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Thread: Triangle Circle Problem

  1. #1
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    Triangle Circle Problem

    Hi,

    I have a small math problem I am trying to figure out, and as it has been a few years since I've done much trig I'm wondering if I could get some help

    I have 2 circles of radius=300m set parallel to each other that are bisecting. I know top bisection is 600m from a north side of a building to the south, and I know a 30-60-90 triangle is formed from the north front of the tower. I want to solve for all 3 distances of this triangle.

    I have drawn a diagram to the best of my ability to give a more clear visual. In this diagram I want to solve for lines A, B, and H. Assume the orange line is right over the red line. I know I could form a new triangle from the circle's origin where the hypotenuse would be 300m to the bisection but I was not sure if that would form a 45-45-90 triangle to solve for length B

    Last edited by Booyaah; Feb 22nd 2018 at 08:24 PM.
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  2. #2
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    Re: Triangle Circle Problem

    from the 30-60-90 triangle

    $A = \dfrac{\sqrt{3}H}{2}$

    $B= \dfrac{H}{2}$

    solving for $B$ in terms of $A$ we get

    $B = \dfrac{A}{\sqrt{3}}$

    now the left circle has center $(-B,A)$ (with the building at the south being the origin), and radius $300$ so it's equation is

    $(x+B)^2 + (y-A)^2 = 300^2$

    and at $x=0,~y=600$ (where the red line terminates) we get

    $B^2 + (600-A)^2 = 300^2$

    Now we have two equations in $A$ and $B$

    These can be solved to obtain

    $A= 450,~B=150\sqrt{3}$

    and thus from above

    $H=2B = 300\sqrt{3}$
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  3. #3
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    Re: Triangle Circle Problem

    By the way, the two circles are "intersecting" not "bisecting".
    Thanks from Booyaah
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  4. #4
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    Re: Triangle Circle Problem

    thank you sir, I did not think about using the ratios to substitute for one of the 2 unknowns in order to use that to solve the pythagorean theorem
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