see the attachment

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- Dec 7th 2008, 04:00 AMhelloyingbearing
see the attachment

- Dec 7th 2008, 06:31 AMskeeter
http://www.mathhelpforum.com/math-he...ing-img013.jpg

$\displaystyle AF = \frac{h}{\tan(30)}

$

$\displaystyle BF = \frac{h}{\tan(50)}$

$\displaystyle (AF)^2 + (BF)^2 = 96^2$

substitute and solve for h ... then use what you get for h to find either AF or BF, then use the appropriate trig ratio to find the requested bearing.