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Two boats leave a marina at the same time. The first boat travels 6 knots at a bearing of N39, and the second boat travels 4 knots at a bearing of S87°W. 2 degrees * E a. How far apart are the boats at the end of 2 hr? b. What is the bearing from the first boat to the second boat at that time?

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Okpalawalter8idn

For the first boat, 6 knots at a bearing of N39, and for the second boat  4 knots at a bearing of S87°W.  the distance of boats at the end of 2 hr and the bearing of 1st to 2nd both is mathematically given as

  • V'=22.93Km
  • B=132°

What is the distance b/w the boats at the end of 2 hr and the bearing of 1st to 2nd boat?

Generally,  the Relative speed  is mathematically given as

V=\sqrt{4^2+6^2+2(4)(6)cos(48)}

V=\sqrt{52+48(0.669)}

X=9.17Knots

Hence, distance b/w the boats

V'=9.17Knots*2

V'=18.34

V'=22.93Km

In conclusion, the bearing of 1st to 2nd boa ist

B=39+90+(90-87)

B=132°

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