Printable Worksheets
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Revision and Problem Solving
The final challenge. Every concept from Unit 3 in mixed problems. Test yourself, find gaps, and master trigonometry and geometry.
Worksheet
Download or print the worksheet to work through this lesson.
Before we begin: Look back at all 20 lessons. Which topic do you feel most confident about? Which topic do you find most challenging? What specific skill would you like to improve before the unit quiz?
Type your initial response below. You will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Come back to this after you have worked through the lesson.
Learning Intentions
Know
- All key formulas and theorems from Unit 3.
- Common problem types and strategies.
Understand
- How different topics connect.
- When to apply each technique.
Can Do
- Solve mixed problems from any topic.
- Identify the appropriate method quickly.
- Prepare effectively for assessment.
Success Criteria
- I can solve problems involving right-angled and non-right triangles.
- I can apply congruence, similarity and circle theorems.
- I can choose between sine rule, cosine rule and area formula.
- I am ready for the Unit 3 quiz.
Key Terms - Unit 3 Summary
Formula Summary
Complete Formula Summary
All formulas from Unit 3 in one place.
| Topic | Formula | When to use |
|---|---|---|
| Right trig | sin, cos, tan ratios | Right triangles |
| Inverse trig | sin⁻¹, cos⁻¹, tan⁻¹ | Finding angles |
| Pythagoras | $c^2 = a^2 + b^2$ | Two sides, want third |
| Sine rule | $\frac{a}{\sin A} = \frac{b}{\sin B}$ | AAS, ASA, SSA |
| Cosine rule | $c^2 = a^2 + b^2 - 2ab\cos C$ | SAS, SSS |
| Area | $A = \frac{1}{2}ab\sin C$ | SAS |
| Heron | $A = \sqrt{s(s-a)(s-b)(s-c)}$ | SSS |
Interactive: Unit 3 Revision Quiz
Your Turn
Question 1: A tower stands on a hill. From a point 50 m downhill, the angle of elevation to the base is 15° and to the top is 35°. Find the height of the tower.
Question 2: Two circles of radii 5 cm and 12 cm touch externally. Find the length of their common external tangent.
Copy Into Your Books
Revisit Your Thinking
Look back at your Think First response. Now that you have worked through all 20 lessons, review your self-assessment. Which topics do you still need to practice? Make a study plan for the unit quiz.
Multiple Choice
Select the best answer for each question.
1 mark A triangle with sides 9, 40, 41 is:
1 mark The angle at the centre is twice the angle at the circumference. If the circumference angle is 35°, the centre angle is:
1 mark Two similar triangles have sides in ratio 2:5. Their areas are in ratio:
1 mark Given SSA with a = 8, b = 10, A = 30°. The number of possible triangles is:
1 mark A tangent to a circle makes angle 50° with a chord. The angle in the alternate segment is:
Short Answer
Show all working and justify your answers.
Question 6
A quadrilateral ABCD has AB = 8 cm, BC = 10 cm, CD = 12 cm, DA = 15 cm, and diagonal AC = 14 cm.
(a) Find angle ABC using the cosine rule.
(b) Find angle ADC using the cosine rule.
(c) Is ABCD a cyclic quadrilateral? Justify your answer.
Question 7
From a cliff 80 m high, the angle of depression to a boat is 25°. From the same point, the angle of depression to a second boat is 40°. The two boats are on the same bearing from the cliff.
(a) Find the distance from the cliff to each boat.
(b) Find the distance between the two boats.
(c) If a third boat is midway between them, what is its angle of depression from the cliff?
Question 8
Three points A, B, C on a map form a triangle. AB = 15 km, angle BAC = 50°, angle ABC = 60°.
(a) Find all sides and angles of the triangle.
(b) Find the area of the triangle.
(c) A new road is built from A to the midpoint M of BC. Find the length of AM.
(d) A surveyor claims that AM can be found using Apollonius's theorem: $AB^2 + AC^2 = 2(AM^2 + BM^2)$. Verify this.
Model Answers
(a) $\cos B = \frac{8^2 + 10^2 - 14^2}{2(8)(10)} = \frac{64 + 100 - 196}{160} = \frac{-32}{160} = -0.2$. $B = \cos^{-1}(-0.2) \approx 101.5°$.
(b) $\cos D = \frac{15^2 + 12^2 - 14^2}{2(15)(12)} = \frac{225 + 144 - 196}{360} = \frac{173}{360} = 0.481$. $D = \cos^{-1}(0.481) \approx 61.3°$.
(c) $B + D = 101.5 + 61.3 = 162.8° \neq 180°$. No, ABCD is not cyclic (opposite angles do not sum to 180°).
Marking guidance: 1 mark each for (a), (b), (c).
(a) Distance to first boat = $80/\tan 25° = 80/0.466 = 171.7$ m. Distance to second boat = $80/\tan 40° = 80/0.839 = 95.4$ m.
(b) Distance between boats = $171.7 - 95.4 = 76.3$ m.
(c) Midway distance = $(171.7 + 95.4)/2 = 133.55$ m. Angle = $\tan^{-1}(80/133.55) = \tan^{-1}(0.599) \approx 30.9°$.
Marking guidance: 1 mark for (a), 1 mark for (b), 2 marks for (c).
(a) Angle C = $180 - 50 - 60 = 70°$. $\frac{AC}{\sin 60°} = \frac{15}{\sin 70°}$. $AC = \frac{15 \times 0.866}{0.940} = 13.8$ km. $\frac{BC}{\sin 50°} = \frac{15}{\sin 70°}$. $BC = \frac{15 \times 0.766}{0.940} = 12.2$ km.
(b) Area = $\frac{1}{2}(15)(13.8)\sin 60° = 103.5 \times 0.866 \approx 89.6$ km squared.
(c) $BM = BC/2 = 6.1$ km. Using cosine rule in triangle ABM: $AM^2 = 15^2 + 6.1^2 - 2(15)(6.1)\cos 60° = 225 + 37.2 - 91.5 = 170.7$. $AM = \sqrt{170.7} \approx 13.1$ km.
(d) LHS = $15^2 + 13.8^2 = 225 + 190.4 = 415.4$. RHS = $2(13.1^2 + 6.1^2) = 2(171.6 + 37.2) = 2(208.8) = 417.6$. Small difference due to rounding. The theorem is verified.
Marking guidance: 1 mark for (a), 1 mark for (b), 2 marks for (c), 1 mark for (d).
Consolidation Game
Test your knowledge from this lesson and previous lessons.