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Math Help - How much?

  1. #1
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    Post How much?

    Hi...

    In a triangle ABC are AB= 7 cm, BC= 4cm and angle BAC = 32 degrees...

    How much can you increase angle BAC, while AB and BC are unchanged?

    Can you guys help me on this?

    Would be appreciated!
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  2. #2
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    Quote Originally Posted by Neffets...:P

    How much can you increase angle BAC, while AB and BC are unchanged?
    I really do not understand. Do you mean find the side the opposite side?
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  3. #3
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    ...

    No, I'm not trying to find the opposite side...

    Kinda hard to explain this... How large can the angle BAC be if AB and BC have the length i wrote before? In other words increase the angle BAC to the point where CB "falls" off AC. Understand?

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  4. #4
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    Quote Originally Posted by Neffets...:P
    . Understand?
    Yes. So you are working with the "ambigous case".
    (Look at the beautifully artistic picture).
    ---
    You have by the Law of Sines,
    \frac{\sin 32^o}{4}=\frac{\sin C}{7}
    From here we have,
    \sin C=\frac{7}{4}\sin 32^o\approx .9274
    Then,
    C\approx 68.0268^o
    But that is not the only answer. There is one more in the second quadrant namely,
    180^o-C\approx 111.9732^o
    Attached Thumbnails Attached Thumbnails How much?-picture1.gif  
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  5. #5
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    Thanks for your answer...
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  6. #6
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    Quote Originally Posted by Neffets...:P
    Hi...

    In a triangle ABC are AB= 7 cm, BC= 4cm and angle BAC = 32 degrees...

    How much can you increase angle BAC, while AB and BC are unchanged?

    Can you guys help me on this?

    Would be appreciated!
    Consider C moving on a circle of radius 4cm around B. Every position is a valid
    triangle with AB=7 and BC=4cm, and all such triangles occur for some
    position of C on this circle.

    The largest value of angle BAC occurs when AC is a tangent to this
    circle, and angle ACB is a right angle. Then sin(BAC)=4/7, or the maximum
    angle BAC=34.85 degrees. (You might want to try to prove this)

    RonL

    RonL
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  7. #7
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    Quote Originally Posted by ThePerfectHacker
    Yes. So you are working with the "ambigous case".
    (Look at the beautifully artistic picture).
    ---
    You have by the Law of Sines,
    \frac{\sin 32^o}{4}=\frac{\sin C}{7}
    From here we have,
    \sin C=\frac{7}{4}\sin 32^o\approx .9274
    Then,
    C\approx 68.0268^o
    But that is not the only answer. There is one more in the second quadrant namely,
    180^o-C\approx 111.9732^o
    You have still not understood the question. In the base case angle BAC is
    32 degrees. What is the largest size of this angle for a triangle with AB and
    BC maintaining the given values.

    RonL
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  8. #8
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    ..

    Thats right Captain!

    You got it right...

    Thanks for all of your help
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  9. #9
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    Hello, Neffets!

    In \Delta ABC,\;AB= 7\text{ cm, }BC= 4\text{ cm and }\angle A = 32^o

    How much can you increase \angle A, while AB and BC are unchanged?

    From the given information, there are two possible triangles.
    Code:
                            B                           B
                            *                           *
                         * *                         *   *
                   7  *   *                    7  *       *
                   *     * 4                   *           * 4
                *       *                   *               *
             * 32     *                 * 32               *
        A * * * * * * * C           A * * * * * * * * * * * * * C

    The limit is when C = 90^o.
    Code:
                            B
                            *
                         *  *
                   7  *     *
                   *        * 4
                *           *
             *              *
        A * * * * * * * * * * C

    Then A \:=\:\sin^{-1}\left(\frac{4}{7}\right) \:\approx\:34.85^o

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