trig
posted by trig .
h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d
(def of tan theta = a^1 o)a = o
opposite = adjacent tan theta
written with respect to the first angle
opposite = (adjacent 1 + adjacent 2) tan theta one
were adjacent 1 is the 50 feet
written for the second angle
opposite = adjacent 2 tan theta two
set them equal to each other
(adjacent 1 + adjacent 2) tan theta one = adjacent 2 tan theta two
distripute
adjacent 1 tan theta one + adjacent 2 tan theta one = adjacent 2 tan theta two
subtract from both sides
adjacent 1 tan theta one = adjacent 2 tan theta two  adjacent 2 tan theta one
factor or whatever
adjacent 1 tan theta one = adjacent 2 (tan theta two  tan theta one)
muliply by inverse to solve for adjacent 2
(adjacent 1 tan theta one = adjacent 2 (tan theta two  tan theta one)) ((tan theta two  tan theta one)^1
adjacent two = (tan theta two  tan theta one)^1 adjacent 1 tan theta one
plug and chug for adjacent two
adjacent two = (tan 32 degrees  tan 40 degrees)^1 (50 ft) tan theta 40 degrees
I got negative 195.8 feet?


that would be the image i put up...
i don\'t understand??? 
If you look at the figure, you will see that θ1 should be smaller than θ2, but you have labelled θ1=40, and θ2=32.
Perhaps the two values have been inverted and perhaps that is the source of your problem.
Could you please check and let us know? 
Yes, I translated the link so that other people could see your image without having to go through the hassle I had to.

hmmmm ok i\'ll try and see what i get thanks

high school Goraul vaishali