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Math Help - Trigonometric Ratio Problems

  1. #1
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    Trigonometric Ratio Problems

    10) Three circles with centres A, B and C are mutually tangent and no circle lies inside another circle. The circle with centre A has a radius of 3 cm and the circle with centre B has a radius of 5 cm.
    a) If angle BAC= 60 degrees, determine the radius of the third circle.
    b) Find the area of triangle ABC

    I drew a picture and went out on a wing and tried to use the cosine law:

    (3+x)^2=8^2+(5+x)^2-2(8)(5+x)cos 60
    9+6x+x^2=64+25+10x+x^2-80-x
    4x=64+25+40-9
    x=10

    Now that's all fine and dandy, but the answer at the back of the book says the answer is 2 cm! I don't know what to do!

    12In any triangle ABC, prove:
    a) a^2-b^2=c(a cos B - b cos A)
    b) c= a cos B + b cos A

    I don't even know where to start...

    Any help would be greatly appreciated!
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  2. #2
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    Lexington, MA (USA)
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    Hello, Dovud!

    Check your diagram . . . You used the wrong angle.


    Three circles with centres A, B and C are externally mutually tangent.
    The circle with centre A has a radius of 3 cm and the circle with centre B has a radius of 5 cm.

    (a) If angle BAC= 60°, determine the radius of the third circle.

    (b) Find the area of triangle ABC.
    Code:
               3           5
        A * * * * * o * * * * * * * B
           * 60°                *
          3 *                 *
             *              * 5
              o           *
               *        o
              x *     *
                 *  * x
                  *
                  C

    Law of Cosines: . (x+5)^2 \;=\;8^2 + (x+3)^2 - 2(8)(x+3)\cos60^o

    . . . . . . . . . x^2 + 10x + 25 \;=\;x^2 + 6x + 9 + 64 - 8x - 24

    And we have: . 12x \:=\:24 \quad\Rightarrow \quad \boxed{x \:=\:2\text{ cm}}


    The triangle looks like this:
    Code:
                8
        A o  *  *  *  o B
           * 60°    *
          5 *     * 7
             *  *
              o
              C

    Formula: . \text{Area} \;=\;\tfrac{1}{2}\,bc\sin A


    Therefore: . A \;=\;\tfrac{1}{2}(5)(8)\sin60^o \;=\;\boxed{10\sqrt{3}\text{ cm}^2}

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  3. #3
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    Thank you!
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