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Math Help - sine rule

  1. #1
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    sine rule

    (Give non-exact answers to 3 significant figures.)

    In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm. Given that the area of the triangle is 1cm2, show that the triangle is right-angled and isosceles.


    Am not sure how to state off, can anyone just start me off . thanks.
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  2. #2
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    Hi

    Using cross product properties we know that the area of the triangle is half the product of 2 lengths of the triangle multiplied by the absolute value of sine of the angle formed by the two sides

    A = \frac12\:||\overrightarrow{AB}\wedge\overrightarro  w{AC}|| = \frac12\:AB\:AC\:|sin(BAC)|

    With this formula you can compute the angle BAC, the length of the other side and conclude
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  3. #3
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     1 = \frac{1}{2} \times 2 \times \sqrt{2} \times sin\theta

     \theta = 45

    where do I go from here? how do I show it is a right angled triangle and isosceles.?
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  4. #4
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    Hello, Tweety

    In a triangle, the largest side has length 2 cm and one of the other sides has length √2 cm.
    Given that the area of the triangle is 1 cm², show that the triangle is right-angled and isosceles.
    Code:
                *
               *  *
           _  *     *
          √2 *        *
            *           *
           * θ            *
          *  *  *  *  *  *  *
                   2

    Formula: . A \;=\;\frac{1}{2}\,bc\sin A


    The area is one-half the product of two sides and the sine of the included angle.


    We have: . \frac{1}{2}\left(\sqrt{2}\right)(2)\sin\theta \:=\:1 \quad\Rightarrow\quad \sin\theta \:=\:\frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta \:=\:45^o


    Hence, the triangle looks like this:

    The right half is identical to the left half.
    Code:
                      *
                    * | *
              _   *   |   *    _
             √2 *     |1    * √2
              *       |       *
            * 45°     |     45° *
          *  *  *  *  *  *  *  *  *
                1           1

    Therefore, the triangle is an isosceles right triangle.

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