Mathematics • Year 9 • Unit 3 • Lesson 11
Inverse Trig — Mixed Challenge
Pull together everything from Unit 3 so far — SOH-CAH-TOA, Pythagoras, exact-value angles, and the three inverse trig functions. Choose the right tool for each problem, spot a plausible Year 9 mistake, and tackle an open-ended challenge.
1. Mixed problems — choose the right tool
Each question uses a different combination of inverse-trig with Pythagoras or angle properties. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 A right triangle has opp = 7 and adj = 24 relative to θ. (a) Find θ to 1 d.p. (b) Use Pythagoras to find the hypotenuse.
1.2 cos θ = 0.5. Find θ exactly (no calculator needed if you can recall the exact-value table) and verify with your calculator.
1.3 A right triangle has hyp = 13 and adj = 5 relative to θ. (a) Find the opposite side using Pythagoras. (b) Find θ using two different ratios and confirm they agree.
1.4 Sam computes sin⁻¹(1.5) on a calculator and gets a "MATH ERROR". Without computing, explain why this answer cannot exist. What is the largest value x can take for which sin⁻¹(x) is defined?
1.5 Find the size of the smaller acute angle in a right triangle whose two legs are 9 and 40. Then find the larger acute angle using the sum-of-angles property (both to 1 d.p.).
1.6 A surveyor stands 80 m from the base of a cliff. The angle of elevation to the top is θ where tan θ = 0.7. (a) Find θ to 1 d.p. (b) Use trig to find the height of the cliff to 2 d.p.
2. Find the mistake
Another Year 9 student is trying to find θ given opp = 4 and hyp = 7. 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 θ given opp = 4, hyp = 7:
Line 1: sin θ = opp / hyp = 4 / 7
Line 2: θ = 1 / sin(4 / 7)
Line 3: Press: 1 ÷ sin(0.5714) = ...
Line 4: θ ≈ 100.06° (?)
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the correct final answer to 1 d.p.
Stuck? The student has written sin⁻¹ as "1 / sin" — but sin⁻¹ is the INVERSE FUNCTION (the angle whose sine is …), not the reciprocal. Also, a right-triangle angle must be between 0° and 90°, so the 100° answer is a giveaway that something has gone wrong.3. Open-ended challenge — design your own triangle
This question has more than one valid answer — there are many different triangles that work. 4 marks
3.1 Find two different right triangles (with integer side lengths) where one of the acute angles is approximately 30° to 1 d.p.
For each triangle you find:
(i) State the three side lengths (opp, adj, hyp).
(ii) Confirm with Pythagoras that they form a right triangle.
(iii) Use inverse trig (any of sin⁻¹, cos⁻¹, tan⁻¹) to show the angle is ≈ 30°.
Bonus: Your two triangles must not be similar to each other (i.e. they must not have the same side-length ratio just scaled).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — opp = 7, adj = 24
(a) TOA: tan θ = 7 / 24, so θ = tan⁻¹(7 / 24) ≈ 16.3°.
(b) Pythagoras: hyp = √(7² + 24²) = √(49 + 576) = √625 = 25. (Famous 7-24-25 Pythagorean triple.)
1.2 — cos θ = 0.5
Exact: cos 60° = 0.5, so θ = 60° exactly (from the 30-60-90 exact-value table).
Calculator check: cos⁻¹(0.5) returns 60° ✓.
1.3 — hyp = 13, adj = 5
(a) Pythagoras: opp = √(13² − 5²) = √(169 − 25) = √144 = 12. (Famous 5-12-13 triple.)
(b) Two ratios — they must agree:
cos⁻¹(5 / 13) ≈ 67.4°.
sin⁻¹(12 / 13) ≈ 67.4° ✓ same answer.
1.4 — sin⁻¹(1.5) error
For any angle θ, sin θ is the ratio opp / hyp in a right triangle. Since the hypotenuse is the LONGEST side, opp ≤ hyp, so sin θ is always between −1 and 1. There is no angle whose sine is 1.5, so sin⁻¹(1.5) is undefined and the calculator returns MATH ERROR. The largest valid input is x = 1, and sin⁻¹(1) = 90°.
1.5 — Legs 9 and 40
Smaller angle is opposite the SHORTER leg (9). tan θ = 9 / 40 = 0.225, so smaller acute angle = tan⁻¹(9 / 40) ≈ 12.7°.
Larger acute angle = 90° − 12.7° = 77.3° (the two acute angles in a right triangle are complementary, from Lesson 12).
Check: hyp = √(9² + 40²) = √(81 + 1600) = √1681 = 41 — another famous Pythagorean triple (9-40-41).
1.6 — Cliff at 80 m
(a) tan θ = 0.7, so θ = tan⁻¹(0.7) ≈ 35.0°.
(b) height = adj × tan θ = 80 × 0.7 = 56.00 m.
Sense check: a cliff height roughly 70% of the horizontal distance gives a moderately steep view — consistent with about 35°.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has written sin⁻¹ as 1 / sin — that's the reciprocal, not the inverse function. The notation sin⁻¹(x) means "the angle whose sine is x" — it is an inverse FUNCTION (use SHIFT-sin on the calculator), not 1 divided by sin.
(c) Corrected working:
sin θ = 4 / 7 ≈ 0.5714.
θ = sin⁻¹(4 / 7).
Press SHIFT → sin → ( 4 ÷ 7 ) → = giving θ ≈ 34.8°.
The student's 100° answer was also a red flag — acute angles in a right triangle are always between 0° and 90°, so anything above 90° must be wrong.
3 — Open-ended challenge (sample solution)
For an acute angle of ≈ 30°, we need opp / hyp ≈ 0.5 (since sin 30° = 0.5) — equivalently, the hypotenuse is roughly twice the opposite.
Triangle 1: opp = 5, hyp = 10. Then adj = √(10² − 5²) = √75 ≈ 8.66.
Check inverse trig: sin⁻¹(5 / 10) = sin⁻¹(0.5) = 30° exactly ✓.
(Note: the adj is irrational, so for fully-integer sides this triangle isn't perfect — a good close pair is (5, 9, ≈10.3) or similar; if integer sides are required, see Triangle 2 below.)
Triangle 2: opp = 7, adj = 12, hyp ≈ √(49 + 144) = √193 ≈ 13.89. Then sin θ = 7 / 13.89 ≈ 0.504, so θ ≈ 30.3° ≈ 30° ✓.
Different ratio from Triangle 1 (7/13.89 ≠ 5/10), so they're not similar.
Other valid approaches: opp = 10, adj = 17, hyp ≈ 19.72 gives sin⁻¹(10/19.72) ≈ 30.5°. Or use the exact 1-√3-2 triangle scaled by any integer. Award full marks for any two distinct, valid triangles where the inverse trig confirms an angle of ≈ 30°.
Marking: 2 marks per valid triangle (1 for stating sides, 1 for confirming with Pythagoras + inverse trig). Up to 4 in total. Strict integer-side requirement is hard to satisfy exactly at 30°; accept "close to 30°" with explicit inverse-trig check.