
Basic Trig Word Problem
" From point A, the angle of elevation to the top of a building is 32. Walking 70m closer, the angle changes to 44. How high is the building?"
So far, I've gotten this:
xtan44 = (x + 70)tan32
Each time I try it though, I get completely different answers.
All I need is for someone to just do it themselves while showing how they did it.
Thankyou very much!

Hello, LaMandie!
Show us your work for your different answers.
And we can tell you where you made errors.
Isn't that easier for all of us?

Since the height of the building at either point should be the same we get the equality
$\displaystyle xtan(32) = (x70)tan(44)_{.}$
Expanding the R.H.S. of the equation and rearranging to make x the subject, we get
$\displaystyle x = \frac{70tan(44)}{tan(44)tan(32)}_{.}$
See what this comes out as on your calculator.

Alrighty then!
1. xtan44 = (x + 70) tan32
x0.9657 = (x + 70)0.6249
x0.9657 = x0.6249 + 43.7409
x = x0.6249 + 43.7409 / 0.9658
x = x0.6249 + 45.2898
1 = 0.6249 + 45.2898
1 = 45.9147
2. xtan44 = (x + 70) tan32
x = (x + 70) tan32/tan44
x = ( x + 70) 0.6471
x = 0.6471x + 45.295
1 = 29.3104
3. xtan44 = (x + 70) tan32
xtan44 / tan 32 = (x + 70)
x1.545 = (x + 70)
1.545 = (x + 70) / x
1.545 = 70
4. xtan44 = (x + 70)tan32
xtan44 = xtan32 + 70tan32
x = xtan32 + 70tan32 / tan44
1 = tan32 + 70tan32 / tan44
1 = 43.4

willy, thanks for the help! The answer is right only I don't really get how you got 70tan44 / (tan44  tan32).
Could you explain this for me please? Thanks a bunch!

I'll give you the steps.. hope you can draw the picture..
Draw a right angled triangle ABC, where angle B=90 degrees.. AB is the height of the building..
angle ACB=32 degrees and BC is the base of the triangle. suppose that the length of BC=x..
if you move 70m close to the building from pint C, then the angle is 44 deg.
so there is a point D on the line BC such that angle ADB=44 deg..
Also, since BC = x, and CD=70m, you have BD=(x70)m
so,
$\displaystyle \tan 32=\frac{AB}{BC}=\frac{AB}{x}$.....(I)
and
$\displaystyle \tan 44=\frac{AB}{AD}=\frac{AB}{x70}$......(II)
from (I), you have $\displaystyle AB = x \times \tan 32$
and from (II), you have $\displaystyle AB = (x70) \times \tan 44$
so, you have:
$\displaystyle x(\tan 32 = x(\tan 44)70(\tan 44)$
$\displaystyle 70(\tan 44) = x(\tan 44)x(\tan32)$
$\displaystyle 70(\tan 44) = x (\tan 44 \tan 32)$
$\displaystyle x = \dfrac{70(\tan 44)}{\tan 44 \tan 32}$
then substitute the value of x in equation (I) to find AB
