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Math Help - Triangle in circle?!

  1. #1
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    Triangle in circle?!

    Triangle ABC has angle B 45 degrees, side AB 9 and side BC 6\sqrt2.

    The area of ABC is:

    And since side AC is:

    ... the circle containing triangle ABC has the radius:

    Would be great to get some insight on this. I'm completely lost as to what to do.
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  2. #2
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    Quote Originally Posted by davidman View Post
    Triangle ABC has angle B 45 degrees, side AB 9 and side BC 6\sqrt2.

    The area of ABC is:

    And since side AC is:

    ... the circle containing triangle ABC has the radius:

    Would be great to get some insight on this. I'm completely lost as to what to do.
    Use the formula

    A = \frac{1}{2}ab\,\sin{C}, where a and b are two sides of the triangle and C is the angle between them.
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    A=\frac{1}{2}absinC=\frac{1}{2}\times9\times6\sqrt  {2}sin45

    and looking at triangles with special angles 45 and 90 tells me that

    sin45=\frac{1}{\sqrt{2}}\:\:\therefore

    27\sqrt{2}\times\frac{1}{\sqrt{2}}=27

    how to get side AC? I only know that one angle...
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  4. #4
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    Quote Originally Posted by davidman View Post
    A=\frac{1}{2}absinC=\frac{1}{2}\times9\times6\sqrt  {2}sin45

    and looking at triangles with special angles 45 and 90 tells me that

    sin45=\frac{1}{\sqrt{2}}\:\:\therefore

    27\sqrt{2}\times\frac{1}{\sqrt{2}}=27

    how to get side AC? I only know that one angle...
    Now to get side c, use the Cosine Rule.

    c^2 = a^2 + b^2 - 2ab\,\cos{C}.


    Also, since you found the area of the triangle before, you need to write = 27\,\textrm{units}^2
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  5. #5
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    AC^2=AB^2+BC^2-2(AB\times{BC})cosB=9^2+(6\sqrt{2})^2-2(9\times{6}\sqrt{2})cos45= 81+36\times{2}-2\times{9}\times{6}\sqrt{2}\times{\frac{1}{\sqrt{2  }}}=45

    AC=\sqrt{45}=\sqrt{9\times5}=3\sqrt{5}

    ok, managed with that somehow... about the circle though, is there a rule for that?
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    Quote Originally Posted by davidman View Post
    Triangle ABC has angle B 45 degrees, side AB 9 and side BC 6\sqrt2.

    The area of ABC is:

    And since side AC is:

    ... the circle containing triangle ABC has the radius:

    Would be great to get some insight on this. I'm completely lost as to what to do.
    Are you saying that the vertices A, B, C need to lie on the circle?
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  7. #7
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    Yes, like

    wasn't sure what it's called in English, but I guess circumscribed circle of a triangle.
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    OK I don't know if there's an easier way, but here goes:

    I placed the length of 9 units along the x axis of a set of cartestian axes, beginning at the origin.


    So that means that two of the vertices lie at (0,0) and (9, 0).


    There is also an angle of 45^\circ made with the x axis, and a length of 6\sqrt{2} units.


    Using some trigonometry, I know that the final vertex must have co-ordinate (x, y) = (6\sqrt{2}\,\cos{45^\circ}, 6\sqrt{2}\,\sin{45^\circ})

    (x, y) = (6, 6).


    So now you have the three vertices as (x, y) co-ordinates.


    Substitute the three co-ordinates into the general equation for the circle

    (x - h)^2 + (y - k)^2 = r^2.


    You will end up with three equations in three unknowns that you will need to solve simultaneously. One of them ( r) is the radius of the circle.
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  9. #9
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    Quote Originally Posted by Wikipedia @ http://en.wikipedia.org/wiki/Law_of_sines#Relation_to_the_circumcircle
    In the equation



    the common value of the three fractions is actually the diameter of the triangle's circumcircle.
    hence;

    \frac{a}{sin\:A}=2R

    or in my case;

    \frac{AC}{sin\:B}=\frac{3\sqrt{5}}{\frac{1}{\sqrt{  2}}}=\frac{3\sqrt{10}}{1}

    R=\frac{3\sqrt{10}}{2}

    so I guess wikipedia helped save the day this time.
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