Solving Projectile Motion: Find x and y with Trig. Identities

[Cursed]

This isn't a problem, but I need help understanding this so I can do problems.

http://img255.imageshack.us/img255/3479/xythingat6.png

So let's say...
1. This is part of the motion of a projectile
2. You're trying to find x and y

There are trig. identities that will give you x and y.

Example:
x = sin(36)*10
y = cos(36)*10

My question is... Which trig. identity/function do you use to solve for x, and why do you use that identity? Also explain the same thing for y. Please and thank you.

 

[mgb_phys]

It's the basic defn of sin and cos.

Sin is the ratio of the opposite side over the hypotonuse
Cos is the ratio of the adjacent side over the hypotonuse

If you draw a line down to the X axis from the end of the slope (opposite the 36deg) you have a triangle.
The hypotonuse is 10units, the adjacent side is the distance in x and the opposite side (the exta line you just drew) the opposite side.

 

[Dick]

And following mgb_phys' instructions you should find x=cos(36)*10 and y=sin(36)*10, the opposite of what you wrote.

 

[Feldoh]

$$sin x = \frac{opposite}{hypotenuse}$$

$$cos x = \frac{adjacent}{hypotenuse}$$

So applying that to your problem:

$$sin(36) = \frac{y}{10}$$

$$cos(36) = \frac{x}{10}$$

From there it's really straight forward for solving for x and y.