# percentage error in estimating the height of the building

• Sep 20th 2009, 02:40 PM
genlovesmusic09
percentage error in estimating the height of the building
A surveyor, standing 40 ft from the base of a building, measures the angle of elevation to the top of the building to be 70 degrees. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 5 percent?

I tried:
$\displaystyle \tan {70^ \circ } = \frac{h} {{40}}$
$\displaystyle h = 40(\tan {70^ \circ }) = 109.899$
but I'm not sure where to go from here
• Sep 20th 2009, 03:55 PM
aidan
Quote:

Originally Posted by genlovesmusic09
A surveyor, standing 40 ft from the base of a building, measures the angle of elevation to the top of the building to be 70 degrees. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 5 percent?

I tried:
$\displaystyle \tan {70^ \circ } = \frac{h} {{40}}$
$\displaystyle h = 40(\tan {70^ \circ }) = 109.899$
but I'm not sure where to go from here

Suppose H is actually the true height of the building
Your calculated value could be 5% high or 5% low.
If h is 5% higher than the true height (H) of the building then
h = H + 0.05H
h = 1.05H
h/1.05 = H = 104.6657
or
If h is 5% lower than the true height of the building then
h = H - 0.05H
h = 0.95H
h/0.95 = H = 115.6832

The angle of elevation must be between:
Angle1: arctan (115.6832/40)
&
Angle2: arctan (104.6657/40)

The question is "How accurately must the angle be measured..."?
Determine half the difference be Angle1 & Angle2.
That is how accurately it must be made.

Most likely you are required to get the angle to the nearest tenth of a degree.
or
it could be plus or minus minutes of arc.

.