Pre-Calculus question! Please Help!
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Pre-Calculus question! Please Help!

[From: ] [author: ] [Date: 12-03-20] [Hit: ]
6)t = +/- 107.14 degreesy = 12 - 2 * sqrt(2)9.6 * sin(t) = 12 - 2 * sqrt(2)sin(t) = (12 - 2 * sqrt(2)) / 9.6t = arcsin(12 - 2 * sqrt(2)) / 9.6t = 107.14 ,......

y[1] = 12 * sin(90) = 12 * 1 = 12
y[2] = 4 * sin(225) = 4 * -sqrt(2)/2 = -2 * sqrt(2)

x[1] = 12 * cos(90) = 12 * 0 = 0
x[2] = 4 * cos(225) = 4 * -sqrt(2)/2 = -2 * sqrt(2)


Overall, y = 12 - 2 * sqrt(2) and x = -2 * sqrt(2)

sqrt(x^2 + y^2) =>
sqrt(8 + 144 - 48 * sqrt(2) + 8) =>
sqrt(160 - 48 * sqrt(2)) =>
sqrt(16) * sqrt(10 - 3 * sqrt(2)) =>
4 * sqrt(10 - 3 * sqrt(2)) =>
9.5977991751281938175426819963171


x = -2 * sqrt(2)
9.6 * cos(t) = -2 * sqrt(2)
cos(t) = -2 * sqrt(2) / 9.6
t = arccos(-2 * sqrt(2) / 9.6)
t = +/- 107.14 degrees

y = 12 - 2 * sqrt(2)
9.6 * sin(t) = 12 - 2 * sqrt(2)
sin(t) = (12 - 2 * sqrt(2)) / 9.6
t = arcsin(12 - 2 * sqrt(2)) / 9.6
t = 107.14 , 72.86 degrees

107.14 degrees (17.14 degrees past the 90 degree mark I set for North; you should note that 342.86 degrees is 17.14 degrees from 360 degrees, so the math is right, I just oriented the bearings incorrectly) at 9.6 mph
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