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Finding height using angle of elevation

  1. Jun 14, 2013 #1

    QuantumCurt

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    1. The problem statement, all variables and given/known data

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



    2. Relevant equations



    3. The attempt at a solution

    I drew out the triangle, and I'm figuring that the adjacent side can be called 13+x. So, I tried using a tangent function to find the length of x, and find the overall length of the adjacent side. Theta measurements are in degrees.

    [tex] tan \ 3.5 = \frac{sin \ 3.5}{13 + x}[/tex]

    Then I solved for x:

    [tex] sin \ 3.5 = 13 + x \ tan \ 3.5[/tex]

    [tex] x = \frac{sin \ 3.5}{tan \ 3.5} - 13 [/tex]

    [tex] x = approximately -12 \ miles [/tex]

    Clearly, -12 miles is not the correct answer for the value of the remaining distance in the adjacent side.

    If I take the positive value of 12 though, and make the adjacent side 25 altogether, the answer seems to come out as something that's potentially feasible...

    [tex]tan \ 3.5 = \frac{opp}{25}[/tex]

    [tex] 25 \ tan \ 3.5 = opp[/tex]

    Then I get opp = approximately 1.5 miles, or 7920 feet in height.


    I'm thinking I did something wrong with the initial equation to find x though. I'm thinking the 9 degree angle of elevation needs to be figured in...but how? We haven't covered any problems even remotely like this in class, I'm just working through the more challenging problems that the professor seems to not want to assign as homework. Any suggestions would be much appreciated!


    edit- I just realized that I should have subtracted the 13 over to the other side before dividing, but that still doesn't solve the problem in either case.
     
  2. jcsd
  3. Jun 14, 2013 #2

    SteamKing

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    Your understanding of trigonometry is faulty. The sine of an angle is the length of the side opposite of the angle divided by the length of the hypotenuse of the triangle. Clearly, in this case, the length of the side opposite of the angle cannot equal sin(3.5 deg.) since the length of the hypotenuse cannot be equal to 1.

    According to the definition of the tangent, your first formula implies that the cos(3.5 deg.) = 13 + x, which is clearly impossible since -1 <= cos (x) <= 1.

    Your solution also disregards the second angle of elevation of 9 degrees, which occurs after traveling 13 miles toward the mountain.

    Reformulate your solution and try again.
     
  4. Jun 14, 2013 #3

    QuantumCurt

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    I just started trigonometry about 2 weeks ago, and we actually just started the actual trig this week, after spending the first week doing prerequisite algebra. So, I'm sure my understanding is quite faulty at this point...lol...But, I just started learning about trig identities and such about 3 days ago.

    This professor has a strange teaching method...she basically just does the example problems straight out of the chapter, which isn't exactly helping. This book doesn't even have a single example remotely similar to this problem.

    I'll give it another try and see what I can come up with.

    Should I be looking at a Pythagorean identity?
     
  5. Jun 14, 2013 #4

    SteamKing

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    Not really. Remember, the object of the problem is to find the height of the mountain. Use the triangle side relationships for sin, cos, and tan to construct your equations.

    sin = opp/hyp
    cos = adj/hyp
    tan = sin/cos = opp/adj
     
  6. Jun 14, 2013 #5

    QuantumCurt

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    [tex] cos \ 3.5 = \frac{x}{r}[/tex]

    [tex] cos \ 9 = \frac{x-13}{r}[/tex]

    Is this setup right? Does the 13 even play into the equation?

    Or do I have to somehow set the sin 3.5 and the sin 9 equal to one another?

    I tried setting it up to find the sin first, but I believe I have to find the cos first, because that's the only one that has any information pertaining to the length. Correct?
     
  7. Jun 14, 2013 #6

    QuantumCurt

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    Wait...I went through and found all of the other angles. The fact that I should have done that seems kind of obvious now. I've got a little more info to work with now.
     
  8. Jun 14, 2013 #7

    Dick

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    Say what your variables mean. The two variables you want to work with are x=initial distance to the mountain and h=height of the mountain. Write two equations relating them using the tangent function, opposite (h) over adjacent (distance to the mountain). Finding the other angles isn't that useful.
     
