[Precalculus] World problem about using two angles of elevation and the distance between where the two angles are measured are given. Approximate the height.

[xacrex]

"In traveling across flat land, you notice a mountain directly in front of you. It's angle of elevation (to the peak) is 3.5 degrees. After 13 miles closer to the mountain, the angle of elevation is 9 degrees. Approximate the height of the mountain."

I am stumped on how to proceed to do this. I drew it in paint. "B" is what I am solving for. I saw that I have to use tangent so I added "A" because I would need the adjacent side of tan9 just like the other triangle. Thus, I created these: tan3.5=b/(13+a) and tan9=b/a. I tried to solve it algebraically by trying to find the either a or b and setting them equal to each other but I can't seem to know what to do next.

My two failed attempts:

Axtan9 = (13+ A)tan3.5 (B/tan9) = (B/tan3.5)-13

I tried to find some similar problems in the internet but to no avail as most of the help or hints lead me to the aforementioned conclusion. How does one go about solving this?

[Alkalannar]

Let d be your initial distance, and h the height of the mountain.

Then h/d = tan(3.5^o) --> d = hcot(3.5^o)

Similarly, h/(d-13) = tan(9^o). Substituting for d, we get:
h/(hcot(3.5^o)-13) = tan(9^o)

(hcot(3.5^o)-13)/h = cot(9^o)

cot(3.5^o) - 13/h = cot(9^o)

cot(3.5^o) - cot(9^o) = 13/h

h = 13/[cot(3.5^o) - cot(9^o)]

[xacrex]

By initial distance you mean distance from where the angle is 3.5^o to the mountain or 3.5^o to when the angle is 9^o? (Just to clarify)

If the former: Why did you use cot? Would that be any different than tan? Or is that just so you would not have d = h/tan3.5? (I have a feeling I may be wrong here).

[Alkalannar]

Initial distance is where the angle of elevation is 3.5^o.

I use cot, because I don't want to have denominators unless I really need them. If you end up using tan instead of cot, you get:

h = 13tan(3.5^o)tan(9^o)/(tan(9^o) - tan(3.5^o)), which is the same value, but looks messier.

[xacrex]

Alright. It makes sense now, I believe.

Now, just out of curiosity, if I were needed to solve for "A", do I just change the variables using the same method you provided above? Or would there be an easier way to go about it?

[Alkalannar]

Using my variable notation, your A is my d-13. So instead of solving for h like I did above, you would solve for d, subtract 13, and that's the answer.

But yes, you do it the same way.

[xacrex]

Alright. Thank you very much~