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Homework Help: Tower Trig Problem

  1. Oct 9, 2004 #1
    Three survey stations X, Y, Z lie in one straight line on the same straight plane. A series of angles of elevation are taken to the top of a chimney which lies to one side of XYZ. The angles of elevation of the top of the chimney measured ay X, Y, Z are 14 degrees 02 ', 26 degrees 34' and 18 degrees 26' respectively. The lengths XY and YZ are 121.92m and 73.15m respectively. Find the height of the chimney.
    Last edited: Oct 9, 2004
  2. jcsd
  3. Oct 9, 2004 #2
    this is a seriosly challenging problem.
  4. Oct 9, 2004 #3
    Did you draw the picture?

    Think law of sines and law os cosines.
  5. Oct 9, 2004 #4
    i use law of sines and law of cosines and i have a picture yes, but its seems like the algebra involved is insane and i still end up with 2 unknowns 1 being a value for an angle and second of course the height which is the solution.
  6. Oct 9, 2004 #5
    Well, what have you done thus far then?
  7. Oct 9, 2004 #6
    I don't know if you noticed this or not but the points from closest to the chimney to the furthest are Y-Z-X. That's if you gave the angles in the correct order in the first post.

    Anyway, lets say the point O is where the chimney intersects the ground (height h) and point H is the top to the chimney the surveyors measured to.

    We'll look at triangle Z-Y-H.

    I don't feel like dealing with the minutes--you can handle that I hope. I'm going to use the degree portion of the angles only.
    Angle z=18,y'=(180-26), H'=(90-z-y')

    y' is the angle of triangle Z-Y-H on the otherside of line segment Y-H from angle y.
    h' is the remaining angle of triangle Z-Y-H lacated at the top of the triangle.

    We now have tree angles of a triangle. You original question also gave is us the length of ZY=73m. You can solve for all sides of triangle Z-Y-H using the law of sines.

    We are in the perfect position to simply employ the definition of cos. [itex]\cos y=\frac{h}{YH}[/itex]. We just found YH, we're given y, and now we have a relation which will tell us what h is.

    Hope this helped. Good luck.
    Last edited: Oct 9, 2004
  8. Oct 9, 2004 #7
    ok i made a diagram cause i can't see ho you came up with your y' cause i think the problem is a little more complicated than that, u do not even need the angle X or the distance XZ from what you are saying and i am certain thy are crucial to correctly visualizing this problem. so here it is:

    Attached Files:

    Last edited: Oct 9, 2004
  9. Oct 9, 2004 #8
    Ahhh, I though XYZ were colinear not spread our like that. Ok then. I'll look this over for a few minutes and get back to you.
  10. Oct 9, 2004 #9
    A few of your equation are wrong...these should be your equations:

    [tex]a = \frac{h}{\tan{14^{\circ}02^{\prime}}}[/tex]

    [tex]b = \frac{h}{\tan{26^{\circ}34^{\prime}}}[/tex]

    [tex]c = \frac{h}{\tan{18^{\circ}28^{\prime}}}[/tex]

    [tex](XY)^2 = a^2 + b^2 - 2ab\cos{\alpha}[/tex]

    [tex](YZ)^2 = b^2 + c^2 - 2ac\cos{\theta}[/tex]

    [tex](XY + YZ)^2 = a^2 + c^2 - 2ac\cos{(\alpha + \theta)}[/tex]

    plug a, b, and c into the lower equation and then you have three equations, three unknowns.
  11. Oct 9, 2004 #10
    Actually [itex]\cos(\alpha+\theta)[/itex] contains four unknows--two of which are already accounted for. [itex]\cos(\alpha+\theta)=\cos{\alpha}\cos{\theta}-\sin{\alpha}\sin{\theta}[/itex]. Granted we can say [itex]\sin \alpha=\sqrt{\cos \alpha}[/itex] but that makes the problem just as difficult. Simply pulling the degree unknowns out of the cosine function is difficult in this problem because you have an unknowns outside the cosine function and unknows within the cosine function.

