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Math Help - triangle with vertices (-5,0), (5,0) and (5cos A , 5sin A )

  1. #1
    Member Veronica1999's Avatar
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    triangle with vertices (-5,0), (5,0) and (5cos A , 5sin A )

    How many triangles have area 10 and vertices at (-5,0), (5,0) and (5cosA, 5sinA) for some angle A?

    I know that the height of the triangle has to be 2.
    I recently studied trig and know how to graph functions.
    Is that enough trig to solve this?


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  2. #2
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    Since, as you say, the height is 2, the third vertex lies on one of the lines y = 2 or y = -2. Also, the point (5cosA, 5sinA) for every A lies on the circle of radius 5 with the center (0, 0).
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  3. #3
    Member Veronica1999's Avatar
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    I think I understand. Am I supposed to know that (5cosA, 5 sinA) lies on a circle because cos squared A + sin squared A=1?
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  4. #4
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    This is one reason. In fact, even without the use of Pythagorean theorem, \cos\alpha is defined (at least it was when I studied it) as the x-coordinate of a point on a unit circle that is rotated \alpha radians counterclockwise from (1, 0) . Similarly, \sin\alpha is the y-coordinate. So a point (\cos\alpha,\sin\alpha) lies on a unit circle by definition.
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  5. #5
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    Hello, Veronica1999!

    Did you make a sketch?


    \text{How many triangles have area 10 and vertices}}
    \text{at }A(-5,0),\;B(5,0),\;C(5\cos\theta, 5\sin\theta)}\text{ for some angle }\theta?
    Code:
                    |
                  * * *
              *     |     *  C
            *       |       o
           *        |  5  * :*
                    |   *   :h
          *         | * @   : *
      - - o - - - - + - - - + o - -
         A*         |         *B
                    |
           *        |        *
            *       |       *
              *     |     *
                  * * *
                    |

    \,C is on a circle with radius 5.

    \Delta ABC has base 10 and height \,h
    . . where: . h \:=\:5\sin\theta \:=\:\pm2

    Hence: . \sin\theta \:=\:\dfrac{\pm2}{5} \quad\Rightarro\quad \theta \:=\:\arcsin(\pm0.4)

    \text{Therefore: }\:\theta \;\approx\;\begin{Bmatrix}23.6^o \\ 156.4^o \\ 203.6^o \\ 336.4^o \end{Bmatrix}

    . . There are four triangles.

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