SOLUTION: A ship A, sailing in a straight line with constant speed, is 10 nautical miles SW of a harbour H from which ship B is just leaving. B sails for two hours at 8 nautical miles per ho

Question 1118988: A ship A, sailing in a straight line with constant speed, is 10 nautical miles SW of a harbour H from which ship B is just leaving. B sails for two hours at 8 nautical miles per hour in the direction 105 degrees true at which time ships A and B collide.
a) show that the distance travelled by A in the two hours is 22.7 nautical miles.
c) find the bearing on which ship A was travelling
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A ship A, sailing in a straight line with constant speed, is 10 nautical miles SW of a harbour H from which ship B is just leaving.
B sails for two hours at 8 nautical miles per hour in the direction 105 degrees true at which time ships A and B collide.
a) show that the distance travelled by A in the two hours is 22.7 nautical miles.
ship B distance from the harbor 2*8 =16 nm
find the interior angle at the harbor. SW = 225 degrees true
225 - 105 = 120 degrees
Use law of cosines, a = side opposite 120 degrees
a^2 = 10^2 + 16^2 - 2(16*10)*cos(120)
a^2 = 356 - (-160)
a^2 = 516
a =
a = 22.7156
:
c) find the bearing on which ship A was travelling
Find the interior angle, at ship A's original position using the law of sines
=
22.7*sin(B) = 16*sin(120)
sin(b) =
B = 37.6 degrees
Using parallel lines and opposite angles
45 + 37.6 = 82.6 degrees true is the bearing of the A's course

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