# Calculate the height of a distant object using estimated angles from two different points.

I knew how to do this long ago, found the exact problem in my old trig book, but I can't seem to work it out.

Say I'm at an unknown distance from a mountain, called point P, and I estimate the angle of elevation to the top of the mountain is 13.5 degrees. Then I move to point N, which is 100 meters closer to the mountain, and I estimate the angle of elevation to be 14.8 degrees. What is the height of the mountain?

I remember this being enough information to solve both triangles, but without the distance to the mountain, or the height of the mountain, I'm at a loss. Hint's would be appreciated.

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is the answer to your problem 30.6261 – user84254 Jun 28 '13 at 12:35

Or...

Use the Law of Sines to find the longest side in the triangle with the 100 m side (you know all the angles).

This longest side is also the hypotenuse of a another, right angle triangle where you know the angle opposite the height you want...

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And that is how I used to do it. Thanks for the memory jog! – Nocturno Feb 13 '13 at 6:06

This picture will give you a good idea. Use the $\tan$ function on both angles, and solve the equation because

\begin{align*} \frac{height}{length+100} &= \tan 13.5^\circ \\ \frac{height}{length} &= \tan 14.8^\circ \end{align*}

It seems that you will get a pair of simultaneous equations. So 2 linear equations and 2 unknowns, pretty easy to solve.

I will respond to your comment right here. What you can do is to solve by cancelling height

Eg. \begin{align*} \\tan 14.8^\circ \times {length} &= \tan 13.5^\circ \times ({length+100}) \end{align*} There are other ways to solve the linear simultaneous equations. I'll leave it to you to figure them all out.

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I failed to mention in my op what I tried. Your answer was almost identical to what I had on my scratch paper, including the picture. Still, your equations made me realize that if I solved for height, I could set both equations equal to each other. Thanks. – Nocturno Feb 13 '13 at 5:44
edited to respond to your comment – bryansis2010 Feb 13 '13 at 6:08

Let $h$ be the height of the mountain (in meters), and $d$ the distance from $P$ to the mountain. Then you have $h/d=\tan(13.5)$, and $h/(d-100)=\tan(14.8)$ (all angles in degrees), which you can solve for $h$ and $d$.

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Good hints. Unfortunately, I had already made it that far and couldn't go any farther. I solved it though! – Nocturno Feb 13 '13 at 5:47
For future reference: when you ask a question (here or elsewhere), tell the people you are asking everything you already know about the question. That way they don't waste their time and yours coming up with answers that just duplicate what you already know. – Gerry Myerson Feb 13 '13 at 6:02

Here's what I've come up with:

α = angle of elevation at P = 13.5 deg
β = angle of elevation at N = 14.8 deg
d = distance between points P and N = 100m
h = height of mountain

h = (d * tan β * tan α) / (tan β - tan α)
h = (100 * tan 14.8 * tan 13.5) / (tan 14.8 - tan 13.5) = 262.8m

I double checked this answer by finding the intersection of bryansis2010's two equations:

x * tan 14.8 = (x + 100) * tan 13.5
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Let $O$ and $M$ the base point and the tip of the mountain, respectively. Then $$NM=100\cdot\frac{\sin13.5}{\sin1.3},$$ and $OM$ is the height of the mountain, where $$OM=NM\cdot\sin14.8\approx262.854.$$ Note that the input of the sine is in degrees.

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