Sin Problem

**detex** · May 9th 2012, 06:31 PM

Hello everyone,
I'm having some trouble solving this equation.

All help would be appreciated

, Thanks!
This is my work that I've done so far...

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

**detex** · May 9th 2012, 06:33 PM

When I plugged back in my 58.91 degrees, I should have gotten 0, but I get something like 82.8

not 0.

**princeps** · May 9th 2012, 08:28 PM

a)

$\begin{cases}90+70\cdot \sin \alpha-130 \cdot \cos \alpha=0 \\\sin^2 \alpha + \cos^2 \alpha=1 \end{cases}$

**srirahulan** · May 9th 2012, 09:53 PM

how do you enter this cos alpha and other maths symbols in this text pad?

**detex** · May 10th 2012, 06:58 AM

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.

**skeeter** · May 10th 2012, 07:56 AM

starting where $\sum F_y = 0$ , 4th line ... one "egregious" algebra error.

note $(a-b)^2 \ne a^2 + b^2$

**detex** · May 10th 2012, 08:10 AM

Ahhhh I see. Thank you

!

**skeeter** · May 10th 2012, 08:32 AM

$a\cos{x} - b\sin{x} = R\cos(x+t)$

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

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

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

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

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

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

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

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