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Here's a tough one for you!

  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.
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  3. Oct 9, 2004 #2


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  4. Oct 9, 2004 #3
    Yes Trig

    Not just as simple as you think.
  5. Oct 10, 2004 #4


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    Rocko, I think I'd approach that problem by finding the length of each of the horizontal plane projections of the lines from X,Y,Y to the top of the chimney. Now these lengths will be in terms of the height h of the chimney.

    L1 = h cot(angle1) and similarly for the other two. h is of course an unknown at this point but the cot()'s are all plain constants. You should find there is just one value for h that allows the lines of length L1 from X, L2 from Y, and L3 from Z to intersect at one point (that point being the base of the chimney). Once you get this point you're home free.
    Last edited: Oct 10, 2004
  6. Oct 10, 2004 #5


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    Just to elaborate on solving for the intersection of those horizontal projection.

    Arbitrarly take points X,Y,Z lying on the x axis with point Y at the origin. Then you get something like this,

    (121.92+x1)^2 + y1^2 = a h^2
    x1^2 + y1^2 = b h^2
    (x1-73.15)^2 + y1^2 = c h^2

    Here, a, b and c are constants (from the cot's squared) and (x1,y1) is the base of the chimney in the coords specified. Three equations and three unknowns = nice :)
    Last edited: Oct 11, 2004
  7. Oct 10, 2004 #6


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    I would start the same way as uart, but then, I'd do the following.

    The projections in the ground plane are a pair of adjacent triangles : XCY and YCZ where C represents the base of the chimney.

    The sides of these triangles are :
    XY = known,
    XC = h cot(angle1)
    YC = h cot(angle2)
    YZ = known
    ZC = h cot(angle3)

    Let angle <ZYC = alpha, then angle <XYC = 180 - alpha

    Use the cosine rule for the two triangles with respect to the angles alpha and 180-alpha.

    You have 2 equations in 2 unknowns.
  8. Oct 10, 2004 #7
    after 3 days and many scrap pages i finally think i come up with a answer thx Goku, you know i tried that method right off the start but thought the algebra would just be to complex to carry thru so i scrapped the idea, but after confirmation i knew it was what needed to be done did you get an answer for h= 34.191m?
  9. Oct 10, 2004 #8


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    Actually, I never did go about solving it. It's the typical mathematicians outlook : once you establish that the solution exists, you walk away from it.

    There's a small fire accidentally started in your house and you have a large bucket of water. What do you do ?

    Engineer : Pour the water over the fire and extinguish it.

    Mathematician : You have a fire; you have the water. A solution exists. (and walks away)
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