Mathematics • Year 10 • Unit 3 • Lesson 3
Finding Unknown Angles — Skill Drill
Build fluency with inverse trig from Lesson 3: form the correct ratio of given sides, apply sin⁻¹, cos⁻¹ or tan⁻¹ on a calculator in DEG mode, then round the angle to the nearest degree (or to 1 d.p. when asked). Verify by substituting the angle back into the ratio.
1. I do — fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. In a right-angled triangle, the hypotenuse is 13 cm and the side opposite an unknown angle θ is 5 cm. Find θ to the nearest degree.
Step 1 — Identify the two sides given.
O = 5, H = 13
Reason: O + H means we will use sine — SOH.
Step 2 — Write the ratio as a fraction.
sin θ = O / H = 5 / 13
Reason: the ratio is what we feed into the inverse function — keep it exact.
Step 3 — Apply inverse sine.
θ = sin⁻¹(5 / 13)
Reason: sin⁻¹ undoes sin and returns "the angle whose sine is …".
Step 4 — Calculate in DEG mode.
θ = sin⁻¹(0.38461…) = 22.6199…°
Reason: feed the fraction directly into the calculator. If your screen shows 0.394 instead, you are in RAD mode — switch back.
Step 5 — Round and verify.
θ ≈ 23°. Check: sin 23° ≈ 0.3907 ≈ 5 / 13 ≈ 0.3846. ✓ Close.
Reason: the question asked for the nearest whole degree.
Answer: θ ≈ 23°.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank. 5 marks
Problem. A right-angled triangle has the side adjacent to angle θ equal to 8, and the side opposite θ equal to 15. Find θ to the nearest degree.
Step 1 — Identify the two sides:
A = ______ , O = ______ . Pair = O + A → ______
Step 2 — Ratio:
tan θ = ______ / ______ = ______
Step 3 — Inverse:
θ = tan⁻¹( ______________ )
Step 4 — Calculate (DEG mode):
θ ≈ ____________° (calculator display)
Step 5 — Round and verify:
θ ≈ ______° (to nearest degree). Check: tan ______° ≈ ______ ≈ 15/8. ✓
3. You do — independent practice
Round every angle to the nearest degree unless the question says otherwise. Show working. Always check your calculator is in DEG mode.
Foundation — single inverse, easy ratios
3.1 sin θ = 1/2. Find θ. (Use the exact-value table — no calculator needed.) 1 mark
3.2 cos θ = √3 / 2. Find θ. (Exact-value table.) 1 mark
3.3 tan θ = 1. Find θ. (Exact-value table.) 1 mark
3.4 A right triangle has O = 7 and H = 10. Find θ (the angle θ is opposite the side of length 7). 1 mark
Standard — from a triangle
3.5 A right triangle has A = 9 and H = 12. Find the angle θ between A and H. 2 marks
3.6 A right triangle has O = 6 and A = 11 (relative to θ). Find θ. 2 marks
Extension — push your thinking
3.7 A right-angled triangle has sides 7, 24, 25 (with 25 as hypotenuse). Find both non-right angles to the nearest degree. (Hint: there are two of them, and they must sum to 90°.) 3 marks
3.8 A right-angled triangle has O = 5 and A = 5 (relative to θ). Find θ without a calculator and explain in one sentence why this triangle is special. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (A = 8, O = 15)
Step 1: A = 8, O = 15. Pair O + A → TAN.
Step 2: tan θ = 15 / 8 = 1.875.
Step 3: θ = tan⁻¹(1.875).
Step 4: θ ≈ 61.9275°.
Step 5: θ ≈ 62°. Check: tan 62° ≈ 1.881 ≈ 15/8. ✓
3.1 — sin θ = 1/2
From the exact-value table, sin 30° = 1/2. So θ = 30°.
3.2 — cos θ = √3 / 2
From the table, cos 30° = √3 / 2. So θ = 30°.
3.3 — tan θ = 1
From the table, tan 45° = 1. So θ = 45°.
3.4 — sin θ = 7 / 10
θ = sin⁻¹(7/10) = sin⁻¹(0.7) = 44.4270° ≈ 44°.
3.5 — cos θ = 9 / 12
9 / 12 = 0.75. θ = cos⁻¹(0.75) = 41.4096° ≈ 41°.
Verify: cos 41° ≈ 0.7547 ≈ 0.75. ✓
3.6 — tan θ = 6 / 11
6 / 11 = 0.5454. θ = tan⁻¹(0.5454) = 28.6105° ≈ 29°.
3.7 — 7-24-25 triangle, both angles
Angle opposite 7: sin α = 7/25 = 0.28, so α = sin⁻¹(0.28) = 16.260° ≈ 16°.
Angle opposite 24: sin β = 24/25 = 0.96, so β = sin⁻¹(0.96) = 73.740° ≈ 74°.
Check: 16° + 74° = 90°. ✓ (Non-right angles in a right-angled triangle must sum to 90°.)
3.8 — O = 5, A = 5
tan θ = 5 / 5 = 1, so θ = 45° (from the exact-value table).
This triangle is the 45-45-90 special triangle from Lesson 1 — when the two legs are equal, both non-right angles are 45°.