Vector problem involving force and direction.?

Question

A force of 50lbs is directed 50 degrees above horizontal, pointing to the right. Determine its horizontal and vertical components.

I dont know where to begin. If you can help me out, that would be fantastic.

In this particular section of this book, we just learned that:

a . b = IIaII IIbII cos theta

IIaII = Sqrt (a1^2, a2^2, a3^2)

Projection of a onto b (P) = [(a . v)/ IIaII^2] * a

Thanks for any ideas.

Thank you emptydoubt for your replies. I will try this.

Answer 1

If you want to use the stuff you've written, one place to start is with this idea:
The horizontal and vertical components are the "projections" of the vector onto the x and y axes (respectively).

However, that may just be confusing... the easiest way:
Draw your vector as the hypotenuse of a triangle, and label what you know (hyp. length = 50 lb, 50 deg angle). Use trigonometry to find the other legs of the triangle, which are the horizontal and vertical components of the vector.

Answer 2

horizontal 50 cos(50)
vertical 50 sin(50)

draw the line and you see that it is just a triangle

Answer 3

rules to remember: horizontal component=Fcos(theta) vertical component=Fsin(theta), where F is the force

so horizontal component=50cos(50degrees)=32.13Ib
vertical component=50sin(50degrees)=38.3lb