# Basic Trig Word Problem

• Nov 21st 2010, 05:32 PM
LaMandie
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!
• Nov 21st 2010, 05:38 PM
Soroban
Hello, LaMandie!

And we can tell you where you made errors.

Isn't that easier for all of us?

• Nov 21st 2010, 05:41 PM
willy0625
Since the height of the building at either point should be the same we get the equality

$xtan(32) = (x-70)tan(44)_{.}$

Expanding the R.H.S. of the equation and rearranging to make x the subject, we get

$x = \frac{70tan(44)}{tan(44)-tan(32)}_{.}$

See what this comes out as on your calculator.
• Nov 21st 2010, 05:48 PM
LaMandie
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
• Nov 21st 2010, 05:59 PM
LaMandie
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!
• Nov 21st 2010, 06:46 PM
harish21
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=(x-70)m

so,
$\tan 32=\frac{AB}{BC}=\frac{AB}{x}$.....(I)

and

$\tan 44=\frac{AB}{AD}=\frac{AB}{x-70}$......(II)

from (I), you have $AB = x \times \tan 32$

and from (II), you have $AB = (x-70) \times \tan 44$

so, you have:

$x(\tan 32 = x(\tan 44)-70(\tan 44)$

$70(\tan 44) = x(\tan 44)-x(\tan32)$

$70(\tan 44) = x (\tan 44 -\tan 32)$

$x = \dfrac{70(\tan 44)}{\tan 44 -\tan 32}$

then substitute the value of x in equation (I) to find AB
• Nov 21st 2010, 07:29 PM
LaMandie
Thanks a bunch! :)