Two forces are acting on a 0.250-kg hockey puck as it slides along the ice. The first force has a magnitude of 0.405 N and points 25.0° north of east. The second force has a magnitude of 0.565 N and points 75.0° north of east. If these are the only two forces acting on the puck, what will be the magnitude and direction of the puck\'s acceleration? Enter the direction as an angle measured in degrees counterclockwise from due east.
The Attempt at a Solution
for the first force I got 0.405cos(25) = 0.367 (right) 0.405sin(25)=0.171 (up)
then for the second force 0.565cos(75)=0.146 (right) 0.565sin(75)=0.546 (up)
So next I would add all the forces 0.367+0.146=0.513N and 0.171+0.546=0.717N
then I don't know where to go from there