Mathematics • Year 10 • Unit 3 • Lesson 3
Finding Unknown Angles — Mixed Challenge
Pull together every idea from Lesson 3: ratio first, then inverse function, then DEG mode, then round. Recall exact-value angles without a calculator, spot the "θ on top of the hypotenuse" inversion error, and design two triangles that share the same angles but different sizes.
1. Mixed problems — choose the right tool
Each question uses a different idea from Lesson 3. Decide which inverse function applies before you start writing. Round to the nearest degree unless told otherwise. 2-3 marks each
1.1 A right-angled triangle has O = 4 and H = 9 relative to an angle θ. Find θ. 2 marks
1.2 A right-angled triangle has A = 13 and H = 18 relative to an angle θ. Find θ. 2 marks
1.3 A right-angled triangle has O = 12 and A = 5 relative to an angle θ. Find θ. 2 marks
1.4 Without a calculator, find θ if cos θ = 1/√2. 2 marks
1.5 A right-angled triangle has the right angle at one corner, an angle of 90° at the opposite corner is impossible — so there are two non-right angles. If one is 38°, find the other without a trig function. Then verify using sin and the side ratios from a 38° triangle of your choice. 2 marks
1.6 A surveyor measures a paddock corner that is supposed to be a right angle. She measures the three sides as 30 m, 40 m, 51 m. Find the actual angle at the corner opposite the 51 m side (between the 30 m and 40 m sides) to the nearest degree. (Hint: rearranged Pythagoras would only test the right-angle. Use the cosine rule? No — that is Lesson 16. Instead, drop a perpendicular and use trig.) Actually, a simpler check: would a true right-angle triangle with legs 30 and 40 have a hypotenuse of 50, not 51? Use this to argue whether the angle is slightly larger or smaller than 90°, then build a right-angled triangle with sides 30 and 40 to find the "expected" hypotenuse, and quote the discrepancy. 3 marks
2. Find the mistake
A Year 10 student tries to find an angle in a right-angled triangle with O = 9 and H = 4. (Yes, the question is sneaky.) Their working is shown. Exactly one line contains a mistake. Spot it, explain why, and re-do correctly. 3 marks
Student's working — find θ where O = 9 and H = 4:
Line 1: sin θ = O / H = 9 / 4 = 2.25
Line 2: θ = sin⁻¹(2.25)
Line 3: Calculator gave ERROR, so I assumed θ = 90°.
(a) Which line contains the fundamental mistake (the source of all the trouble)?
(b) Explain in one or two sentences why the input data is impossible for a right-angled triangle.
(c) State what the student should do when given numbers like this.
Stuck? The hypotenuse is always the longest side — by definition. If you are told H = 4 and O = 9, the labels themselves are wrong (or the data is bogus).3. Open-ended challenge — design two similar triangles
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design two right-angled triangles with different sizes that both have an angle of approximately 53°. (Hint: 53° comes from a 3-4-5 triangle — tan⁻¹(4/3) ≈ 53.13°.)
For each triangle:
(i) State the three integer side lengths and verify Pythagoras.
(ii) Use tan⁻¹ to confirm the angle opposite the longer leg is ≈ 53° (to nearest degree).
(iii) Confirm that the other non-right angle is ≈ 37°, so the angles in your two triangles match.
Bonus: Explain in one sentence why this is guaranteed to work for any pair of scaled 3-4-5 triangles.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — sin θ = 4 / 9
θ = sin⁻¹(4/9) = sin⁻¹(0.4444) = 26.387° ≈ 26°.
1.2 — cos θ = 13 / 18
θ = cos⁻¹(13/18) = cos⁻¹(0.7222) = 43.836° ≈ 44°.
1.3 — tan θ = 12 / 5
θ = tan⁻¹(12/5) = tan⁻¹(2.4) = 67.380° ≈ 67°.
This is the larger of the two non-right angles in a 5-12-13 triangle.
1.4 — cos θ = 1 / √2
From the exact-value table, cos 45° = 1/√2. So θ = 45°.
1.5 — Other non-right angle when one is 38°
Without a trig function: angles in a triangle sum to 180°. With a right angle of 90°, the two non-right angles sum to 90°. So the other angle = 90° − 38° = 52°.
Verify by building a 38° triangle: if H = 10, then O = 10 × sin 38° = 6.16, A = 10 × cos 38° = 7.88. Other angle has sin = 7.88 / 10 = 0.788, so it equals sin⁻¹(0.788) = 51.99° ≈ 52°. ✓
1.6 — Surveyor's discrepancy
True right-angle hypotenuse for legs 30 and 40: √(30² + 40²) = √(900 + 1600) = √2500 = 50 m.
Measured side is 51 m, which is 1 m longer than the true right-angle hypotenuse.
Since the side opposite the corner is longer than the right-angle would predict, the angle at the corner must be slightly larger than 90° — it is not a true right angle.
(For interest: using the cosine rule from Lesson 16, the actual angle is about 92.3°. But that calculation is outside Lesson 3's scope; what matters here is the comparison.)
2 — Find the mistake
(a) The mistake is in the setup itself — the labels in Line 1 are physically impossible. You cannot have O = 9 and H = 4 in a right-angled triangle.
(b) The hypotenuse is always the longest side of a right-angled triangle (it is opposite the right angle). So any side called H must be longer than any side called O or A. The data O = 9, H = 4 violates that rule, which is why sin θ = 9/4 comes out greater than 1, and why sin⁻¹ then errors.
(c) When given numbers like this, the student should re-read the question: either they have mis-labelled which side is the hypotenuse, or there is a typo in the problem. In a right-angled triangle, the longest side must be the hypotenuse.
This is the misconception flagged in Lesson 3 § "Spot the Trap" — never accept an input ratio greater than 1 for sin or cos.
3 — Open-ended (sample solution)
Triangle A — 3-4-5: 3² + 4² = 9 + 16 = 25 = 5². ✓
Angle opposite 4: tan θ = 4/3 = 1.333, θ = tan⁻¹(1.333) = 53.13° ≈ 53°. ✓
Other angle = 90° − 53° = 37°. ✓
Triangle B — 6-8-10 (scale ×2): 6² + 8² = 36 + 64 = 100 = 10². ✓
Angle opposite 8: tan θ = 8/6 = 1.333, θ = tan⁻¹(1.333) = 53.13° ≈ 53°. ✓
Other angle = 90° − 53° = 37°. ✓
Both triangles have the same angles (53°, 37°, 90°) but different sizes — they are similar.
Bonus: Any scaled 3-4-5 triangle has identical side ratios, so the trig ratios (and therefore the angles, given by the inverse functions) are also identical. The angles depend only on the ratios, not on the size.
Marking: 1 for each triangle's Pythagoras check + correct 53° (× 2 = 2 marks), 1 for verifying both angle pairs match, 1 for the similarity explanation.