Mathematics • Year 9 • Unit 3 • Lesson 11
Finding Angles with Inverse Trig
Build fluency with sin⁻¹, cos⁻¹ and tan⁻¹ — the inverse trig functions. Walk from a fully worked example through one guided fill-in to eight independent practice problems that find an unknown angle from two known sides.
1. I do — fully worked example
Read every step. Each line has a short reason on the right so you see why, not just what. Make sure your calculator is in DEG mode before starting.
Problem. A right triangle has opposite = 4 m and hypotenuse = 7 m relative to θ. Find θ to 1 d.p.
Step 1 — Pick the right ratio.
We know opp and hyp, so use sine. SOH: sin θ = opp / hyp.
Reason: SOH-CAH-TOA — opp + hyp pair → sin.
Step 2 — Substitute the known values.
sin θ = 4 / 7 ≈ 0.5714
Reason: keep the fraction first; rounding too early loses precision.
Step 3 — Apply the inverse (the "undo" step).
θ = sin⁻¹(4 / 7)
Reason: sin⁻¹ is the INVERSE FUNCTION — the angle whose sine is 4/7. It is NOT 1/sin.
Step 4 — Press the keys: SHIFT → sin → ( 4 ÷ 7 ) → =
θ ≈ 34.8°
Reason: SHIFT (or 2nd) turns the sin key into sin⁻¹.
Step 5 — Sanity check.
0 < 34.8° < 90° ✓ — acute angle for a right triangle ✓
Reason: in any right triangle, both non-right angles are acute (between 0° and 90°).
Answer: θ ≈ 34.8°.
2. We do — fill in the missing steps
Same structure as Section 1, but the working is faded. Fill in each blank. 4 marks
Problem. In a right triangle, adjacent = 5 cm and hypotenuse = 8 cm relative to θ. Find θ to 1 d.p.
Step 1 — Pick the ratio: we know adj and hyp, so use ________ . (Hint: CAH.)
Step 2 — Substitute:
cos θ = ____ / ____ = ____
Step 3 — Apply the inverse:
θ = cos⁻¹( ____ )
Step 4 — Calculator: SHIFT → cos → ( 5 ÷ 8 ) → =
θ ≈ ________ °
Step 5 — Sanity check: is your answer between 0° and 90°? ________
3. You do — independent practice
Show working under each problem. Problems 3.1–3.4 are foundation (single inverse, given values). Problems 3.5–3.6 are standard (you must first identify which ratio). Problems 3.7–3.8 are extension (a word problem and a "find n" question).
Foundation — apply one inverse
3.1 sin θ = 0.5. Find θ. 1 mark
3.2 cos θ = 0.8. Find θ to 1 d.p. 1 mark
3.3 tan θ = 1. Find θ. 1 mark
3.4 tan θ = 2. Find θ to 1 d.p. 1 mark
Standard — pick the ratio yourself
3.5 opp = 5, hyp = 13. Find θ to 1 d.p. 2 marks
3.6 adj = 4, hyp = 5. Find θ to 1 d.p. 2 marks
Extension — push your thinking
3.7 A 7 m ladder leans against a wall with its base 2 m from the wall. Find the angle the ladder makes with the ground (1 d.p.). 3 marks
3.8 Sketch a right triangle with opp = 8 and adj = 6 relative to θ. (a) Find θ to 1 d.p. (b) Use Pythagoras to find the hypotenuse. (c) Verify your angle by also computing sin⁻¹(opp / hyp). 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded cos θ = 5/8)
Step 1: use cos (adj + hyp → cos).
Step 2: cos θ = 5 / 8 = 0.625.
Step 3: θ = cos⁻¹(0.625).
Step 4: SHIFT → cos → (5 ÷ 8) → = gives θ ≈ 51.3°.
Step 5: yes — 51.3° is between 0° and 90° ✓.
3.1 — sin θ = 0.5
θ = sin⁻¹(0.5) = 30°. (A classic exact value to remember.)
3.2 — cos θ = 0.8
θ = cos⁻¹(0.8) ≈ 36.9°.
3.3 — tan θ = 1
θ = tan⁻¹(1) = 45°. (Another exact-value angle.)
3.4 — tan θ = 2
θ = tan⁻¹(2) ≈ 63.4°.
3.5 — opp = 5, hyp = 13
SOH: sin θ = 5/13. θ = sin⁻¹(5/13) ≈ 22.6°. (Famous 5-12-13 angle.)
3.6 — adj = 4, hyp = 5
CAH: cos θ = 4/5. θ = cos⁻¹(4/5) ≈ 36.9°. (Same triangle as the 3-4-5 right triangle.)
3.7 — Ladder leaning on wall
Relative to the ground angle: adj = 2 (base distance), hyp = 7 (ladder). CAH: cos θ = 2/7. θ = cos⁻¹(2/7) ≈ 73.4°. (Real-world note: this sits right in the safe 70–80° range.)
3.8 — opp = 8, adj = 6
(a) TOA: tan θ = 8/6 = 4/3. θ = tan⁻¹(4/3) ≈ 53.1°.
(b) Pythagoras: hyp = √(8² + 6²) = √100 = 10.
(c) Check: sin⁻¹(8/10) = sin⁻¹(0.8) ≈ 53.1° ✓ (same angle — three ways into one triangle).