    Also, [itex]\cos \alpha+\cos \theta\neq\cos{\alpha+\theta}[/itex]

    this is a rather tough problem I must say.

    More to come.
  12. Oct 9, 2004 #11
    you can find alpha and theta in terms of h from the above two equations...then you will have one really ugly equation with only one unknown, h....granted that it will not be pretty at all...bu hey, that is why the made calculators :)
  13. Oct 9, 2004 #12
    When you do that you're putting unknowns into a trig function with unknowns of the same variables outside the trig function as well. It wont help IMO.
  14. Oct 9, 2004 #13
    Yes i posted the equation wrong. thats not what i was using in fact i just had to re-arrange my letters for clarity and mixed them up. I do however believe the identity Cos(180-theta)= -Cos(theta) is to be used in this problem. I will rewrite my equations and see if you can see if they help.
  15. Oct 9, 2004 #14
    ok i re - worked some stuff but still see a dead end wall of algebra damn, I am in a surveying program and this is a bonus question for my maths course. Apparently this method is used it practical situations so i would like to learn how it is solved. I have completed every bonus question thus far so i really wanna solve this by Monday. here's my new outlook, anyone can throw in their $0.02.

    Attached Files:

  16. Oct 9, 2004 #15
    any trig genius's ?

    come on all you brainiacs lemme know how to do this one.
  17. Oct 9, 2004 #16


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    Homework Helper

    All right, I think I have a solution, but it's getting too late for me to be willing to figure it out all the way. I'm going to have to redefine some of your variables, and I apologize profusely for that.

    First off, it's obvious if you think about it that the point at which the line on which the towers are located most closely approaches the chimney is between X and Y. Let's call that point, wherever it is, the origin. The distance between that and the chimney is d, and the angle of elevation from that point to the chimney is q. It's trivial, then, that

    tan q = h/d

    Now: measure a point along the line of the towers a distance x' from the origin. The distance from that point to the chimney is r. There is a right triangle between x, d and r, such that

    x'^2 + d^2 = r^2

    The angle of elevation from that point - call it e - can be found by

    tan e = h/r

    We have three angles of elevation: Call the distance from the unknown point of closest approach to X, a. To Y, call it b. The distance between Y and Z is known, so the distance from the origin to Z is b + 73.15 m.

    We know, then, that

    tan 14 2' = h/sqrt(d^2 + a^2) with h, d and a unknown

    tan 26 34' = h/sqrt(d^2 + b^2) with h, d and b unknown

    tan 18 26' = h/sqrt(d^2 + (b + 73.15)^2) with h, d and b unknown

    and we know that a + b = 121.92m

    Four equations, four unknowns. In principle, this is solvable.

    I did get a start on it - taking the first two equations and solving them for d^2 let's you eliminate d from the mix all together. The remaining unknowns are b and h. Taking the the second and third equations and solving for d^2 should let you eliminate it there, as well, resulting in another equation with only h and b, from which you can solve for whichever is most convenient, but I'm afraid that's where my head started hurting.

    Let me know if this bears fruit. I'll try it again tomorrow after church.
  18. Oct 10, 2004 #17
    Ok I worked it out, I think now i have no idea if i am right or not but i found the height to be about 67.6437m if anyone gets a number anything close to this please let me know. thx diane, i found your method hard to follow at first but i got it. funny thing is i never had to use cosine law or sine law and thats what i have been studying so...... i guess that method is a form of cosine law derivation. Anyway i am done for the night.
  19. Oct 10, 2004 #18
    3 days struggle

    I guess i errored somewehre along the waybut i reworked it with the last diagrams and just brute force plugged the numbers through and i believe i got the correct answer now i used cosine law and the identity:
    Cos(180-X)=-Cos(X) so if you care to test your answer vs mine i am fairly confident the height of the chimney is 34.191m.
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