Question

Two ships leave a port at the same time. The first ship sails
at 38 knots on a bearing of 042º and the second ship sails at
25 knots on a bearing of 315º. How far apart are the ships two
hours later? [1 knot is a speed of 1 nautical mile per hour. ]​

Answer 1

Given :- Two ships leave a port at the same time. The first ship sails at 38 knots on a bearing of 042º and the second ship sails at 25 knots on a bearing of 315º. How far apart are the ships two hours later ? [1 knot is a speed of 1 nautical mile per hour. ]

Solution :-

Let us assume that, both ships leave the port A at the same time.

→ speed of first ship = 38 knots = 38 nautical mile/ hour.

→ Time = 2 hours.

so,

→ Distance travelled by first ship (AB) = speed * time = 38 * 2 = 76 nautical miles.

similarly,

→ speed of second ship = 25 knots = 25 nautical mile/ hour.

→ Time = 1 hours.

so,

→ Distance travelled by first ship (AC) = speed * time = 25 * 2 = 50 nautical miles.

now,

→ Angle between both ships after 1 hour = 42° + (360 - 315)° = 42° + 45° = 87° = ∠BAC .

therefore, using cosine rule in ∆ABC we get,

→ BC² = AB² + AC² - 2 * AB * AC * cos (∠BAC)

→ BC² = 76² + 50² - 2 * 76 * 50 * cos 87°

→ BC² = 5776 + 2500 - 7600 * 0.052

→ BC² = 8276 - 395.2

→ BC² = 7880.8

→ BC ≈ 88.77 nautical miles (Ans.)

Hence, the ships are 88.77 nautical miles apart 2 hours laters.

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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