Is My Answer Correct for Boat Crossing a River Perpendicularly?

[Omid]

Let me know if the following statement is wrong :
If a boat is going to cross a river directly ( i.e perpendicularly ) the x, horizontal, component of the boat velocity must be cancelled, that's be equal, by the water stream vector.

I followed this idea in a problem but my answer is not true according to the solution manual included in the textbook.

 

[e(ho0n3]

The statement looks right to me (per my understanding). Maybe you have an error somewhere in your calculations. What is the solutions manual's explanation?

 

[Omid]

Here is both the problem and explanation :
If the water flows at 20 mi/h, and the boat at 30 mi/h in what angle would you head the boat to cut directly across the river?

v_BE cuts directly across the river. Head at v_BW somewhat upstream at an angle Theta such that v_BE = v_BW + v_WE, sin (Theta) = (20 mi/h)/(30 mi/h) = 0.667 so Theta = 42 degrees.


I assumed the horizontal component of v_boat to be 20 mi/h, so the vertical component became 22 mi/h then the sin(Theta) = (22 mi/h)/(30 mi/h) which leads to a Theta equal to 48.
What is wrong ?

 

[Doc Al]

Since the water moves east (I presume) at 20 mi/h, the boat must move at an angle so that its east-west component is 20 mi/h west. Thus $30 cos\theta = 20$, which gives the boat's needed direction as $\theta = 48\deg$ north of west.

 

[Omid]

Thus $30 cos\theta = 20$, which gives the boat's needed direction as $\theta = 48\deg$ north of west.

This is my answer too, but the answer in the solutions manual is 42 deg, so you mean that's wrong ?

 

[Doc Al]

This is my answer too, but the answer in the solutions manual is 42 deg, so you mean that's wrong ?

42 deg with respect to what? Realize that 48 deg N of W = 42 deg W of N.

 

[Omid]

sorry

. Realize that 48 deg N of W = 42 deg W of N.