Mathematics • Year 9 • Unit 3 • Lesson 10

Mixed Trigonometry — Mixed Challenge

Tie together everything from L6–L10: pick the right ratio for any side-pair, combine ratios with Pythagoras, work with exact values for the special angles 30°, 45° and 60°, spot a ratio-choice mistake and design a "find every side" problem of your own.

Master · Mixed Challenge

1. Mixed problems — every problem could be any ratio

Identify the side-pair, name the ratio, rearrange, compute. Round to 2 d.p. unless told. 3 marks each

1.1 A right triangle has θ = 50°, hyp = 9. Find opp.

1.2 A right triangle has θ = 25°, adj = 14. Find opp.

1.3 A right triangle has θ = 70°, opp = 18. Find hyp. Then find adj using Pythagoras.

1.4 Without a calculator, find opp for a right triangle with θ = 30° and hyp = 12. (Use sin 30° = 0.5 exactly.) Then find adj using cos 30° ≈ 0.8660.

1.5 A right triangle has θ = 45°, adj = 7. Find opp (use tan 45° = 1 exactly) and hyp (use cos 45° = √2/2 ≈ 0.7071).

1.6 Two right triangles share the angle θ = 60°. Triangle A has opp = 4 cm. Triangle B has hyp = 14 cm. Find the missing measurements: hyp in A, and both legs in B. (Use sin 60° ≈ 0.8660, cos 60° = 0.5, tan 60° ≈ 1.7321.)

Stuck on 1.6? Triangle A: opp + θ → use sin to find hyp = 4/sin 60°. Triangle B: hyp + θ → use sin to find opp, cos to find adj.

2. Find the mistake

A student tries to find the adjacent side of a right triangle given the angle and the hypotenuse. Exactly one line contains a Year 9 ratio-choice mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks

Problem: A right triangle has θ = 35° and hyp = 20. Find adj.

Student's working:

Line 1:   Known: hyp = 20, θ = 35°. Wanted: adj.

Line 2:   Sides involved: adj and hyp.

Line 3:   Ratio: tan (since I want to find adj).

Line 4:   tan 35° = adj / 20

Line 5:   adj = 20 × tan 35° ≈ 20 × 0.7002 ≈ 14.00

(a) Which line contains the mistake?

(b) Explain in one or two sentences why that line is wrong. Focus on the SIDE-PAIR → RATIO rule from L10.

(c) Write out the corrected working from that line onward, including the corrected final answer.

Stuck? The pair adj + hyp matches COS, not tan. Tan is the leg-only ratio (opp + adj, no hyp). The student picked tan because it was the ratio they used most recently, ignoring the side-pair.

3. Open-ended challenge — design a "find every side" problem

This question has many valid answers. 4 marks

3.1 Design a complete worked example that uses BOTH sin and cos in a single real-world scenario. Your design must include:

(i) A real-world scenario (e.g. ladder, ramp, kite, tree, slide) in one sentence.
(ii) A right triangle with a clearly stated angle θ (choose 25° < θ < 75°) and a clearly stated hypotenuse length (choose 5 m ≤ hyp ≤ 30 m).
(iii) The side-pair, ratio name and worked computation for the opposite side (use sin).
(iv) The side-pair, ratio name and worked computation for the adjacent side (use cos).
(v) A Pythagoras check: opp² + adj² ≈ hyp² (state the actual numbers and confirm).
(vi) A "sanity sentence" — a one-line check that your answer makes physical sense (e.g. "the ladder reaches about 4 m up the wall — a sensible height for a 5 m ladder leaning at 60°").

Stuck? Start with: "A 12 m fire-truck ladder is set at 60° to the ground. Find the height it reaches and the foot distance from the wall." Then write the two trig steps and the Pythagoras check.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — θ = 50°, hyp = 9

opp + hyp → sin. opp = 9 sin 50° ≈ 9 × 0.7660 ≈ 6.89.

1.2 — θ = 25°, adj = 14

opp + adj → tan. opp = 14 tan 25° ≈ 14 × 0.4663 ≈ 6.53.

1.3 — θ = 70°, opp = 18

opp + hyp → sin. hyp = 18 / sin 70° ≈ 18 / 0.9397 ≈ 19.16.
Pythagoras: adj² ≈ 19.16² − 18² ≈ 367.10 − 324 ≈ 43.10, so adj ≈ 6.57.

1.4 — θ = 30°, hyp = 12 (no calculator for sin)

opp = 12 sin 30° = 12 × 0.5 = 6 exactly.
adj = 12 cos 30° ≈ 12 × 0.8660 ≈ 10.39.

1.5 — θ = 45°, adj = 7

opp = 7 tan 45° = 7 × 1 = 7 exactly (45° → opp = adj).
hyp = 7 / cos 45° ≈ 7 / 0.7071 ≈ 9.90.
Exactly: hyp = 7√2 ≈ 9.899.

1.6 — Two triangles, θ = 60°

Triangle A (opp = 4): hyp = 4 / sin 60° ≈ 4 / 0.8660 ≈ 4.62 cm. adj = 4 / tan 60° ≈ 4 / 1.7321 ≈ 2.31 cm.
Triangle B (hyp = 14): opp = 14 sin 60° ≈ 14 × 0.8660 ≈ 12.12 cm. adj = 14 cos 60° = 14 × 0.5 = 7 cm exactly.

2 — Find the mistake

(a) The mistake is on Line 3.
(b) The side-pair was correctly identified on Line 2 as adj + hyp. The L10 decision rule says adj + hyp → cos, not tan. Tangent is the leg-only ratio (opp + adj, with NO hyp). The student appears to have picked tan by association with "want to find adj" instead of by the side-pair rule.
(c) Corrected from Line 3:
Ratio: cos (adj + hyp → cos, CAH).
cos 35° = adj / 20
adj = 20 × cos 35°
≈ 20 × 0.8192
16.38.
Sense-check: for θ < 45°, adj > opp (and adj close to hyp). 16.38 is close to 20 ✓. The student's wrong answer of 14.00 is the opp, not the adj.

3 — Design a "find every side" problem (sample solution)

(i) Scenario: A 12 m fire-truck ladder is set at 60° to the ground against a building wall.

(ii) Triangle: θ = 60°, hyp = 12 m, opp = height up the wall, adj = foot distance from the wall.

(iii) Opposite (height up wall): side-pair opp + hyp → sin. opp = 12 sin 60° ≈ 12 × 0.8660 ≈ 10.39 m.

(iv) Adjacent (foot from wall): side-pair adj + hyp → cos. adj = 12 cos 60° = 12 × 0.5 = 6.00 m.

(v) Pythagoras check: opp² + adj² = 10.39² + 6.00² ≈ 107.95 + 36.00 ≈ 143.95 ≈ 12² = 144 ✓ (tiny rounding error).

(vi) Sanity sentence: A 12 m ladder at 60° reaches about 10.4 m up the wall with its foot 6 m out — a steep, tall reach. That matches what a fire crew uses to reach a 3rd-storey window.

Marking: 1 mark for parts (iii) and (iv) computed correctly with side-pair + ratio name; 1 mark for the Pythagoras check; 1 mark for a sensible scenario and sanity sentence; 1 mark for full presentation.