Mathematics Standard • Year 12 • Trigonometry • Lesson 6
The Sine Rule, Skill Drill
Build fluency in the two arrangements of the sine rule: unknown side on top, and unknown angle found by flipping the rule. Then practise spotting when the third angle has to be found first.
1. Quick recall
Answer in the space provided. 1 mark each
Q1.1 Write the sine rule.
Q1.2 What is a "matched pair", and why does the sine rule need one?
Q1.3 The three angles of any triangle add to ____________.
2. Worked example, unknown side
Follow each line. Every step has its reason.
Problem. In triangle ABC, A = 40°, B = 75° and a = 12. Find b.
Step 1, find the matched pair.
a = 12 sits opposite A = 40°. That pair is complete.
Reason: without a side paired to its own opposite angle, the rule cannot start.
Step 2, put the unknown on top.
b / sin 75° = 12 / sin 40° → b = 12 sin75° / sin40°
Reason: with b on top the answer is a multiply and a divide, with nothing to rearrange.
Step 3, evaluate and sense check.
b = 12 × 0.965926 / 0.642788 = 18.03
Reason: 75° is bigger than 40°, so b must be longer than a. It is. The bigger angle always faces the bigger side.
3. Unknown side
Each of these already has a matched pair. Give answers to 2 decimal places. 1 mark each
Q3.1 A = 45°, B = 70°, a = 9. Find b.
Q3.2 B = 65°, C = 40°, b = 15. Find c.
Q3.3 C = 80°, A = 55°, c = 20. Find a.
4. Unknown angle
Flip the rule so the sines are on top, then take the inverse sine. 2 marks each
Q4.1 a = 7, A = 40°, b = 9. Find the acute angle B, to 2 decimal places and to the nearest minute.
Q4.2 b = 8, B = 50°, c = 6. Find the acute angle C, to 2 decimal places.
5. Third angle first
Here the known side has no known opposite angle. Find it before you substitute. 2 marks each
Q5.1 In triangle ABC, A = 38° and B = 72°. Find C.
Q5.2 In the same triangle, c = 14. Find a, to 2 decimal places.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The sine rule
a / sin A = b / sin B = c / sin C, where each lower-case side is opposite its capital-letter angle.
Q1.2, Matched pair
A side and the angle opposite it, both known. The rule equates fractions of that shape, so without one complete fraction there is nothing to equate the unknown to.
Q1.3, Angle sum
180°. This is what lets you manufacture a matched pair when the question does not hand you one.
Q3.1, A = 45°, B = 70°, a = 9
b = 9 sin70° / sin45° = 9 × 0.939693 / 0.707107 = 11.96. Check: 11.96 × sin45° = 8.46 and 9 × sin70° = 8.46.
Q3.2, B = 65°, C = 40°, b = 15
c = 15 sin40° / sin65° = 15 × 0.642788 / 0.906308 = 10.64. Since 40° is smaller than 65°, c is shorter than b, as expected.
Q3.3, C = 80°, A = 55°, c = 20
a = 20 sin55° / sin80° = 20 × 0.819152 / 0.984808 = 16.64.
Q4.1, a = 7, A = 40°, b = 9
sin B = 9 sin40° / 7 = 9 × 0.642788 / 7 = 0.826442 [1]. B = inverse sin(0.826442) = 55.73°, and 0.7349 × 60 = 44.1, so 55° 44′ [1]. Check: sin(55.73°) = 0.8264.
Q4.2, b = 8, B = 50°, c = 6
sin C = 6 sin50° / 8 = 6 × 0.766044 / 8 = 0.574533 [1]. C = inverse sin(0.574533) = 35.07° [1].
Q5.1, Third angle
C = 180° − 38° − 72° = 70°. Now c pairs with C, and the rule can start.
Q5.2, Find a
a = 14 sin38° / sin70° = 14 × 0.615661 / 0.939693 = 9.17 [2]. Check: 9.17 × sin70° = 8.62 and 14 × sin38° = 8.62.