cos 60° = 0.5 exactly.) 1 mark
3.3 adj = 9, θ = 35°. Find hyp. 1 mark
3.4 adj = 6, θ = 45°. Find hyp. (Use cos 45° ≈ 0.7071.) 1 mark
Standard — practical contexts
3.5 A 6 m ladder leans against a wall at 70° to the ground. How far is the foot of the ladder from the wall? 2 marks
3.6 A sliding door rail runs at 8° below horizontal. The horizontal distance covered by the door is 1.8 m. Find the actual length of the rail (the hyp). 2 marks
Extension — push your thinking
3.7 A 4 m ladder leans against a wall at 75°. Find the foot-of-ladder distance from the wall, then find the vertical height the ladder reaches up the wall (use sin θ = opp/hyp, since you know the hyp). 3 marks
3.8 Two guy wires support a tower, both anchored 8 m from the base. Wire P makes a 50° angle with the ground; wire Q makes a 70° angle. Find the length of each wire (2 d.p.) and state which is shorter. Why is the steeper wire shorter? 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (adj = 15, θ = 40°)
Step 1: adj = 15, θ = 40°, ratio is cos.
Step 2: cos 40° = 15 / hyp.
Step 3: hyp = 15 / cos 40°.
Step 4: hyp ≈ 15 / 0.7660 ≈ 19.58.
Step 5: hyp (19.58) > adj (15) ✓ — hyp should always be the longest side.
3.1 — hyp = 14, θ = 30°
adj = 14 cos 30° ≈ 14 × 0.8660 ≈ 12.12.
3.2 — hyp = 20, θ = 60°
adj = 20 cos 60° = 20 × 0.5 = 10 exactly.
3.3 — adj = 9, θ = 35°
hyp = 9 / cos 35° ≈ 9 / 0.8192 ≈ 10.99.
3.4 — adj = 6, θ = 45°
hyp = 6 / cos 45° ≈ 6 / 0.7071 ≈ 8.49.
3.5 — 6 m ladder at 70°
Foot distance (adj) = 6 cos 70° ≈ 6 × 0.3420 ≈ 2.05 m.
3.6 — Sliding door rail, 1.8 m horizontal, 8° below horizontal
adj = 1.8, want hyp. hyp = 1.8 / cos 8° ≈ 1.8 / 0.9903 ≈ 1.82 m.
Sense check: the rail is barely longer than the horizontal because 8° is a shallow angle.
3.7 — 4 m ladder at 75°
Foot distance: adj = 4 cos 75° ≈ 4 × 0.2588 ≈ 1.04 m.
Height up wall: opp = 4 sin 75° ≈ 4 × 0.9659 ≈ 3.86 m.
Pythagoras check: 1.04² + 3.86² ≈ 1.08 + 14.90 ≈ 15.98 ≈ 4² ✓.
3.8 — Two guy wires from anchor 8 m out
Wire P (50°): hyp = 8 / cos 50° ≈ 8 / 0.6428 ≈ 12.45 m.
Wire Q (70°): hyp = 8 / cos 70° ≈ 8 / 0.3420 ≈ 23.39 m.
Wire P is shorter. The steeper wire (Q at 70°) needs much more length because it has to travel further UP the tower — the anchor distance is fixed, so a steeper angle means the wire goes mostly vertically and the hyp (slant) becomes long.