# Sin Problem

• May 9th 2012, 07:31 PM
detex
Sin Problem
Hello everyone,
I'm having some trouble solving this equation.
Attachment 23826
All help would be appreciated :), Thanks!
This is my work that I've done so far...
Attachment 23827

PS I know this might not be the right section for a statics problem, but I'm really only having trouble with the trig part so yeah... lol:)
• May 9th 2012, 07:33 PM
detex
Re: Sin Problem
When I plugged back in my 58.91 degrees, I should have gotten 0, but I get something like 82.8 :( not 0.
• May 9th 2012, 09:28 PM
princeps
Re: Sin Problem
a)

$\begin{cases}90+70\cdot \sin \alpha-130 \cdot \cos \alpha=0 \\\sin^2 \alpha + \cos^2 \alpha=1 \end{cases}$
• May 9th 2012, 10:53 PM
srirahulan
Re: Sin Problem
how do you enter this cos alpha and other maths symbols in this text pad?
• May 10th 2012, 07:58 AM
detex
Re: Sin Problem
princeps, I tried that already (if you look in the picture of the work I showed), and when I solved for alpha I got 58.91 degrees. However, upon plugging it back into the problem I don't get 0.
• May 10th 2012, 08:56 AM
skeeter
Re: Sin Problem
starting where $\sum F_y = 0$ , 4th line ... one "egregious" algebra error.

note $(a-b)^2 \ne a^2 + b^2$
• May 10th 2012, 09:10 AM
detex
Re: Sin Problem
Ahhhh I see. Thank you :)!
• May 10th 2012, 09:32 AM
skeeter
Re: Sin Problem
$a\cos{x} - b\sin{x} = R\cos(x+t)$

$R = \sqrt{a^2+b^2}$ , $\tan{t} = \frac{b}{a}$

let x = your " $\alpha$" ...

$13\cos{x} - 7\sin{x} = 9$

$R = \sqrt{13^2 + 7^2}$ , $t = \arctan\left(\frac{7}{13}\right)$

$R \cos(x + t) = 9$

$x+t = \arccos\left(\frac{9}{R}\right)$

$x = \arccos\left(\frac{9}{R}\right) - \arctan\left(\frac{7}{13}\right) \approx 24.14^\circ$