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.

Build · I Do / We Do / You Do

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.

Opposite and hypotenuse known → sin θ = 5/13, so θ = sin⁻¹(5/13).

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°.

Stuck? Revisit lesson § "Words You Need", sin⁻¹ is the inverse sine button (often above sin on the calculator, accessed with SHIFT).

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. ✓

Stuck? O + A means tangent. 15 / 8 = 1.875. tan⁻¹(1.875) ≈ 61.93°.

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

Stuck on 3.8? tan θ = 5/5 = 1, and the exact-value table says tan 45° = 1.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers, Do not peek before attempting

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°.