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Math Help - Finding height and base of building with 2 given angles.

  1. #1
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    Unhappy Finding height and base of building with 2 given angles.

    hey, ive been working on this question for a while now; i just cant seem to get it.

    A building is in the shape of a square prism with base edge m and height h metres. It stands on level ground. A base diagonal AC is extended to point K.
    From K, a surveyor finds that the angles of elevation of F and G are 30 degrees and 45 degrees respectively. Find an exact value for the ratio h/m.

    The top square base is read with vertices (left to right to bottom right to bottom left) H, G, F, E.
    the bottom square is read with vertices ( left to right to bottom right to bottom left) D, C, B, A.
    Last edited by mr fantastic; June 18th 2010 at 03:44 AM. Reason: Restored deleted question.
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  2. #2
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    Quote Originally Posted by MATHSMATHSAMATHS View Post
    hey, ive been working on this question for a while now; i just cant seem to get it.

    A building is in the shape of a square prism with base edge m and height h metres. It stands on level ground. A base diagonal AC is extended to point K.
    From K, a surveyor finds that the angles of elevation of F and G are 30 degrees and 45 degrees respectively. Find an exact value for the ratio h/m.
    Hi MATHSMATHSAMATHS , welcome to MHF.

    If d is the distance between K and C, hen KC = h, because angle of elevation to G is 45 degrees.

    Distance between B and K is \sqrt{(m^2 + h^2)}

    So tan(30) = \frac{h}{\sqrt{(m^2 + h^2)}}

    Now proceed.
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  3. #3
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    Quote Originally Posted by MATHSMATHSAMATHS View Post
    hey, ive been working on this question for a while now; i just cant seem to get it.

    A building is in the shape of a square prism with base edge m and height h metres. It stands on level ground. A base diagonal AC is extended to point K.
    From K, a surveyor finds that the angles of elevation of F and G are 30 degrees and 45 degrees respectively. Find an exact value for the ratio h/m.
    Try seeing from our perspective. You haven't told us what F and G are. How are we supposed to know? Please give all information available.

    Edit: I guess it's unambiguous after all, but still requires too much inferring from context for my taste. sa-ri-ga-ma's post will lead you to the answer.
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Hi MATHSMATHSAMATHS , welcome to MHF.

    If d is the distance between K and C, hen KC = h, because angle of elevation to G is 45 degrees.

    Distance between B and K is \sqrt{(m^2 + h^2)}

    So tan(30) = \frac{h}{\sqrt{(m^2 + h^2)}}

    Now proceed.
    thankyou; there was a diagram involved to help with the question but seeing as though i am a new member i did not know how to add it.
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    Hi MathsMathsAMaths,

    I have the same problem, here is the diagram:



    you need to add in point K just imagine the line AC continuing outwards

    I hope that helps people do it, can anyone do it showing more steps than the above solution?

    thanks
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    Quote Originally Posted by sa-ri-ga-ma View Post
    Hi MATHSMATHSAMATHS , welcome to MHF.

    If d is the distance between K and C, hen KC = h, because angle of elevation to G is 45 degrees.

    Distance between B and K is \sqrt{(m^2 + h^2)}

    So tan(30) = \frac{h}{\sqrt{(m^2 + h^2)}}

    Now proceed.

    Sa-ri-ga-ma

    thats incorrect, the angle BCK is actually 135 degrees, so pythagoras does not apple, and if you use the cos rule it is acutally rather complex. Any ideas?
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  7. #7
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    Quote Originally Posted by SxcMathsBoy View Post
    Sa-ri-ga-ma

    thats incorrect, the angle BCK is actually 135 degrees, so pythagoras does not apple, and if you use the cos rule it is acutally rather complex. Any ideas?
    Hi SxcMathsBoy, you are right.
    The problem is really complicated. The observer sees the perspective view of the prism.
    Since the height of the observer is not given, we have to assume that the eye sight is along the ground level. So the observer sees BCD in the same level.
    When the observer moves from the center of face BC to the line along BC, side m shrinks from m to zero. When he is along AC, m' = mcos(45)
    Because of the perspective view, h also shrinks. Since angle of elevation of G is 45 degrees and that of F is 30 degrees, h' = h - m'tan(15) and KC = h

    The apparent distance KB = \sqrt{h^2 + m'^2}

    Now tan(30) = \frac{h'}{KB} = \frac{h-m'tan(15)}{\sqrt{(h^2 +m'^2)}}

    After simplification and squaring , you get

    h^2 + m'^2 = 3(h^2 + m'^2tan^2(15) - 2m'htan(15))

    Substituting the value of m' and dividing h^2 on both side, solve the quadratic to find m/h.
    Last edited by sa-ri-ga-ma; June 6th 2010 at 05:27 PM.
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