bearing

• December 7th 2008, 05:00 AM
helloying
bearing
see the attachment
• December 7th 2008, 07:31 AM
skeeter
http://www.mathhelpforum.com/math-he...ing-img013.jpg

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

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

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

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.