# Percent Error (Simple Problem?)

• Oct 22nd 2008, 10:56 AM
jackbunny
Percent Error (Simple Problem?)
Hi, I'm new here, but I was looking for some help with a Calc problem and here I am. Odd thing about this problem is we worked it out in class, but when checking the answer it seems wrong. I talked to my calc teacher and she's stumped too, so can anyone see what's wrong here?

A surveyor is standing 50 feet from the base of a tree. He measures his angle of elevation to the top of the tree is 71.5 degrees. How accurately must he measure his angle of elevation to measure the within a 6% error.

This is what I did. Set up a triangle and get

$h = 50tan(a)$

Differentiating the equation I wind up with

$dh/da = 50sec^2(a)$

Because we want dh/h to be less than 6% I set dh = .06h
A little algebra gives me :

$da = dh/(50sec^2(a)) = .06h/(50*sec^2(a))$
$
da = (.06*50tan(a))/(50sec^2(a))
$

Cancelling the 50's and converting the

$sec^2$ to $1/cos^2$

I get

$
da = .06tan(a)cos^2(a)
$

Exact height at 71.5 degrees would come out to be 149.4342.

Plugging in 71.5 my da comes out to be roughly .0181
At first I thought this was degrees which gives no where near a 6% differences in the height. Then I thought it was radians. Converting .0181 from radians to degrees gives 1.03 degree difference. Applying this gives a 6.3% difference in height.. which is close, but it's still over my tolerance.

I know it's rubbish but i did discover that 1 - .0181 = .9819 which added to my 71.5 degrees and applied to the height equation gives me the 6% exactly, but I can't think of any logical reason I would want to subtract my result from 1.

Any help on this (admittedly long winded) post would be greatly appreciated. Thanks.
• Oct 23rd 2008, 05:55 AM
jackbunny
I hate to do it, but I'm going to give this a little bump in hopes someone will be able to help. Thanks.