Mathematics • Year 10 • Unit 3 • Lesson 20
Unit 3 Revision — Formula Drill
Build fluency across the whole Unit 3 toolkit: SOHCAHTOA, Pythagoras, sine rule, cosine rule, area formula, Heron's formula, congruence (SSS / SAS / AAS / RHS) and similarity (SSS / SAS / AA), and the core circle theorems. One formula at a time, then a mini-chain.
1. I do — fully worked example
Read every step. The point is to demonstrate which formula to reach for at each piece of given information.
Problem. A cyclic quadrilateral ABCD has AB = 8, BC = 10, ∠ABC = 70°. (a) Find AC. (b) State ∠ADC.
Step 1 — Triangle ABC: identify the pattern.
Two sides (8, 10) + included angle (70°) → SAS → cosine rule.
Step 2 — Apply the cosine rule for AC.
AC² = 8² + 10² − 2(8)(10) cos 70° = 64 + 100 − 160 × 0.342 = 164 − 54.7 = 109.3
AC = √109.3 ≈ 10.5
Step 3 — Cyclic quadrilateral theorem.
Opposite angles of a cyclic quadrilateral sum to 180°.
∠ADC = 180° − ∠ABC = 180° − 70° = 110°
Reason: the inscribed-angle pair theorem from Unit 3's circle theorems.
Answers: AC ≈ 10.5, ∠ADC = 110°.
2. We do — fill in the missing steps
SOHCAHTOA + Pythagoras review. Fill the blanks. 5 marks
Problem. A right-angled triangle has the right angle at C. The side opposite ∠A is 5, the side adjacent to ∠A is 12, and the hypotenuse is unknown.
Step 1 — Find the hypotenuse (Pythagoras):
hyp² = 5² + 12² = ______ + ______ = ______
hyp = √______ = ______
Step 2 — Trig ratios for ∠A (SOHCAHTOA):
sin A = opp / hyp = ______ / ______ = ______
cos A = adj / hyp = ______ / ______ = ______
tan A = opp / adj = ______ / ______ = ______
Step 3 — Find ∠A using inverse trig:
A = tan⁻¹(______) ≈ ______°
Step 4 — Find ∠B:
B = 90° − A ≈ ______°
3. You do — independent practice
Each question targets a different Unit 3 idea. Name the formula or theorem you are using before you calculate. Round to 1 dp / 0.1°.
Foundation — single-formula warm-ups
3.1 (Pythagoras) A right-angled triangle has legs 9 and 40. Find the hypotenuse. 1 mark
3.2 (SOHCAHTOA) A right-angled triangle has opposite 7 and hypotenuse 25 relative to angle θ. Find θ to 0.1°. 2 marks
3.3 (Sine rule) Triangle with ∠A = 40°, ∠B = 75°, a = 8. Find side b. 2 marks
3.4 (Cosine rule) Triangle with a = 7, b = 10, ∠C = 65°. Find c. 2 marks
Standard — area, congruence, similarity
3.5 (Area formula) Find the area of a triangle with sides 9 and 12 and included angle 50°. 2 marks
3.6 (Heron's) A triangle has sides 7, 9, 12. Find its area. 2 marks
3.7 (Similarity) Two similar triangles have side ratio 2 : 5. (a) What is the ratio of their areas? (b) Of their perimeters? 2 marks
Extension — circle theorem and mini-chain
3.8 (Circle theorem) The angle subtended at the centre of a circle by a chord is 110°. What is the angle subtended at any point on the major arc by the same chord? Justify with the theorem name. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (5-12-? right triangle)
hyp² = 25 + 144 = 169 → hyp = √169 = 13. (5-12-13 Pythagorean triple.)
sin A = 5/13 ≈ 0.3846; cos A = 12/13 ≈ 0.9231; tan A = 5/12 ≈ 0.4167.
A = tan⁻¹(5/12) ≈ 22.6°.
B = 90° − 22.6° = 67.4°.
3.1 — legs 9, 40
hyp = √(81 + 1600) = √1 681 = 41. (9-40-41 Pythagorean triple.)
3.2 — opposite 7, hyp 25
sin θ = 7/25 = 0.28. θ = sin⁻¹(0.28) ≈ 16.3°.
3.3 — Sine rule with ∠A = 40°, ∠B = 75°, a = 8
b / sin 75° = 8 / sin 40°. b = 8 × 0.9659 / 0.6428 ≈ 12.0.
3.4 — Cosine rule a = 7, b = 10, C = 65°
c² = 49 + 100 − 140 × cos 65° = 149 − 140 × 0.4226 = 149 − 59.2 = 89.8. c = √89.8 ≈ 9.5.
3.5 — Area, sides 9 and 12, included 50°
A = ½ × 9 × 12 × sin 50° = 54 × 0.766 ≈ 41.4.
3.6 — Heron's on 7, 9, 12
s = (7 + 9 + 12)/2 = 14. A = √(14 × 7 × 5 × 2) = √980 ≈ 31.3.
3.7 — Similar triangles ratio 2 : 5
(a) Area ratio = (linear)² = 2² : 5² = 4 : 25.
(b) Perimeter ratio = linear ratio = 2 : 5. (All linear measurements scale by the same factor.)
3.8 — Centre angle 110°
Angle at circumference (major arc) = ½ × 110° = 55°. (Circle theorem: the angle at the centre is twice the angle at the circumference standing on the same arc.)