A surveyor,standing 30ft from the base of a building,measures the angle of elevation to the top of the building to be 75 degree.How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%?

2. Originally Posted by turkeyy
A surveyor,standing 30ft from the base of a building,measures the angle of elevation to the top of the building to be 75 degree.How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%?
Well the height is:

$\displaystyle h=30 \tan(\theta)$

where theta is the angle of elevation of the top of the building. Now suppose that we measure the elevation with error $\displaystyle \Delta \theta$, and then compute a height $\displaystyle h+\Delta h$ where $\displaystyle \Delta h$ is the error in the height. so:

$\displaystyle h+\Delta h = 30 \tan(\theta + \Delta \theta)\approx 30 \tan(\theta)+\Delta \theta \frac{d}{d \theta}[30 \tan(\theta)]$ $\displaystyle =h+\Delta \theta \frac{d}{d \theta}[30 \tan(\theta)]$

Can you take it from there?

CB

3. Hello, turkeyy!

A different interpretation . . . a different approach.

A surveyor, standing 30ft from the base of a building,
measures the angle of elevation to the top of the building to be 75°.
How accurately must the angle be measured for the percentage error
in estimating the height of the building to be less than 4%?
The wording is somewhat vague.
I will assume that 75° is the exact angle of elevation.

Let $\displaystyle H$ = the exact height of the building.

We have: .$\displaystyle \tan75^o \:=\:\frac{H}{30} \quad\Rightarrow\quad H \:=\:30\tan75^o$ .[1]

Suppose he measures an angle of elevation of $\displaystyle \theta$ degrees.

Then his estimated height $\displaystyle h$ is: .$\displaystyle h \:=\:30\tan\theta$ .[2]

Divide [2] by [1]: .$\displaystyle \frac{h}{H} \:=\:\frac{30\tan\theta}{30\tan75^o} \quad\Rightarrow\quad \frac{h}{H} \:=\:\frac{\tan\theta}{\tan75^o}$

$\displaystyle \frac{h}{H}$ is the ratio of the estimated height to the exact height.
. . We want this to be between 96% and 104%.

. . $\displaystyle \begin{array}{ccccc} 0.96 & < & \dfrac{\tan\theta}{\tan75^o} & < & 1.04 \\ \\[-3mm] 3.582768775 & < & \tan\theta & < & 3.88133284 \\ \\[-3mm] 74.40^o & < & \theta & < & 75.55^o\end{array}$

Therefore, the angle must be measured to ±0.5°.

4. Originally Posted by Soroban
Hello, turkeyy!

A different interpretation . . . a different approach.

The wording is somewhat vague.
I will assume that 75° is the exact angle of elevation.
It does not matter if the 75 degrees is exact or approximate, to the accuracy involved in the calculation they give the same result.

Also the required accuracy form your calculation must be ~= +/- 0.6 degrees.

CB