Finding the Distance to a Building with Trigonometry

[RayDonaldPratt]

Homework Statement

"A woman standing on a hill sees a building that she knows is 55 feet tall. The angle of depression to the bottom of the building is 27°, and the angle of elevation to the top of the building is 35°. Find the straight line distance from the woman to the building."

Relevant Equations

((35/62)*55)/Adj. = Tan 35°; so, Adj.= ((35/62)*55)/Tan 35°, and so Adj. = ≈44.34169214.
Similarly:
((27/62)*55)/Adj. = Tan 27°; so, Adj.= ((27/62)*55)/Tan 27°, and so Adj. = ≈47.00768711.
It's the same adjacent side, so why are the computed lengths so different?

I'm doing self-study out of a free .PDF book entitled, Trigonometry, by Richard W. Beveridge (©June 18, 2014).
The problem I'm interested in is as follows:

"A woman standing on a hill sees a building that she knows is 55 feet tall. The angle of depression to the bottom of the building is 27°, and the angle of elevation to the top of the building is 35°. Find the straight line distance from the woman to the building."

I drew a diagram and placed my stick figure woman standing above the ground at an unknown height, but roughly lower than half of the building, and I drew a line from her eyes to the top of the building (≈35°), another line straight from her eyes perpendicular to the building, and then the final line from her eyes to the bottom of the building (≈27°). My task was to find the distance of the line straight from her eyes perpendicular to the building.

My first conundrum was how to correctly divide the 55' building into two parts so that I could get the lengths of the opposite sides of the two triangles formed by the lines from her eyes to the side of the building. I figured out that her entire field of view is 35° + 27° = 62° and that the measure of the upper half of the building would be 35/62 of 55' and that the lower half would be 27/62 of 55'.

From there, I could use my calculator's trigonometry features to find the distance of the line perpendicular to the building from her eyes using either triangle since both triangles have the same adjacent side from her eyes. However, I got thrown for a loop: the same side for either triangle computed out to seriously different lengths: ≈44.34' for the 35° triangle, and ≈47.01' for the 27° triangle. Worse, the book gives the answer of ≈45.5'. The only way that I get close to that answer is if I add both of my full answers together and then divide by two: (44.34169214 + 47.00768711) / 2 = 45.67468962.

If trigonometry is that bad, and I did this correctly, then I am okay with that; but, if I did something wrong, I would like to know what it is. Please.

 

[BvU]

Hi,

I drew a diagram

There already is a diagram provided in the exercise that I found when googling:

So the exercise wants you to solve $d$ from $d \tan 35^\circ + d \tan 27^\circ = 55'$.
The book answer is correct.

entire field of view is 35° + 27° = 62° and that the measure of the upper half of the building would be 35/62 of 55' and that the lower half would be 27/62 of 55'.

This is utterly wrong.

my calculator's trigonometry features to find the distance of the line perpendicular to the building from her eyes using either triangle since both triangles have the same adjacent side from her eyes. However, I got thrown for a loop: the same side for either triangle computed out to seriously different lengths: ≈44.34' for the 35° triangle, and ≈47.01' for the 27° triangle.

I don't understand what you are doing here. Can you post your diagram ?

The trig is perfect and therefore you did something wrong. What did you do, exactly ? Show the steps.

$\$

 

[PeroK]

If trigonometry is that bad, and I did this correctly, then I am okay with that; but, if I did something wrong, I would like to know what it is. Please.

Your mistake was to assume that the heights of the building above and below the horizontal were proportional to the angles, rather than the tangent of the angles.

So, you got two different values for $d$. One was an overestimate and one an underestimate. By taking the average of these two wrong answers you got something close to the right answer!

The important lesson is that the height of a building is not directly proportional to the angle, but to the tangent of the angle. If you are standing a distance $d$ from two buildings, one of height $h$ and one of height $2h$, then the angles from the horizontal to the top of each building is given by:
$$\tan \theta_1 = \frac h d, \ \tan \theta_2 = \frac{2h}{d}$$And we see that$$\tan \theta_2 = 2\tan \theta_1$$This does not mean that $\theta_2 = 2 \theta_1$. In fact, if $\theta_1 > 45$ degrees, then it would be impossible for $\theta_2$ to be twice this.

This, in a way, is what trigonometry is all about and why we need $\sin, \cos$ and $\tan$ and can't just use angles directly.

 

[RayDonaldPratt]

Hi,There already is a diagram provided in the exercise that I found when googling:
View attachment 316553
So the exercise wants you to solve $d$ from $d \tan 35^\circ + d \tan 27^\circ = 55'$.
The book answer is correct.This is utterly wrong.I don't understand what you are doing here. Can you post your diagram ?

The trig is perfect and therefore you did something wrong. What did you do, exactly ? Show the steps.

$\$

BvU, thank you, your formula was easy to transform algebraically to solve for d, and it makes sense.

 

[RayDonaldPratt]

Your mistake was to assume that the heights of the building above and below the horizontal were proportional to the angles, rather than the tangent of the angles.

So, you got two different values for $d$. One was an overestimate and one an underestimate. By taking the average of these two wrong answers you got something close to the right answer!

The important lesson is that the height of a building is not directly proportional to the angle, but to the tangent of the angle. If you are standing a distance $d$ from two buildings, one of height $h$ and one of height $2h$, then the angles from the horizontal to the top of each building is given by:
$$\tan \theta_1 = \frac h d, \ \tan \theta_2 = \frac{2h}{d}$$And we see that$$\tan \theta_2 = 2\tan \theta_1$$This does not mean that $\theta_2 = 2 \theta_1$. In fact, if $\theta_1 > 45$ degrees, then it would be impossible for $\theta_2$ to be twice this.

This, in a way, is what trigonometry is all about and why we need $\sin, \cos$ and $\tan$ and can't just use angles directly.

PeroK, thank you. It seemed so intuitive to me that the angles would span the building in an equal fashion, and it didn't occur to me that the trig ratios of the different angles would produce different values for d. I had actually thought about how chords in a circle will not produce twice the angle when a chord length is doubled, but I could not relate it to trigonometry. I think that what you explained involves a similar idea, and I appreciate your help in getting me to understand that. And, again, hats off to BvU for giving me the simple formula for solving the problem. Both of you did great!