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

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    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.

    COuld anyone please help explain how to figure this problem out?
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    Re: 123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help

    Fb = 35.551
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    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?
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    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 View Post
    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.

    COuld anyone please help explain how to figure this problem out?
    \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*}

    Now plug this into your calculator.
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    Re: 123.5 tan(68.3º) – 4.862 FB = 6.53 sin(75.0º) FB – 86.75 SOlve for FB Help

    KK i see where i went wrong. thank you.
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