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## Homework Statement

280m wide river, destination 120m upstream, river current is 1.35 m/s downstream and the boat speed in still water is 2.70 m/s. What should the boat's heading angle be (relative to the shore)?

## Homework Equations

[itex]V_x = Vcosθ[/itex]

[itex]V_y = Vsinθ[/itex]

[itex]x = V_0 t[/itex] (starting points set to 0 for x and y, no acceleration involved)

[itex]V_{BS} = V_{BW} + V_{WS}[/itex]

## The Attempt at a Solution

y component of velocity unaffected by current, so [itex]V_{BS} = V_{BW} + V_{WS}[/itex] will be relevent for x component only.

[itex]V_x = Vcosθ = 2.7 cosθ[/itex]

[itex]V_y = Vsinθ = 2.7 sinθ[/itex]

[itex]V_{BS} = V_{BW} + V_{WS} = (2.7 cosθ) - 1.35[/itex]

[itex]y = V_0 t = 280 = (2.7 sinθ)t[/itex]

[itex]t = 280/(2.7 sin)[/itex]

[itex]x = V_0 t = 120 = ((2.7 cosθ) - 1.35) t[/itex]

[itex]t = 120/((2.7 cosθ)-1.35)[/itex]

time it takes to go across river (y component) equals time to travel upstream (x component)

[itex]t = t = 280/(2.7 sin) = 120/((2.7 cosθ)-1.35)[/itex]

[itex]280((2.7 cosθ) - 1.35) = 120 (2.7 sinθ)[/itex]

[itex](756 cosθ) - 378 = (324 sinθ)[/itex]

[itex](14 cosθ) - 7 = (6 sinθ)[/itex]

[itex]7((2 cosθ) - 1) = (6 sinθ)[/itex]

[itex](2 cosθ) - 1 = (6/7) sinθ[/itex]

[itex]((2 cosθ) - 1)/sinθ = 6/7[/itex]

[itex](2(cosθ / sinθ)) - (1/sinθ) = 6/7[/itex]

[itex]2 cotθ - cscθ = 6/7[/itex]

I then plotted this using a graphing calculator and found where [itex]y = 2 cot(x) - csc(x)[/itex] crossed [itex]y = 6/7[/itex] and found θ = 39.5°. This answer does seem to check, but I have a feeling I'm doing something wrong, and that I should be able to solve for θ without using a graphing calculator. I'm not sure if the law of sines or law of cosines will come into play, or if I'm missing a certain trigonometric identity. Any insight would be greatly appreciated. Thanks.

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