Three Villages: A Proposed Road and a 13 km Mast
Two villages, Bako and Chika, are surveyed from a third village, Audu. From Audu, Bako lies 12 km away on a bearing of 050°, and Chika lies 9 km away on a bearing of 128°. The three villages are joined by earth roads. A contractor proposes to build a direct road from Bako to Chika and claims the new road will be 'under 13 km, saving fuel'. A radio mast at Audu has a reliable range of 13 km, and the district officer wants to know whether the whole of the proposed road would fall inside that range. The officer also needs the bearing of Chika from Bako so that a direction signpost can be erected at Bako. All answers must be justified with written reasoning, distances to two decimal places and angles to one decimal place.
1. Calculate the distance from Bako to Chika.
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Find the included angle at Audu. Both bearings are measured from Audu, so the angle BAC is the difference between them: ∠BAC = 128° − 50° = 78°.
Apply the cosine rule to triangle ABC with AB = 12 km and AC = 9 km: BC² = 12² + 9² − 2(12)(9)cos 78° = 144 + 81 − 216(0.2079) = 225 − 44.91 = 180.09
BC = √180.09 ≈ 13.42 km.
Sanity check. A right angle at A would give √(144 + 81) = 15 km; the angle is only 78°, so the third side must be shorter than 15 km but longer than 12 − 9 = 3 km. 13.42 km fits. Note also that the direct road is longer than either existing leg from Audu, which is possible because Audu is not on the line BC - a common misjudgement is to assume BC must be less than 12 km.
2. Find the bearing of Chika from Bako, giving the reason for choosing between the two possible bearings.
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Angle at Bako (ABC). By the sine rule: sin(ABC) ÷ 9 = sin 78° ÷ 13.42 sin(ABC) = 9 × 0.9781 ÷ 13.42 = 8.8029 ÷ 13.42 = 0.6560 ∠ABC ≈ 41.0° (the acute value; the obtuse 139° cannot sit alongside the 78° already at Audu).
Then ∠ACB = 180° − 78° − 41.0° = 61.0°, and all three angles are acute, which confirms the choice.
Set up the bearing at Bako. The bearing of Audu from Bako is the back bearing: 050° + 180° = 230°.
Choose the direction. Chika lies 41° away from that line, giving candidates 230° + 41° = 271° (west) or 230° − 41° = 189° (south-southwest). Chika is south of Bako: Bako's northing from Audu is 12 cos 50° ≈ +7.71 km, while Chika's is 9 cos 128° ≈ −5.54 km, so the vector from Bako to Chika points southwards. Hence the bearing must be the southerly candidate.
Answer: bearing of Chika from Bako ≈ 189.0°.
Check. From Bako, Audu bears 230° and Chika bears 189°; the difference is 41°, matching ∠ABC. A signpost pointing to 271° would send travellers due west into open bush.
3. Evaluate the contractor's 'under 13 km' claim and determine whether the mast at Audu covers the entire proposed road.
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The claim. The road is 13.42 km, so it is not under 13 km; the contractor's figure is short by about 0.42 km, roughly 3%. Accepting the claim would understate fuel and construction costs, so the district officer should reject it.
Mast coverage - endpoints. AB = 12 km < 13 km and AC = 9 km < 13 km, so both villages are inside range.
Mast coverage - the road in between. The distance from Audu to a point on BC is shortest at the perpendicular foot and longest at an endpoint, provided the foot falls inside BC. Find the perpendicular distance h using area two ways: Area = ½(12)(9)sin 78° = 54 × 0.9781 ≈ 52.82 km² Also Area = ½ × BC × h → h = 2(52.82) ÷ 13.42 ≈ 7.87 km. Since ∠ABC = 41.0° and ∠ACB = 61.0° are both acute, the foot lies inside BC, so the greatest distance from Audu to any point on the road is the larger endpoint distance, 12 km.
Conclusion: 12 km < 13 km, so the mast covers the whole of the proposed road, including the midpoint (about 8.2 km from Audu). The contractor is wrong about the length but the road is still fully within radio range.