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Math Help - Triangle Question

  1. #1
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    Triangle Question

    Hi,
    Please, what is the formula to determine the length of the two longest sides on an isosceles triangle with the angles of 36, 72, 72, and the base length of 550 feet? I'm not a trig guy so please provide the answer too because I may not be able to work the formula correctly.
    Thank you very much.
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  2. #2
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    Quote Originally Posted by billionj View Post
    Hi,
    Please, what is the formula to determine the length of the two longest sides on an isosceles triangle with the angles of 36, 72, 72, and the base length of 550 feet? I'm not a trig guy so please provide the answer too because I may not be able to work the formula correctly.
    Thank you very much.
    law of sines ...

    L = leg length

    \frac{550}{\sin(36^\circ)} = \frac{L}{\sin(72^\circ)}
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  3. #3
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    Hello, billionj!

    Determine the length of the two longest sides on an isosceles triangle
    with the angles of 36, 72, 72, and the base length of 550 feet.
    This is probably not an acceptable solution to this problem.

    But it is a fascinating bit of mathematical trivia . . . hope you enjoy it.



    Code:
                  *
                 / \
                /36\
               /     \
              /       \ φ
             /         \
            /           \
           / 72     72 \
          * - - - - - - - *
                  1

    This is a very special isosceles triangle.
    The ratio of the equal side to the base is: . \phi \:=\:\frac{1+\sqrt{5}}{2} . . . the Golden Ratio.



    Hence, with this triangle:


    Code:
                  *
                 / \
                /36\
               /     \
            x /       \
             /         \
            /           \
           / 72     72 \
          * - - - - - - - *
                 550
    we have: . \frac{x}{550} \:=\:\phi \quad\Rightarrow\quad x \;\approx\;890


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~



    The Fibonacci sequence: . 1,1,2,3,5,8,13,21,34,55,89,144,233, 377, \hdots
    . . where each term is the sum of the preceding two terms.

    The ratios of consecutive terms approach the Golden Mean.

    . . \begin{array}{ccc}<br />
\phi & = & 1.618 033 989 \\ \hline \\[-3mm]<br />
\dfrac{55}{34} &=& 1.617 647 059 \\ \\[-2mm]<br />
\dfrac{233}{144} &=& 1.618 025 751 \\ \\[-2mm]<br />
\dfrac{987}{610} &=& 1.618 032 787 \\ \\[-2mm]<br />
\dfrac{2584}{1597} &=& 1.618 033 813 \\ <br />
\vdots && \vdots \end{array}


    Hence, any pair of consecutive terms of the Fibonacci sequence
    . . will form a triangle very close to the 36-72-72 triangle.


    Example:

    Code:
                  A
                  *
                 / \
                /   \
               /     \
          377 /       \ 377
             /         \
            /           \
           /             \
        B * - - - - - - - * C
                 233

    The vertex angle is: . A \;\approx\;36.0002^o

    The base angles are: . B \:=\:C \;\approx\;71.9999^o

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