Mathematics • Year 9 • Unit 4 • Lesson 3
Solving Right-Angled Triangles
Build fluency choosing the right trig ratio, setting up the equation, rearranging to make the unknown the subject, and using inverse trig to find angles. One step at a time, from a fully worked example through guided practice to independent problems.
1. I do, fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. A right-angled triangle has opposite = 8 cm and adjacent = 6 cm. Find the unknown angle θ to 1 dp.
Step 1, Identify which sides you know and which you want.
Known: opposite = 8, adjacent = 6. Want: the angle θ.
Reason: writing this down first stops you picking the wrong ratio later.
Step 2, Pick the ratio that uses ONLY the sides you know.
Opposite + adjacent → TOA → tan.
Reason: don't drag the hypotenuse into the equation when you don't need it.
Step 3, Set up the equation.
tan θ = opp / adj = 8 / 6 = 4/3
Reason: simplify the fraction if you can, easier to read.
Step 4, Use the inverse function to solve for θ.
θ = tan⁻¹(4/3) = tan⁻¹(1.3333…)
Reason: tan⁻¹ means "the angle whose tangent is …". It UNDOES tan.
Step 5, Evaluate on the calculator (degree mode), keep full precision until the final answer.
θ ≈ 53.1301° → round to 53.1° (1 dp)
Reason: keep all decimals on the calculator screen; only round the FINAL answer.
Answer: θ ≈ 53.1°. (This is one of the angles in the 3-4-5 / 6-8-10 right-angled triangle.)
2. We do, fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. A right-angled triangle has angle 25°, hypotenuse 10 cm. Find the adjacent side to 2 dp.
Step 1, Known and wanted: known = angle and ____________; wanted = ____________ side.
Step 2, Pick the ratio. Hypotenuse + adjacent → ________ → cos.
Step 3, Set up:
cos 25° = ________ / 10
Step 4, Rearrange to make adjacent the subject:
adjacent = 10 × ________ ° = 10 × ____________ ≈ ____________ cm
Step 5, Round to 2 dp:
adjacent ≈ ____________ cm
3. You do, independent practice
Show your working in the space under each problem. The first four are foundation (single ratio). The middle two are standard (combine 2 ideas or use inverse trig). The last two are extension (multi-step + check).
Foundation, single ratio
3.1 Find x to 2 dp: angle = 30°, hypotenuse = 14, x = opposite. 1 mark
3.2 Find x to 2 dp: angle = 45°, adjacent = 9, x = opposite. 1 mark
3.3 Find the angle θ to 1 dp: opposite = 5, hypotenuse = 13. 1 mark
3.4 Find the angle θ to 1 dp: adjacent = 12, hypotenuse = 15. 1 mark
Standard, solve for the unknown
3.5 Find the HYPOTENUSE to 2 dp: angle = 35°, adjacent = 8 cm. (Hint: rearrange cos = adj/hyp to make hyp the subject, you'll need to DIVIDE.) 2 marks
3.6 A ramp rises 2 m vertically over a horizontal distance of 10 m. Find the angle of inclination to 1 dp. 2 marks
Extension, push your thinking
3.7 In a right-angled triangle, the opposite side to one acute angle is 5 cm and the adjacent is 12 cm. Find the angle, the hypotenuse, and the OTHER acute angle. 3 marks
3.8 A student calculates the opposite side of a triangle (angle 40°, hyp 10) and gets 6.43 cm. They want to check by computing the adjacent side and then using Pythagoras. Carry out the check and state whether the answer is consistent. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2, We do (cos 25°, hyp 10, find adj)
Step 1: known = angle and hypotenuse; wanted = adjacent.
Step 2: hyp + adj → CAH → cos.
Step 3: cos 25° = adjacent / 10.
Step 4: adjacent = 10 × cos 25° = 10 × 0.9063 ≈ 9.063 cm.
Step 5: adjacent ≈ 9.06 cm.
3.1, Angle 30°, hyp 14, find opp
sin 30° = x/14 → x = 14 × 0.5 = 7.00 cm.
3.2, Angle 45°, adj 9, find opp
tan 45° = x/9 → x = 9 × tan 45° = 9 × 1 = 9.00 cm.
3.3, Opposite 5, hyp 13, find angle
sin θ = 5/13 = 0.3846 → θ = sin⁻¹(0.3846) ≈ 22.6°. (5-12-13 triangle.)
3.4, Adjacent 12, hyp 15, find angle
cos θ = 12/15 = 0.8 → θ = cos⁻¹(0.8) ≈ 36.9°. (9-12-15 = 3×(3-4-5) triangle.)
3.5, Angle 35°, adj 8, find hyp
cos 35° = 8 / hyp → hyp = 8 / cos 35° = 8 / 0.8192 ≈ 9.77 cm.
Sanity check: hyp > adj (9.77 > 8). ✓
3.6, Ramp rise 2 m over 10 m
tan θ = 2/10 = 0.2 → θ = tan⁻¹(0.2) ≈ 11.3°. (A gentle wheelchair-friendly ramp slope.)
3.7, Opposite 5, adjacent 12
tan θ = 5/12 = 0.4167 → θ ≈ 22.6°.
Hypotenuse by Pythagoras: √(25 + 144) = √169 = 13 cm.
Other acute angle = 90° − 22.6° = 67.4° (since the two acute angles sum to 90°).
3.8, Check the opposite is 6.43
Adjacent: cos 40° = adj/10 → adj = 10 × cos 40° ≈ 10 × 0.7660 = 7.66 cm.
Pythagoras check: 6.43² + 7.66² ≈ 41.3 + 58.7 = 100 = 10² ✓.
The answer is consistent, the triangle works.
This is the cross-check technique from card 3 of the lesson.