math
posted by lulu .
a hotair balloon is seen in the sky simultaneously by two observers standing a t two different points on the ground. (the 2 people are facing each other and the balloon is in between them in the air.)they are 1.75 miles apart on level ground. The angles of elevation are 33 and 37 respectively. how high above the ground is the balloon?

make a diagram
let the base be BC, where angleB= 37 and angle C = 33°
From the top vertex A drop an altitude to meet BC at D
let AD = h, BD = x, then DC = 1.75x
you now have two rightangled triangles
from ABD,
tan 37 = h/x
h = xtan37
from ADC
tan 33 = h/(1.75x)
h = tan33(1.75x)
then x tan37 = 1.75tan33  xtan33
xtan37  xtan33 = 1.75tan33
x(tan37tan33) = 1.75tan33
x = 1.75tan33/(tan37tan33)
once you have that evaluated, put that x value into
h = xtan37
you do all the buttonpushing. 
i got up to the very last step but when i plugged my x value into h=xtan37 i just got decimals, like .97131.. what is the answer supposed to be?