River width by sine rule and bridge costing
A surveyor needs the width of a river. She stands at point A on one bank. A point B on the same bank is 60 m from A, measured along the straight bank. A landmark T on the opposite bank is visible from both A and B. Using a theodolite, she measures ∠TAB = 75° and ∠TBA = 62°. The bank is assumed straight and the ground level. A footbridge is to be built from the point D on the bank closest to T. The bridge costs MWK 85,000 per metre plus a fixed design fee of MWK 250,000. The local council requires the bridge landing to be at least 5 m from A and at least 5 m from B. The surveyor must decide whether the measurements support a feasible and affordable bridge.
1. Calculate the distances AT and BT using the sine rule.
Read the worked answer
Angle at T = 180° − 75° − 62° = 43°.
By the sine rule:
AT/sin 62° = AB/sin 43° = 60/sin 43°.
AT = 60 × sin 62° ÷ sin 43° = 60 × 0.88295 ÷ 0.68200 = 77.68 m.
BT = 60 × sin 75° ÷ sin 43° = 60 × 0.96593 ÷ 0.68200 = 84.98 m.
2. Calculate the perpendicular distance from T to the bank, which is the river width.
Read the worked answer
Drop perpendicular TD to the bank. In right triangle ATD, angle TAD = 75°, so:
TD = AT sin 75° = 77.68 × 0.96593 = 75.03 m.
Check using BT: TD = BT sin 62° = 84.98 × 0.88295 = 75.04 m.
The river width is about 75.0 m.
3. Find the position of D along the bank and check the 5 m clearance conditions. Then calculate the bridge cost.
Read the worked answer
D is the foot of the perpendicular, so:
AD = AT cos 75° = 77.68 × 0.25882 = 20.10 m.
DB = 60 − 20.10 = 39.90 m.
Both 20.10 m and 39.90 m exceed 5 m, so the council clearance condition is met.
Cost = 75.03 × 85,000 + 250,000 = 6,377,550 + 250,000 = MWK 6,627,550.
Judgement: the bridge is feasible on clearance, but the cost is about MWK 6.63 million. Because the span is sensitive to the measured angles, the survey should be repeated before committing to the fixed design fee.