Mathematics Standard • Year 12 • Trigonometry • Lesson 6
The Sine Rule in Context, Application
Find the third angle before substituting, handle an angle the question tells you is obtuse, and use the sine rule on the surveying and height problems the HSC actually asks.
1. Third angle, then the side
Neither of these hands you a matched pair. Build one first. 3 marks each
Q1.1 In triangle ABC, A = 47°, B = 68° and a = 18. Find c, to 2 decimal places.
Q1.2 In triangle ABC, B = 105°, C = 32° and b = 25. Find a, to 2 decimal places.
2. Unknown angles, acute and obtuse
Read each question carefully. Only one of them tells you the angle is obtuse. 3 marks each
Q2.1 a = 10, A = 33°, b = 13. Find the acute angle B, to 2 decimal places.
Q2.2 a = 8, A = 24°, b = 15, and B is obtuse. Find B to 2 decimal places, then show that the triangle closes.
3. Surveying across an obstacle
The classic use of the rule: a distance nobody can pace out. 4 marks
Q3.1 A surveyor at P wants the distance QR across a gorge. She measures PQ = 800 m, the angle at P to be 55°, and the angle at R to be 40°. Find QR, correct to 2 decimal places.
4. A leaning tower
Draw the triangle before you touch the calculator. 4 marks
Q4.1 A pole leans away from an observer. From a point 60 m from its base, the angle of elevation to the top is 35°, and the angle the pole makes with the ground on the far side is 105°. Find the length of the pole, to 2 decimal places.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, A = 47°, B = 68°, a = 18
C = 180° − 47° − 68° = 65° [1]. The matched pair is a = 18 with A = 47°, so c = 18 sin65° / sin47° = 18 × 0.906308 / 0.731354 [1] = 22.31 [1]. Check: 22.31 × sin47° = 16.31 and 18 × sin65° = 16.31.
Q1.2, B = 105°, C = 32°, b = 25
A = 180° − 105° − 32° = 43° [1]. a = 25 sin43° / sin105° = 25 × 0.681998 / 0.965926 [1] = 17.65 [1]. The 105° angle causes no trouble: sin105° is a perfectly ordinary positive number.
Q2.1, a = 10, A = 33°, b = 13
sin B = 13 sin33° / 10 = 13 × 0.544639 / 10 = 0.708031 [2]. B = inverse sin(0.708031) = 45.07° [1]. Check: sin(45.07°) = 0.7080.
Q2.2, a = 8, A = 24°, b = 15, B obtuse
sin B = 15 sin24° / 8 = 15 × 0.406737 / 8 = 0.762632 [1]. The calculator returns the acute value 49.70° [1]. Because B is obtuse, B = 180° − 49.70° = 130.30° [1].
The triangle closes: 24° + 130.30° = 154.30°, which is under 180°, leaving about 25.70° for C. Had the sum exceeded 180°, the obtuse reading would have been impossible.
Q3.1, Across the gorge
The angle at Q is 180° − 55° − 40° = 85°, but it is not needed. PQ = 800 pairs with the angle at R = 40°, and the unknown QR pairs with the angle at P = 55° [2].
QR = 800 sin55° / sin40° = 800 × 0.819152 / 0.642788 = 1019.50 m [2]. Check: 1019.50 × sin40° = 655.32 and 800 × sin55° = 655.32. The answer being longer than the 800 m baseline is right, because 55° is the larger angle.
Q4.1, The leaning pole
The triangle has the 60 m distance, the 35° elevation at the observer, and 105° where the pole meets the ground. The third angle, at the top of the pole, is 180° − 35° − 105° = 40° [1].
The 60 m side is opposite the 40° angle, and the pole is opposite the 35° angle [1], so pole = 60 sin35° / sin105° = 60 × 0.573576 / 0.965926 [1] = 35.63 m [1].