  9. Jun 14, 2013 #8

    QuantumCurt

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    I found out after a minute that finding the other angles wasn't helpful at all. I went back to using a tangent function after I realized that, and this is where I've gotten.


    [tex]mountain height= h \ \ \ \ distance \ to \ mountain = d[/tex]

    [tex]tan \ 3.5=\frac{h}{d}[/tex]
    [tex]h \ = \ d \ tan \ 3.5[/tex]

    [tex] tan \ 9=\frac{h}{d-13}[/tex]
    [tex] h \ = \ (d-13) \ tan \ 9[/tex]


    [tex] \ d \ tan \ 3.5 \ = \ (d-13) \ tan \ 9[/tex]


    So, now my question is. Do I distribute the tan 9 into the (d-13) like you would with a normal expression? If so, that gives me-

    [tex]d \ tan \ 3.5 \ = \ d \ tan \ 9 \ - \ 13 \ tan \ 9[/tex]

    Now I can't seem to solve this in a way that won't cancel the d's out.

    edit-Is there any difference between (d-13)tan 9, and d tan 9 - 13 tan 9?
     
    Last edited: Jun 14, 2013
  10. Jun 14, 2013 #9

    Dick

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    No difference. But you want to solve for d and ultimately find h. Rearrange it to put all of the d's on one side. So d*tan(9)-d*tan(3.5)=13*tan(9). Solve for d. Don't forget the angle numbers are degrees if you are using a calculator. Your 'guess' in the first post is not terribly bad.
     
    Last edited: Jun 15, 2013
  11. Jun 15, 2013 #10

    QuantumCurt

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    Ok...so picking up where I left off-

    [tex]d \ tan \ 3.5 \ = \ d \ tan \ 9 \ -13 \ tan \ 9[/tex]
    [tex]d(tan \ 3.5 \ - \ tan \ 9) \ = \ -13 \ tan \ 9[/tex]
    [tex]-.09722d \ = \ \frac{-13 \ tan \ 9}{-.09722}[/tex]
    [tex]d \ = \ 21.1787[/tex]


    [tex]tan \ 3.5 \ = \frac{h}{21.1787}[/tex]
    [tex]h \ = \ (tan \ 3.5)(21.1787)[/tex]
    [tex]h \ = \ approximately \ 1.295 \ miles \ or \ 6838 \ feet[/tex]

    Please tell me this is right...lol
     
  12. Jun 15, 2013 #11

    Dick

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    You might have mistyped a few numbers in that. But you've got the right idea and the final numbers are good. Good job. Do I have to say lol? Not so hard, was it? lol.
     
    Last edited: Jun 15, 2013
  13. Jun 15, 2013 #12

    QuantumCurt

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    haha...No, you didn't have to say the "lol" part, but I sure do appreciate it!

    What numbers are you referring to? I just double checked, and they're all the same as the numbers I've got on paper here.

    Thanks for the help! It's much appreciated!!:biggrin:
     
  14. Jun 15, 2013 #13

    Dick

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    I was just talking about this $$-.09722d \ = \ \frac{-13 \ tan \ 9}{-.09722}$$. There's an extra -0.09722 in it. You're welcome!
     
  15. Jun 15, 2013 #14

    verty

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    There is a way to find the height using those angles but with laws you will learn only later; one can use simultaneous equations as you did or rate of change of width versus height.

    Trigonometry was my preferred branch of school math because most questions can be answered in more than one way.
     
  16. Jun 15, 2013 #15

    QuantumCurt

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    Ahh...I didn't notice that. Yes, that should have been broken apart into two steps.

    So far trigonometry seems pretty cool. We're still on the basics, but I've previewed some of the material that comes later in the class, and it looks like there's going to be some cool stuff coming up. Right now we're just going over identities and how to use them. The professor doesn't seem to want to assign very difficult problems, although that may change as the course moves along. This problem was really the only one that stumped me. There were a couple problems that involved using identities to transform one side of an equation into the other side that kind of threw me off, but only momentarily.
     
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