# Math Help - resultant vectors

1. ## resultant vectors

Hi... I'm having a little trouble with this problem,

A power boat crossing a wide river has a compass heading of 25 degrees and speed relative to the water of 15 mph. The river is flowing in the direction of 135 degrees at 3.9 miles per hour. What is the boats actual velocity, that is, what is the speed and direction relative to the ground.

I know how to set this problem up, and then find the resultant vector using law of cosines. the thing I'm having trouble with is finding the angle measures in the triangle from the information given in the problem. I get the how to do the problem, but I'm just getting hung up on this one step.
Thanks!!

2. Originally Posted by seagirt24
Hi... I'm having a little trouble with this problem,

A power boat crossing a wide river has a compass heading of 25 degrees and speed relative to the water of 15 mph. The river is flowing in the direction of 135 degrees at 3.9 miles per hour. What is the boats actual velocity, that is, what is the speed and direction relative to the ground.

I know how to set this problem up, and then find the resultant vector using law of cosines. the thing I'm having trouble with is finding the angle measures in the triangle from the information given in the problem. I get the how to do the problem, but I'm just getting hung up on this one step.
Thanks!!
Well you can just break the vectors down into components

for the boat
$x_b=15\cos(25^\circ)\approx 13.6 \mbox{ and } y_b=15\sin(25^\circ) \approx 6.3$

so the vector of the boat is
$v_b=(13.6,6.3)$

doing the same for the water we get

$v_w=(-2.76,2.76)$

so now we can add the vectors to get the resultant

$v=(10.84,9.06)$

so the angle is given by

$\theta=\tan^{-1}(y/x)=\tan^{-1}\left( \frac{9.06}{10.84}\right)\approx 39.9^\circ$

and the velocity is the magnitue of the vector

$v=\sqrt{(10.84)^2+(9.06)^2} \approx 14.13$

I hope this helps