# Thread: Simple Statics / Trig problem

1. ## Simple Statics / Trig problem

Okay I can't seem to think of how to solve this

I basically need to have the y force be zero... currently I have

90 + 70 sin(x) - 130 cos(x) = 0

I feel like there is an inverse tan in my future but I can't get there...

2. Hello msuMath311
Originally Posted by msuMath311
Okay I can't seem to think of how to solve this

I basically need to have the y force be zero... currently I have

90 + 70 sin(x) - 130 cos(x) = 0

I feel like there is an inverse tan in my future but I can't get there...
This question doesn't make much sense. Do you need to solve this equation for $\displaystyle x$? If so, do this:

$\displaystyle 130\cos x - 70\sin x = 90$

Let $\displaystyle 130\cos x - 70 \sin x = R\cos(x + \alpha) = R\cos x\cos\alpha-R\sin x\sin\alpha$

$\displaystyle \Rightarrow R^2 = 130^2 + 70^2$ and $\displaystyle \tan\alpha = \frac{70}{130}$

Can you complete it from here?

3. ## Problem Edit

I'm sorry I realized I mixed and matched problems from my whiteboard....

I actually need to find the tensions on A and B, (I have the angles)

so...

A*cos(25) + B*cos(40) = -70*cos(10)

4. Hello msuMath311
Originally Posted by msuMath311
I'm sorry I realized I mixed and matched problems from my whiteboard....

I actually need to find the tensions on A and B, (I have the angles)

so...

A*cos(25) + B*cos(40) = -70*cos(10)
Still doesn't make sense! You'll need a second equation in order to solve for A and B.