Mathematics • Year 9 • Unit 4 • Lesson 3
Solving Triangles — Mixed Challenge
Pull together every trig skill from Lessons 1-3: labelling sides, picking the right ratio, rearranging, using inverse trig, and verifying with Pythagoras or angle sums. You'll spot a mistake in someone else's working and tackle an open-ended challenge.
1. Mixed problems — choose the right tool
Each question uses a different combination of the ideas from Lessons 1-3. Decide which ratio applies before you start writing. Show your working. 3 marks each
1.1 Find the hypotenuse to 2 dp: angle = 28°, opposite = 9 cm.
1.2 Solve the triangle completely (find ALL sides and ALL angles) given one acute angle is 55° and the adjacent side to it is 6 cm.
1.3 A right-angled triangle has sides 9, 40 and 41 (a Pythagorean triple). Find the two acute angles to 1 dp.
1.4 A ramp 8 m long has a vertical rise of 2 m. Find (a) the angle of inclination to 1 dp, and (b) the horizontal length covered to 2 dp (use Pythagoras).
1.5 An isosceles right-angled triangle has a hypotenuse of 12 cm. Find the lengths of the two equal sides EXACTLY (no decimals) and state the two acute angles.
1.6 A 5 m long ladder leans on a wall at 65° to the ground. Find (a) the height up the wall, (b) the distance the foot of the ladder is from the wall, and (c) verify the ladder length using Pythagoras on your two answers.
2. Find the mistake
Another student has tried to find the angle in a right-angled triangle with adjacent = 6 and opposite = 8. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — find the angle θ in a triangle with opposite = 8, adjacent = 6:
Line 1: Known: opposite = 8, adjacent = 6. Want: angle θ.
Line 2: Opposite + adjacent → TOA → tan θ = opp/adj.
Line 3: tan θ = 8/6 = 1.333…
Line 4: θ = 1 / tan(1.333) ≈ 1 / 0.0233 ≈ 42.9°
Line 5: So θ ≈ 42.9°.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong. (Hint: revisit Lesson 3 "Common Misconceptions" — what does tan⁻¹ actually mean?)
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? tan⁻¹ is the INVERSE TANGENT — the button that gives you back the angle from a tangent value. It is NOT 1 divided by tan. Re-check on your calculator.3. Open-ended challenge — design a triangle from constraints
This question has more than one valid answer. 4 marks
3.1 Design two different right-angled triangles that BOTH have one acute angle of exactly 30°. One of your triangles must have ALL side lengths as exact values (using surds is fine), and the other must have at least one side that is a whole number when rounded.
For each triangle you design:
(i) State the three angles (one is 90°, one is 30°, the third you must work out).
(ii) Give the lengths of all three sides — labelled opposite/adjacent/hypotenuse relative to the 30° angle.
(iii) Verify all three sides satisfy Pythagoras' theorem (opp² + adj² = hyp²).
(iv) State one real-world scenario where your triangle could appear.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Angle 28°, opposite 9, find hyp
sin 28° = 9 / hyp → hyp = 9 / sin 28° = 9 / 0.4695 ≈ 19.17 cm.
1.2 — Angle 55°, adjacent 6 cm, solve fully
Other acute angle = 90° − 55° = 35°.
Opposite: tan 55° = opp / 6 → opp = 6 × tan 55° = 6 × 1.4281 ≈ 8.57 cm.
Hypotenuse: cos 55° = 6 / hyp → hyp = 6 / cos 55° = 6 / 0.5736 ≈ 10.46 cm.
Pythagoras check: 6² + 8.57² ≈ 36 + 73.4 ≈ 109.4 ≈ 10.46². ✓
1.3 — Triangle 9-40-41
9-40-41 is a Pythagorean triple: 81 + 1600 = 1681 = 41². ✓
Smaller acute angle: sin θ = 9/41 → θ = sin⁻¹(9/41) ≈ 12.7°.
Larger acute angle = 90° − 12.7° = 77.3° (or independently sin⁻¹(40/41) ≈ 77.3°).
1.4 — Ramp 8 m long, rise 2 m
(a) sin θ = 2/8 = 0.25 → θ = sin⁻¹(0.25) ≈ 14.5°.
(b) Horizontal = √(8² − 2²) = √(64 − 4) = √60 ≈ 7.75 m.
1.5 — Isosceles right-angled triangle, hyp 12 cm
The two equal sides each have length: opp = adj = 12 × sin 45° = 12 × (√2/2) = 6√2 cm (≈ 8.49 cm).
The two acute angles are both 45° (since the triangle is isosceles right-angled).
Check: (6√2)² + (6√2)² = 72 + 72 = 144 = 12². ✓
1.6 — Ladder 5 m at 65°
(a) Height = 5 × sin 65° = 5 × 0.9063 ≈ 4.53 m.
(b) Distance from wall = 5 × cos 65° = 5 × 0.4226 ≈ 2.11 m.
(c) Pythagoras check: 4.53² + 2.11² ≈ 20.52 + 4.45 ≈ 24.97 ≈ 25 = 5². ✓ (Tiny rounding error.)
2 — Find the mistake
(a) The mistake is on Line 4.
(b) The student wrote θ = 1 / tan(1.333) — this is WRONG. The inverse tangent (tan⁻¹) is NOT the same as the reciprocal 1/tan. tan⁻¹ is the inverse function (the "shift tan" button on a calculator) that returns the angle whose tan is the given value.
(c) Corrected working:
tan θ = 8/6 = 1.333…
θ = tan⁻¹(1.333…)
θ ≈ 53.1°.
This is the 6-8-10 = 2 × (3-4-5) triangle — the angle opposite the 8 is about 53.1°, exactly as in the worked example of this worksheet.
3 — Open-ended challenge (sample solutions)
Many valid answers; here are two:
Triangle A (exact, using surds): Angles 30°, 60°, 90°. Relative to the 30° angle: opp = 1 cm, adj = √3 cm, hyp = 2 cm. Pythagoras: 1² + (√3)² = 1 + 3 = 4 = 2². ✓ Real-world scenario: a set square used in technical drawing.
Triangle B (multiplied so sides are nearly whole numbers): Angles 30°, 60°, 90°. Sides scaled by 10: opp = 10 cm, adj = 10√3 ≈ 17.32 cm, hyp = 20 cm. Pythagoras: 10² + (10√3)² = 100 + 300 = 400 = 20². ✓ The opposite (10) and hypotenuse (20) are whole numbers. Real-world scenario: the cross-bracing of a fence post leaning at 30° from vertical.
Other valid approaches: any multiple of the 1 : √3 : 2 ratio. Scale by 6 to get 6 : 6√3 : 12.
Marking: 2 marks per triangle (1 for correct sides + Pythagoras check, 1 for correct angles + real-world scenario). Up to 4 in total.