At 12 noon a ship observes a lighthouse at a distance of 15 km in a direction of N50°E. It sails at 15 km/h in a direction S35°W. How do you find the distance and direction of the lighthouse from the ship at 3 pm?

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At 12 noon a ship observes a lighthouse at a distance of 15 km in a direction of N50°E. It sails at 15 km/h in a direction S35°W. How do you find the distance and direction of the lighthouse from the ship at 3 pm?

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Answer 1 · 2015-09-19T22:25:01

Find distance from boat to lighthouse

Explanation:

Call A the boat, B the lighthouse, and C the new position of the boat.
We have: AB = 15 ; AC = 15 km x 3 hrs = 45
Angle A = (180 - 50) + 35 = 165 deg.
To find the side BC, apply the trig identity:
$a^{2} = b^{2} + c^{2} - 2 b c . cos A$ $a^2 = b^2 + c^2 - 2bc.cos A$
$a^{2} = {(15)}^{2} + {(45)}^{2} - 2 (45) (15) . cos 165 =$ $a^2 = (15)^2 + (45)^2 - 2(45)(15).cos 165 =$
$a^{2} = 225 + 2025 - 1350 (- 0.97) = 3559.5$ $a^2 = 225 + 2025 - 1350(-0.97) = 3559.5$

$a = B C = 59.66$ $a = BC =59.66$ Km

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At 12 noon a ship observes a lighthouse at a distance of 15 km in a direction of N50°E. It sails at 15 km/h in a direction S35°W. How do you find the distance and direction of the lighthouse from the ship at 3 pm?

1 Answer

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