Question 144643
Two boats leave a marina at the same time. Boat A travels at 20km/hr in a direction of 65 degrees. Boat B travels at 12.5km/hr in a direction of 145 degrees. How far apart are the boats after 2 hours. (2) In what direction would the skipper of Boat A have to look to see Boat B. I have this.
Boat A 20km x 2 hrs = 40 km
Boat B 12.5 km x 2 hrs = 25 km.
They are offset by an angle of 145-65 = 80 degrees. Using the Law of Cosines I get
d^2 = 40^2 + 25^2 - 2(40)(25)cos(80 degrees) = 1877.70. so d = 43.33 km. 
The boats are 43.33 km apart after 2 hours.
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(2) In what direction would the skipper of Boat A have to look to see Boat B.
Find the angle (theta) formed by the 43.33 and the 25 km sides:
cos(theta) = (25^2+43.33^2-40^2)/(2*25*43.33) = 0.416565...
theta = 65.3826.. degrees
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Assuming you have drawn a picture of the problem on an x/y coordinate
system, draw a line parallel to the x-axis thru the vertex of the point
where boat B is.  The 25 km side is a trasversal between the two parallel
lines.
The total angle at B is 65.3826 degrees; the alternate interior angles
formed by the transversal are 180-145 = 35 degrees.  
So the line of sight from point B to point A is 65.38-35 = 30.38 degrees.
The line of sight from point A to B is 180 + 30.38 = 210.38 degrees
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Cheers,
Stan H.

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