Checkpoint 3: Combined Topics
Test your understanding of sine rule, cosine rule, area formula and mixed problem solving. Covers Lessons 15 through 20.
Multiple Choice
Select the best answer for each question. Each question is worth 1 mark.
1 mark The sine rule is used when you know:
1 mark The cosine rule is used when you know:
1 mark A triangle has area 20, sides 8 and 7. The included angle is:
1 mark In triangle ABC, a = 6, b = 8, angle A = 30°. How many triangles?
1 mark Heron's formula for sides 5, 12, 13 gives area:
1 mark A triangle with sides 9, 12, 15 is:
1 mark The area of a triangle with sides 10, 10 and included angle 120° is:
1 mark Given a = 5, b = 7, c = 9, the largest angle is:
1 mark A ship sails 30 km on bearing 045°, then 40 km on bearing 135°. The direct distance back is:
1 mark The formula $A = \frac{1}{2}ab\sin C$ requires:
Short Answer
Show all working and justify your answers.
Question 11
In triangle ABC, angle A = 40°, angle B = 60°, and side c = 12 cm.
(a) Find angle C.
(b) Find sides a and b.
(c) Find the area of the triangle.
(d) Verify your area using Heron's formula.
(e) A point D is on AB such that CD is perpendicular to AB. Find CD.
Question 12
Three towns A, B, C form a triangle. AB = 20 km, BC = 25 km, angle ABC = 70°.
(a) Find AC.
(b) Find angle BAC.
(c) Find the area of the triangle ABC.
(d) A new town D is to be built equidistant from A, B, and C. Describe how to find this location and calculate the distance from D to each town.
Model Answers
(a) $C = 180 - 40 - 60 = 80°$.
(b) $\frac{a}{\sin 40°} = \frac{12}{\sin 80°}$. $a = \frac{12 \times 0.643}{0.985} = 7.8$ cm. $\frac{b}{\sin 60°} = \frac{12}{\sin 80°}$. $b = \frac{12 \times 0.866}{0.985} = 10.5$ cm.
(c) Area = $\frac{1}{2}(7.8)(10.5)\sin 80° = 40.95 \times 0.985 = 40.3$ cm squared. Or $\frac{1}{2}(12)(10.5)\sin 40° = 63 \times 0.643 = 40.5$ cm squared.
(d) $s = (7.8 + 10.5 + 12)/2 = 15.15$. $A = \sqrt{15.15(7.35)(4.65)(3.15)} = \sqrt{1634} \approx 40.4$ cm squared. Verified.
(e) $CD = \frac{2 \times \text{Area}}{AB} = \frac{2 \times 40.3}{12} = 6.7$ cm.
Marking guidance: 1 mark each for (a), (b), (c), (d), (e).
(a) $AC^2 = 20^2 + 25^2 - 2(20)(25)\cos 70° = 400 + 625 - 1000 \times 0.342 = 1025 - 342 = 683$. $AC = \sqrt{683} \approx 26.1$ km.
(b) $\frac{\sin A}{25} = \frac{\sin 70°}{26.1}$. $\sin A = \frac{25 \times 0.940}{26.1} = 0.900$. $A = \sin^{-1}(0.900) \approx 64.2°$.
(c) Area = $\frac{1}{2}(20)(25)\sin 70° = 250 \times 0.940 = 235$ km squared.
(d) D is the circumcentre. First find circumradius R: $R = \frac{abc}{4 \times \text{Area}} = \frac{20 \times 25 \times 26.1}{4 \times 235} = \frac{13050}{940} = 13.9$ km. D is at distance 13.9 km from each town. To construct: find perpendicular bisectors of AB and BC; their intersection is D.
Marking guidance: 1 mark for (a), 1 mark for (b), 1 mark for (c), 2 marks for (d).