Boat Velocity in a River: Angle for 4.0km/h & 1.8km/h

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To determine the angle a boat must be pointed upstream to cross a river with a current, the boat's speed in still water (4.0 km/h) and the current speed (1.8 km/h) are used. The relationship between these speeds is analyzed using trigonometric functions, specifically the sine function. By applying the equation sin-1(1.8/4), the angle calculated is 27 degrees upstream. This angle compensates for the current, allowing the boat to travel straight across the river. Understanding this vector relationship is key to solving similar problems involving motion in currents.
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A boat that can travel at 4.0 km/h in still water crosses a river with a current of 1.8 km/h. At what angle must the boat be pointed upstream to travel straight across the river? In other words, in what direction is the velocity of the boat relative to the water?

I have already found the answer due to the equation: sin-1(1.8/4)

The answer is 27 degrees upstream.

Reading through my book, I understand this has something to do with the boat's velocity equaling the vector sum of it's velocity in relation to the water?

Something like that. Can any of you clarify the relationship between the boat and water speeds, and why this equation yielded a correct answer? Thanks!
 
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the boat is trying to go straight across, but the water is pushing it to the side at 1.8km/h. so to compensate for the current, the boat has to travel at 27 degrees upstream. (picture a triangle).

~Amy
 
I understand that, but how do figure out that sin-1(1.8/4) is the method of reaching the 27 degree conclusion?
 
if you draw a right angled triangle, you ll see that the hypotenuse is 4 km/h and the opposite side is 1.8 km/h.

so using the trig thing SOH CAH TOA (let me know if you're not familiar with this), you'll see that you only have the hypo. side and the opposite side so the only one you can use is the sine function (SOH). you want to find the opposite angle.

sin = opposite/hypo. = 1.8/4 = 0.45 sin-1 = 27 degrees.

~Amy
 
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