Mathematics • Year 10 • Unit 3 • Lesson 6
Pythagoras and Trigonometry Combined — Skill Drill
Build fluency with the Lesson 6 decision rule: two sides only → Pythagoras (a² + b² = c²); one angle + one side → trigonometry (SOH CAH TOA); two sides + need an angle → both, in sequence. One step at a time, from a fully worked example to independent practice.
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 legs of 6 cm and 8 cm meeting at the right angle. Find the angle opposite the 6 cm side.
Step 1 — Spot what you have.
Two sides given (6 and 8). One angle wanted.
Reason: two sides + need an angle → Pythagoras first, then inverse trig.
Step 2 — Find the hypotenuse (Pythagoras).
c² = 6² + 8² = 36 + 64 = 100 → c = 10 cm
Reason: classic 6-8-10 (a doubled 3-4-5 triple).
Step 3 — Pick the trig ratio.
Want angle θ opposite the 6 cm side. Known: opposite = 6, hypotenuse = 10.
Reason: opposite + hypotenuse pair with sine (SOH).
Step 4 — Solve for the angle.
sin θ = 6/10 = 0.6 → θ = sin⁻¹(0.6) ≈ 36.87°
Reason: inverse sine turns the ratio back into an angle.
Step 5 — Sanity check.
Other acute angle = 90° − 36.87° = 53.13°. Both angles between 0 and 90 ✓
Answer: The angle opposite the 6 cm side is ≈ 36.9°.
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 hypotenuse 13 cm and one other side 5 cm. Find the remaining side.
Step 1 — Spot what you have: two ____________ are given; no angle is mentioned (other than 90°).
Step 2 — Choose the tool.
Two sides only → use ____________________ theorem.
Step 3 — Substitute into a² + b² = c² (where c is the hypotenuse).
a² + ______² = ______²
Step 4 — Rearrange and evaluate.
a² = ______ − ______ = ______ → a = ______ cm
Step 5 — Verify with the Pythagorean triple.
5, ____, 13 is a famous triple ✓
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (single tool). The middle two are standard (deeper choice). The last two are extension (both tools combined).
Foundation — pick the tool, one step
3.1 Legs of 9 cm and 12 cm meet at a right angle. Find the hypotenuse. 1 mark
3.2 A right-angled triangle has hypotenuse 25 cm and one leg 7 cm. Find the other leg. 1 mark
3.3 An isosceles right triangle has equal sides of 5 cm. Find the hypotenuse in exact (surd) form. 1 mark
3.4 An isosceles right triangle has equal sides of length a. State the hypotenuse and the two acute angles. 1 mark
Standard — choose between Pythagoras and trig
3.5 A right-angled triangle has hypotenuse 20 cm and one acute angle of 40°. Find the side opposite the 40° angle (to 2 d.p.). 2 marks
3.6 A right-angled triangle has one leg 24 cm and hypotenuse 26 cm. Find sin θ where θ is opposite the unknown leg. (Hint: Pythagoras first.) 2 marks
Extension — both tools, in sequence
3.7 A right-angled triangle has legs of 5 cm and 12 cm. Find the angle opposite the 12 cm side (to 1 d.p.). Show each step: hypotenuse first, then inverse trig. 3 marks
3.8 A square has diagonal 10 cm. Find the side length in exact form, and state the two acute angles of either triangle formed by the diagonal. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (5-12-13 triangle)
Step 1: two sides.
Step 2: Pythagoras.
Step 3: a² + 5² = 13².
Step 4: a² = 169 − 25 = 144 → a = 12 cm.
Step 5: 5, 12, 13 ✓.
3.1 — Hypotenuse from 9 and 12
c² = 9² + 12² = 81 + 144 = 225 → c = 15 cm. (3-4-5 triple scaled by 3.)
3.2 — Missing leg, hypotenuse 25, leg 7
a² = 25² − 7² = 625 − 49 = 576 → a = 24 cm. (7-24-25 triple.)
3.3 — Isosceles right triangle with legs 5
c² = 5² + 5² = 50 → c = √50 = 5√2 cm (≈ 7.07 cm).
3.4 — Generic isosceles right triangle
Hypotenuse = a√2. Acute angles = 45° and 45°.
3.5 — Hypotenuse 20, angle 40°, opposite side
sin 40° = opp/20 → opp = 20 × sin 40° ≈ 20 × 0.6428 ≈ 12.86 cm.
3.6 — Leg 24, hypotenuse 26, find sin θ
Other leg: a² = 26² − 24² = 676 − 576 = 100 → a = 10 cm.
sin θ = opposite/hypotenuse = 10/26 = 5/13 ≈ 0.385.
3.7 — Legs 5 and 12, angle opposite 12
Step 1 (Pythagoras): c² = 5² + 12² = 169 → c = 13 cm.
Step 2 (inverse trig): sin θ = 12/13 → θ = sin⁻¹(12/13) ≈ 67.4°.
Check: the other acute angle is ≈ 22.6°, and 67.4 + 22.6 + 90 = 180 ✓.
3.8 — Square diagonal 10 cm
Diagonal = side × √2, so side = 10/√2 = 10√2/2 = 5√2 cm (≈ 7.07 cm).
The two acute angles are 45° and 45° (every square diagonal makes a 45-45-90 triangle).