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  1. #1
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    square tower question

    A square tower stands upon a horizontal plane.From a point in this plane, from which three of its upper corners are visible , their angular elevations are respectively 45 degrees,60 degrees and 45 degrees .Show that the height of the tower is to the breadth of one of its sides as \sqrt6(\sqrt5 + 1) to 4
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  2. #2
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    Hello jashansinghal
    Quote Originally Posted by jashansinghal View Post
    A square tower stands upon a horizontal plane.From a point in this plane, from which three of its upper corners are visible , their angular elevations are respectively 45 degrees,60 degrees and 45 degrees .Show that the height of the tower is to the breadth of one of its sides as \sqrt6(\sqrt5 + 1) to 4
    Look at the diagram I have attached.

    Because two of the angles are equal ( 45^o) the point in the plane, P, lies on the diagonal produced of the square base of the tower.


    Therefore, in the diagram, \angle CDP = 135^o.


    Also, \angle DPA = 60^o,\; \angle CPB = 45^o.


    Therefore, if the height and the breadth of the tower are h and b respectively:
    DP = \frac{h}{\tan60^o}=\frac{h}{\sqrt3}
    and
    CP = \frac{h}{\tan45^o}=h
    and
    DC = b
    Therefore, using the Cosine Rule on \triangle DPC:
    h^2=\frac{h^2}{3}+b^2-2\cdot\frac{h}{\sqrt3}\cdot b\cdot\cos135^o
    =\frac{h^2}{3}+b^2+2\cdot\frac{h}{\sqrt3}\cdot b\cdot\frac{1}{\sqrt2}
    Multiply both sides by 3\sqrt3:
    \Rightarrow 3\sqrt3h^2=\sqrt3h^2+3\sqrt3b^2+3\sqrt2bh

    \Rightarrow 2\sqrt3h^2-3\sqrt2bh-3\sqrt3b^2=0

    Treating this as a quadratic in h, and using the formula:
    \Rightarrow h = \frac{3\sqrt2\pm\sqrt{18+72}}{4\sqrt3}\cdot b
    =\frac{3\sqrt2+\sqrt{90}}{4\sqrt3}\cdot b, since the other root is negative

    =\frac{3\sqrt2+3\sqrt2\sqrt5}{4\sqrt3}\cdot b


    =\frac{3\sqrt2(1+\sqrt5)}{4\sqrt3}\cdot b


    =\frac{\sqrt6(1+\sqrt5)}{4}\cdot b

    \Rightarrow h:b = \sqrt6(\sqrt5+1):4
    Grandad
    Attached Thumbnails Attached Thumbnails square tower question-untitled.jpg  
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