Law of Sine or Cosine application problem? - sailing cosine
A ship leaves the port at 8.00 clock and runs at 10 miles per hour over a course of S E. 20degree An hour later, a ship leaves the port and sail 12 miles per hour over a course of S W. 60degree about how far you have lunch.
Wednesday, January 27, 2010
Sailing Cosine Law Of Sine Or Cosine Application Problem?
2 comments:
I hope that this is true, has been a while.
Ship 1: Travel 40 miles in 4 hours
Boat 2: Travel 36 miles in 3 hours
Angle between two boats: (20 +60) = 80
c ^ 2 = a ^ 2 + b ^ 2 - (2 * a * b * cos (angle))
c ^ 2 = 40 ^ 2 + 36 ^ 2 - (2 * 40 * 36 * cos (80))
c ^ 2 = 1600 + 1296 - (2 * 40 * 36 * 0.1736)
c ^ 2 = 2896 to 499.97 (or 500)
c ^ 2 = 2396
C = 48.95
At noon on the first ship has traveled 40 miles or 4x10.
The coordinates are:
40 sine 20 ° = 40 x 34 = 13.7 kilometers east of
40 cos 20 ° = 40 x 94 = 37.6 miles south
Around noon, the second boat 3x12 or 36 miles and is
36 sine 60 ° = 36 x 87 = 31.2 miles to the west
36 cos 60 ° = 36 x 0.5 = 18.0 miles south
The distance between the ships is the square root of the squared horizontal distance plus the vertical distance squared.
Horizontal distance = 13.7 + 31.2 = 44.9
Green = distance from 37.6 to 18.0 = 19.6
Square numbers: 2016.0 + 383.7
Add the squares: 2399.7
The square root is 49.0
The distance between the two ships
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