Mathematics Standard • Year 12 • Trigonometry • Lesson 7
The Cosine Rule and the Area Rule in Context
Apply both rules to land, navigation and construction problems, where nobody labels the triangle for you. The first job in every question below is deciding which rule the given information allows.
1. Choosing the rule
For each situation, name the rule you would use and say what makes it the right one. Do not calculate. 1 mark each
Q1.1 Two roads leave a junction at a known angle. You know how far along each road two towns sit, and want the direct distance between them.
Q1.2 A triangular block of land has all three boundaries measured, and you want the angle at one corner.
Q1.3 You know one side of a triangular sail, the angle opposite that side, and one more angle, and you want a second side.
Q1.4 A council needs the area of a triangular reserve. Two boundaries and the angle between them are on the plan.
2. Navigation
Draw a diagram first, with north marked, before writing any rule. 3 marks each
Q2.1 Two ships leave the same port at the same time. Ship A sails 18 km on a bearing of 030°T. Ship B sails 25 km on a bearing of 105°T. How far apart are they, correct to 2 decimal places?
Q2.2 Three towns are joined by straight roads. A to B is 45 km, B to C is 62 km and A to C is 88 km. Find the angle at B, correct to the nearest degree, and say how you knew the answer would be obtuse before finishing.
3. Land and construction
4 marks each
Q3.1 A triangular block of land has two street frontages of 42 m and 58 m meeting at a corner of 112°.
(a) Find the length of the third boundary, correct to 2 decimal places.
(b) Find the area of the block, correct to the nearest square metre.
Q3.2 A triangular park has sides of 120 m, 150 m and 190 m. Find its largest angle correct to the nearest minute, and then its area correct to the nearest square metre.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1
Cosine rule, side form. Two sides with the angle between them is SAS, and no side is paired with the angle opposite it, so the sine rule cannot start.
Q1.2
Cosine rule, angle form. Three sides and no angle is SSS. With no angle at all there is nothing for the sine rule to work from.
Q1.3
Sine rule. A side with the angle opposite it is a matched pair, and the second angle gives a second complete fraction. The cosine rule is not needed and would not help, because only one side is known.
Q1.4
Area rule, ½ ab sin C. Two sides with the angle between them is exactly what the rule takes, and no perpendicular height has to be found or measured.
Q2.1, Two ships
The angle at the port is the difference of the bearings, 105° − 030° = 75° [1]. That is two sides and the angle between them, so d² = 18² + 25² − 2(18)(25) cos75° = 949 − 900(0.258819) = 949 − 232.9371 = 716.0629 [1]. So d = 26.76 km [1]. Sense check, the ships diverge by less than a right angle, so they end up closer than √949 = 30.81 km apart.
Q2.2, Three towns
You knew it was obtuse before finishing because AC = 88 km is longer than √(45² + 62²) = √5869 = 76.6 km, so the angle facing it must be bigger than a right angle [1]. cos B = (45² + 62² − 88²) / (2 × 45 × 62) = (5869 − 7744) / 5580 = −1875 / 5580 = −0.336022 [1], and the negative confirms it. B = 110° to the nearest degree [1].
Q3.1, Triangular block
(a) d² = 42² + 58² − 2(42)(58) cos112° = 5128 − 4872(−0.374607) = 5128 + 1825.09 = 6953.09 [1], so the third boundary is 83.39 m [1]. The obtuse corner makes cos112° negative, so the correction term is added and the boundary is longer than √5128 = 71.61 m.
(b) Area = ½(42)(58) sin112° = 1218 × 0.927184 = 1129.31, which is 1129 m² [2]. The same two sides and the same angle serve both parts, which is why the area is one line once (a) is done.
Q3.2, Triangular park
The largest angle faces the longest side, 190 m. cos θ = (120² + 150² − 190²) / (2 × 120 × 150) = (36900 − 36100) / 36000 = 800 / 36000 = 0.022222 [1]. So θ = 88.7267°, and 0.7266 × 60 = 43.6, giving 88° 44′ [1]. The cosine is only just positive, which says the park is very nearly right-angled at that corner.
Area = ½(120)(150) sin88.7267° = 9000 × 0.999753 = 8997.78, which is 8998 m² [2]. Sense check, at a full 90° the area would be exactly 9000 m², and 88° 44′ is barely under that.