Mathematics • Year 10 • Unit 3 • Lesson 2
Finding Unknown Sides — Skill Drill
Build fluency with the side-finding process from Lesson 2: spot the marked angle + given side + unknown side, choose the correct ratio, rearrange so the unknown is the subject, then evaluate in DEG mode and round. Round all answers to 2 decimal places unless told otherwise.
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. A right-angled triangle has hypotenuse 12 cm and one angle of 35°. Find the length of the side opposite the 35° angle, correct to 2 decimal places.
Step 1 — Identify the three ingredients.
Angle = 35°, Given side = H = 12, Unknown = O = x
Reason: Lesson 2 § "Parts of the Whole" — every problem has angle + given + unknown.
Step 2 — Match the pair to the ratio.
O + H → SIN, so use sin 35° = O / H
Reason: Opposite + Hypotenuse means we use sine (the S of SOH).
Step 3 — Make the unknown the subject.
sin 35° = x / 12 → x = 12 × sin 35°
Reason: unknown is on top → multiply both sides by 12.
Step 4 — Calculate (calculator in DEG mode).
x = 12 × 0.57358… = 6.8829…
Reason: sin 35° ≈ 0.57358. If you got 0.0367, you are in RAD mode — switch back to DEG.
Step 5 — Round and write the unit.
x ≈ 6.88 cm
Reason: the question asked for 2 decimal places. The unit (cm) is part of the answer.
Answer: the opposite side is 6.88 cm (to 2 d.p.).
2. We do — fill in the missing steps
Same five-step 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 a 40° angle equal to 9 cm. Find the length of the hypotenuse, correct to 2 decimal places.
Step 1 — Ingredients:
Angle = ______ ° , Given side = ______ = 9, Unknown = ______ = x
Step 2 — Pair → ratio:
A + H → ______ , so use ______ 40° = ______ / ______
Step 3 — Subject (the unknown is on the bottom):
cos 40° = 9 / x → x × cos 40° = 9 → x = 9 / ______
Step 4 — Evaluate (DEG mode):
x = 9 / ____________ ≈ ____________
Step 5 — Round and unit:
x ≈ ______________ cm (to 2 d.p.)
Sanity check: the hypotenuse must be longer than 9 cm. Is your answer larger than 9? ____ ✓
3. You do — independent practice
Round every answer to 2 decimal places. Show working. Always check your calculator is in DEG mode.
Foundation — single ratio, unknown on top
3.1 Hypotenuse 20 cm, marked angle 28°. Find the opposite side. 1 mark
3.2 Hypotenuse 15 m, marked angle 62°. Find the adjacent side. 1 mark
3.3 Adjacent 7 cm, marked angle 50°. Find the opposite side. 1 mark
3.4 Hypotenuse 30 cm, marked angle 18°. Find the adjacent side. 1 mark
Standard — unknown on the bottom
3.5 Opposite side 15 cm, marked angle 50°. Find the hypotenuse. 2 marks
3.6 Opposite side 8 m, marked angle 25°. Find the adjacent side. 2 marks
Extension — push your thinking
3.7 A right-angled triangle has a 40° angle and a hypotenuse of 25 cm. Find both the opposite and the adjacent sides, then check your two answers with Pythagoras (you should get back to 25 cm). 3 marks
3.8 A right-angled triangle has one angle of 38° and the side opposite that angle is 11 m. Without finding the hypotenuse first, find the side adjacent to the 38° angle using a single trig ratio. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (adjacent 9 cm, 40°, find H)
Step 1: Angle = 40°, Given = A = 9, Unknown = H = x.
Step 2: A + H → COS, so cos 40° = A / H = 9 / x.
Step 3: x = 9 / cos 40°.
Step 4: x = 9 / 0.7660 ≈ 11.7480.
Step 5: x ≈ 11.75 cm (to 2 d.p.).
Sanity: 11.75 > 9 ✓ (hypotenuse must be longer than the adjacent).
3.1 — sin 28°, H = 20
sin 28° = x / 20 → x = 20 × sin 28° = 20 × 0.4695 = 9.3893 ≈ 9.39 cm.
3.2 — cos 62°, H = 15
cos 62° = x / 15 → x = 15 × cos 62° = 15 × 0.4695 = 7.0418 ≈ 7.04 m.
3.3 — tan 50°, A = 7
tan 50° = x / 7 → x = 7 × tan 50° = 7 × 1.1918 = 8.3422 ≈ 8.34 cm.
3.4 — cos 18°, H = 30
cos 18° = x / 30 → x = 30 × cos 18° = 30 × 0.9511 = 28.5317 ≈ 28.53 cm.
3.5 — sin 50°, O = 15, find H
sin 50° = 15 / x → x = 15 / sin 50° = 15 / 0.7660 = 19.5811 ≈ 19.58 cm.
Unknown on the bottom: rearrange before calculating.
3.6 — tan 25°, O = 8, find A
tan 25° = 8 / x → x = 8 / tan 25° = 8 / 0.4663 = 17.1543 ≈ 17.15 m.
3.7 — 40°, H = 25 cm, find O and A
O = 25 × sin 40° = 25 × 0.6428 = 16.07 cm.
A = 25 × cos 40° = 25 × 0.7660 = 19.15 cm.
Pythagoras check: 16.07² + 19.15² = 258.24 + 366.72 = 624.96 ≈ 625 = 25². ✓
The tiny rounding gap is expected — use 2 d.p. inputs and you get a 2 d.p.-level rounded answer.
3.8 — tan 38°, O = 11, find A directly
tan 38° = O / A = 11 / x.
x = 11 / tan 38° = 11 / 0.7813 = 14.0788 ≈ 14.08 m.
Using tangent skips finding H entirely — a single ratio saves a step.