The Triangular Shade Sail: Cosine Rule, Area and a Contractor's Claim
The school is fitting a triangular shade sail over a corner of the courtyard. It is fixed to three steel posts at A, B and C. Post A is the corner post; AB is 9.0 m, AC is 7.0 m, and the angle between them at A is 62°.
The third side BC takes a steel edge cable, and the cloth must cover the whole triangle with 8% extra allowed for hems and stretching. Shade cloth costs P95 per square metre and is sold only in whole square metres; cable is sold in 5 m coils at P42 each, with 5% extra needed for ties and wastage. The contractor describes the sail as 'near-equilateral, with no angle greater than 70°'. The principal wants that claim checked before signing, and the order must go in on Friday. Two of the posts are already set in concrete, so the shape cannot be changed - only the order quantities and the wording of the contract are still open.
1. Use the cosine rule to find the length of BC, then calculate the length of cable required and the number of coils to order, with the cost.
Read the worked answer
Cosine rule. BC² = AB² + AC² − 2 × AB × AC × cos A = 9.0² + 7.0² − 2 × 9.0 × 7.0 × cos 62°.
= 81 + 49 − 126 × 0.4695 = 130 − 59.15 = 70.85, so BC = √70.85 = 8.42 m (8.417 m unrounded).
Perimeter. 9.0 + 7.0 + 8.42 = 24.42 m. Add 5% for ties and wastage: 24.42 × 1.05 = 25.64 m.
Coils. 25.64 ÷ 5 = 5.13, so 5 coils (25 m) are not enough; order 6 coils = 30 m, costing 6 × P42 = P252.
Judgement. Rounding BC to 8.4 m changes nothing here (24.4 × 1.05 = 25.62 m still needs 6 coils), but rounding the 5% away would give 24.42 m and only 5 coils - a 1.4 m shortfall discovered on site. The wastage allowance, not the geometry, decides the order, and it must be applied before rounding up to whole coils.
2. Calculate the area of the sail and the cost of the shade cloth, allowing 8% extra and buying whole square metres.
Read the worked answer
Area (two sides and included angle). Area = ½ × AB × AC × sin A = ½ × 9.0 × 7.0 × sin 62°.
= 31.5 × 0.8829 = 27.81 m².
Add 8%. 27.81 × 1.08 = 30.04 m². Cloth is sold in whole square metres, so Neo must order 31 m².
Cost. 31 × P95 = P2,945.
Judgement. This is a case where rounding is decisive. Ordering 30 m² (P2,850) leaves the sail 0.04 m² short - the cloth cannot be joined invisibly and a second order would cost far more than the P95 saved. The 8% allowance must be applied to the calculated area before rounding up, not added as 8% of the rounded area (27.81 → 28 → 28 × 1.08 = 30.24, which still gives 31 m², so the answer is robust, but the order of operations should be stated). Using sin instead of cos, or using 62° as a side, are the errors to guard against; the area formula needs the included angle.
3. Find the largest angle of the triangle and decide whether the contractor's description ('near-equilateral, no angle greater than 70°') can be accepted.
Read the worked answer
Largest angle is opposite the longest side. The sides are AB = 9.0 m, AC = 7.0 m, BC = 8.42 m, so the longest is AB and the largest angle is C (opposite AB).
Cosine rule for C. cos C = (AC² + BC² − AB²) ÷ (2 × AC × BC) = (49 + 70.85 − 81) ÷ (2 × 7.0 × 8.42) = 38.85 ÷ 117.88 = 0.3296.
C = cos⁻¹(0.3296) = 70.8°.
Check. The angle at B: cos B = (81 + 70.85 − 49) ÷ (2 × 9.0 × 8.42) = 102.85 ÷ 151.56 = 0.6786, giving B = 47.3°. Angles sum to 62 + 47.3 + 70.8 = 180.1°, correct to rounding.
Verdict. The claim is wrong on both counts. The sides 7.0 m, 8.42 m and 9.0 m differ by up to 2 m - roughly 29% of the shortest side - so 'near-equilateral' exaggerates the symmetry; a near-equilateral sail would have all angles close to 60°. And the largest angle is 70.8°, which exceeds 70°, so the stated limit fails, though only marginally. The principal should require the wording to be corrected (e.g. 'largest angle 70.8°') rather than accept a figure that is contradicted by the measured geometry.