123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help

• Sep 12th 2012, 08:10 PM
gramgilb
123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help
I am currently doing homework and have been struggilng over this question for quite some time trying many different things.
Everytime i get an answer i plug it into both equations only to find it is wrong.

• Sep 12th 2012, 08:28 PM
MaxJasper
Re: 123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help
Fb = 35.551
• Sep 12th 2012, 08:29 PM
gramgilb
Re: 123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help
OK but how did you find that?
• Sep 12th 2012, 08:32 PM
Prove It
Re: 123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help
Quote:

Originally Posted by gramgilb
I am currently doing homework and have been struggilng over this question for quite some time trying many different things.
Everytime i get an answer i plug it into both equations only to find it is wrong.

\displaystyle \displaystyle \begin{align*} 123.5\tan{\left( 68.3^{\circ}\right)} - 4.862FB &= 6.53\sin{\left( 75^{\circ} \right)}FB - 86.75 \\ 123.5\tan{\left( 68.3^{\circ} \right)} + 86.75 &= 6.53 \sin{\left( 75^{\circ} \right)}FB + 4.862FB \\ 123.5\tan{\left( 68.3^{\circ} \right)} + 86.75 &= \left[ 6.53 \sin{\left( 75^{\circ} \right)} + 4.862 \right] FB \\ \frac{123.5\tan{\left( 68.3^{\circ} \right)} + 86.75}{6.53\sin{\left( 75^{\circ} \right)} + 4.862} &= FB \end{align*}