# Finding the height of a skyscraper

I am trying to solve an optional self-assessment geometry problem in MIT's 18.01x course, but am struggling somewhat with the geometry. I don't know how to fully convey the problem without a picture, so I've included a screenshot below. The problem statement is:

To estimate the height of a skyscraper $$1$$ km in the distance, Jenny finds that if her friend Steve stands $$2.5$$ meters away, the top of his head just lines up with the top of the building. Steve is $$2$$ meters tall, and Jenny's eye is $$1.5$$ meters from the ground. How high is the building?

I think I found the answer, but I'm stumbling around with justifying it. Here is what I have. I put the part I am especially confused about in bold.

Consider the smaller triangle formed from Jenny's eye level with the top of the skyscraper, the vertical distance from Jenny's head to Steve's head, and the horizontal from Jenny's eye level to Steve's back. If $$\theta$$ is the angle made between Jenny's eye level to the top of the skyscraper and the horizontal, then we get $$\tan \theta = \frac{2 - 1.5}{2.5} = \frac{1}{5}$$. The angle from Steve's eye level to the top of the skyscraper is also $$\theta$$. (Adjacent angles? I'm not sure why.) Consider the larger right triangle formed from the top of the skyscraper, the horizontal from Steve's eye, and the view from Steve's eye level to the top. Call the opposite side $$x$$. As $$\tan \theta = \frac{1}{5}$$ and the adjacent side is equal to $$1 \text{km} = 1000 \; \text{m}$$, we get $$\frac{1}{5} = \frac{x}{1000}$$ and so $$x = 200$$. Adding the vertical distance of $$2$$, representing Steve's height, gives that the skyscraper is $$202$$ m tall.

I'm not sure if this argument is right, but my biggest line of confusion is the part in bold. In other to reason about the larger right triangle, I think I need to make some reference to the smaller right triangle. I don't think I have a way of invoking similar triangles. The argument seems to work if the angle above Steve's eye level is also $$\theta$$, but I'm not sure why that is true.

• But there are similar triangles: the three points (Jenny's eye, Steve's head, the top of the skyscraper) are on the same straight line, this straight line makes the same $\theta$ with the common horizon, and Steve and the skyscraper are vertical (hopefully). Assuming the Earth is flat. Commented May 18 at 0:26
• From context, I figured out that when you wrote "Jenny's eye level with the top of the skyscraper" you intended it to mean "Jenny's line of sight to the top of the skyscraper." Without context I would have no idea what you meant. I can't think of any place I've seen a non-horizontal line called a "level." Commented May 18 at 3:26
• (It's not steve's eye level but the top of his head... but that's not relevant). But there are three parallel lines: the pavement; a line parallel to the pavement but 1.5 meters that hit's Jenny in the eye, and a line parallel to the top of steve head that brushes his scalp. The angle $\theta$ is the same either by similar triangles (Jenny eye. point 1.5 meters on Steve's back. Top of steve's head$\sim$ Top of steves head, 2 meters on sky scraper, top of sky scraper) but even more fundamental it is an interior angle that transverses two parrallel lines. Commented May 18 at 3:40
• In case of doubt it is a good practice to use the simplest words possible. In this case you can say "the line from Jenny's eye to the top of the skyscraper" and there will be no doubt what you meant. That is also a direct description of the diagram, in which you can identify the point where Jenny's eye is, the point at the top of the skyscraper, and the line that connects those two points. That's even clearer than "line of sight" (the term I used in the previous comment). Commented May 18 at 3:43
• Can I link to an image here? i.sstatic.net/H3ZWIbwO.png I don't want to post an answer, but I think this image would really help you. Commented May 18 at 3:59

You are close, but there is some muddiness in the thinking. From the observer's eye, you have a triangle with height:length of $$1:5$$, and since the head is in line with the skyscraper $$1000$$ meters away, the height is also in the same ratio of $$1:5$$ so height $$200$$. BUT, the triangles start from the observer's eye, which is $$1.5$$m high, so the skyscraper is $$201.5$$m tall, not $$202$$.
• In practical terms the answer is $200$ meters because the input data are unlikely to have three decimal digits of accuracy. If we suppose the measurements are precise, we have to ask which distance is $1$ km. In the question there is a diagram that says Steve is $1000$ meters from the building and Jenny is $1002.5$ meters from the building. This answer assumes that Steve is $997.5$ meters from the building and Jenny is $1000$ meters from it, which is the most plausible interpretation of "$1$ km in the distance" if we assume that measurement is exact. Commented May 18 at 3:38