How to Calculate Boat Angle for Straight River Crossing?

[kboykb]

Homework Statement

A: A boat capable of making 11.0 km/h in still water is used to cross a river flowing at speed of 3.0 km/h. At what angle (in o) must the boat be directed (from the perpendicular to the shore) so that its motion will be straight across the river?

B: What is the resultant speed relative to the shore?

Homework Equations

I'm not sure of the Exact Equations.. Possibly
Ax = A*cos(Theta) Ay = A*sin(Theta)
Bx = B*cos(Theta) By = B*sin(Theta)
Ax+Bx=Cx .. or Ax-Bx=Cx

(I'm at work right now, I forgot the last equation .. Ay/Ax = The tangent of Theta ? By/Bx)

The Attempt at a Solution

For my solution I drew a vector triangle. One directly North, one at the head of this vector going to the right (east) .. and then a resultant line connecting these two vectors. Mainly I just divided 11/3 and took the tangent of this, but couldn't find the correct answer. I can't really attempt the 2nd problem without the way to do the first.

Thanks in advance for the help :)

 

[Spinnor]

The vector of length 11 is the hypotenuse of the right triangle so you don't want the tangent(11/3). The tangent(3/11) would be closer but that is still wrong.

Draw a picture roughly to scale so you have an idea of the correct answer.

 

[Spinnor]

The vector of length 11 is the hypotenuse of the right triangle so you don't want the tangent(11/3). The tangent(3/11) would be closer but that is still wrong.

Draw a picture roughly to scale so you have an idea of the correct answer.

That should be arc-tangent(3/11) above, hope that was obvious.

 

[kboykb]

Okay, that does give me a better visual of everything. I have something that looks like this (Ignore the Text)

with 11 as the hypotenuse. If the top vector is 3, would the straight North line be 10.8? (By the pythagorean theorem)

I'm not sure what formulas to use here. I'm pretty sure I need to break it down into x and y components though