Trig Help

[pointintime]

1. The problem statement, all variables and given/known data

56. Finding the Height of a Building
To measure the height of a building, two sightings are taken a distance of 50 feet apart. if the first angle of elevation is 40 degrees and the seconds is 32 degrees, what is the height of the building?

2. Relevant equations

defintion of tan theta

3. The attempt at a solution

http://img40.imageshack.us/img40/148/asdasdye.jpg

(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?

[pointintime]

nevermind