A is alpha B is omega , not like it matters , but help is much appreciated thanks

http://i8.photobucket.com/albums/a48...athproblem.jpg

if the picture is not showing , the url is " http://i8.photobucket.com/albums/a48...athproblem.jpg "

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- May 4th 2008, 07:23 PMerockdontstoplaw of sines
A is alpha B is omega , not like it matters , but help is much appreciated thanks

http://i8.photobucket.com/albums/a48...athproblem.jpg

if the picture is not showing , the url is " http://i8.photobucket.com/albums/a48...athproblem.jpg " - May 4th 2008, 07:42 PMo_O
Let x be the bottom side of the smaller right triangle. Let H1 be the hypotenuse of the big right triangle and H2 be the hypotenuse of the smaller right triangle. Now, you can represent sinA, sinB, cosA, and cosB in terms of all these sides. So, we plug them into your right hand side (the double angle identity of sin(A - B) is needed to isolate all of the above):

$\displaystyle \text{RHS} = \frac{l\sin A \sin B}{\sin A \cos B - \cos A \sin B}$

$\displaystyle = \frac{l \left(\frac{h}{H_{1}}\right)\left(\frac{h}{H_{2}}\ right)} {\left(\frac{h}{H_{1}}\right) \left(\frac{x}{H_{2}}\right) - \left(\frac{l+x}{H_{1}}\right)\left(\frac{h}{H_{2} }\right)}$

Now if this was really representative of h, then this entire expression should simplify to h. Can you show that?