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Math Help - finding largest value of tan angle

  1. #1
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    finding largest value of tan angle

    Triangle ABC has angle C = 60 degrees and BC = 4. Point D is the midpoint of BC. What is the largest possible value of tan ( angle BAD ) ?

    I have been working on this problem for over a week and I understand it is way over my level, but I can't give up. It's frustrating because the problem seems to be so simple. ( but now i see it looked simple because of my lack of knowledge)

    Hoping to get some ideas, I have been studying Trigonometry and Geometry( triangle section only).

    So far, I was looking for a way to make DA as small as I can.
    Wouldn't that make tan (angle BAD ) the largest?
    Or am I hopelessly off?


    Vicky.
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  2. #2
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    Smile

    Quote Originally Posted by Vicky1997 View Post
    Triangle ABC has angle C = 60 degrees and BC = 4. Point D is the midpoint of BC. What is the largest possible value of tan ( angle BAD ) ?

    I have been working on this problem for over a week and I understand it is way over my level, but I can't give up. It's frustrating because the problem seems to be so simple. ( but now i see it looked simple because of my lack of knowledge)

    Hoping to get some ideas, I have been studying Trigonometry and Geometry( triangle section only).

    So far, I was looking for a way to make DA as small as I can.
    Wouldn't that make tan (angle BAD ) the largest?
    Or am I hopelessly off?


    Vicky.
    hi
    you have \tan(\widehat{BAD})=\frac{BD}{AD}
    what is the largest value of \frac{BD}{AD} ?
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  3. #3
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    Quote Originally Posted by Raoh View Post
    hi
    you have \tan(\widehat{BAD})=\frac{BD}{AD}
    what is the largest value of \frac{BD}{AD} ?

    I am not sure how to get it.
    Could you please explain to me how to find the largest value?

    Vicky.
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  4. #4
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    Hello, Vicky!

    I hope I understand the problem . . .


    Triangle ABC has: . \angle C = 60^o,\;BC = 4.
    Point D is the midpoint of BC.
    What is the largest possible value of \tan(\angle BAD) ?
    Code:
                    C
                    o
                   *  *
                  * 60 * 2
                 *        *
                *           *   D
               *              o
              *           *   | *
             *        *       |   * 2
            *     *           |     *
           *  *  θ            |       *
        A o - - - - - - - - - o - - - - o B
                              E
    We have: . \angle C = 60^o,\;CD = DB = 2
    Let \theta = \angle BAD.
    Draw DE \perp AB.

    And we want to maximize: . \tan\theta \,=\,\frac{DE}{AE}

    Is this correct?

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, Vicky!

    I hope I understand the problem . . .

    Code:
                    C
                    o
                   *  *
                  * 60 * 2
                 *        *
                *           *   D
               *              o
              *           *   | *
             *        *       |   * 2
            *     *           |     *
           *  *  θ            |       *
        A o - - - - - - - - - o - - - - o B
                              E
    We have: . \angle C = 60^o,\;CD = DB = 2
    Let \theta = \angle BAD.
    Draw DE \perp AB.

    And we want to maximize: . \tan\theta \,=\,\frac{DE}{AE}

    Is this correct?

    Thanks to you, I think I am starting to understand the problem better.
    But all I have is the answer.
    Would your solution get root 3 over 4 root 2 minus 3 ?

    Vicky.
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  6. #6
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    Hello Vicky1997

    This is quite tricky - I've been puzzling over it for some time. But I have a solution - and it agrees with the one you've been given.

    In the attached diagram, I have called the length of AC, x cm, and I've drawn the perpendiculars from C and M (the mid-point of BC) to AB, meeting it at E and D.

    Here's my solution.

    The Cosine Rule on \triangle ABC gives:
    AB^2 = x^2-4x+16 (1)
    The area of \triangle ABC can be written in two ways, using the formulae \tfrac12AC.BC\sin \angle ACB and \tfrac12AB.EC. This gives:
    \tfrac124x\sin60^o=\sqrt3x=\tfrac12AB.EC

    \Rightarrow EC = \frac{2\sqrt3x}{AB}

    \Rightarrow DM = \frac{\sqrt3x}{AB}, using the similar triangles BMD, BCE (2)
    Next, using Pythagoras on \triangle BMD:
    DB^2=MB^2-DM^2
    = 4 - \frac{3x^2}{AB^2}, from (2)

    = \frac{4AB^2-3x^2}{AB^2}

    =\frac{x^2-16x+64}{AB^2}, from (1)

    =\frac{(8-x)^2}{AB^2}
    \Rightarrow DB = \frac{8-x}{AB} (3)
    You'll see that I've taken this sign, rather than (x-8), since x>8 will clearly not maximise the value of \tan\theta.

    Now:
    AD = AB-DB
    = \frac{AB^2-(8-x)}{AB}, from (3)

    =\frac{x^2-3x+8}{AB}, from (1)
    and:
    \tan\theta = \frac{MD}{AD}
    = \frac{\sqrt3x}{x^2-3x+8} (4)
    All the hard work is done now. It's a simple matter to differentiate this function, and equate the result to zero for a maximum or minimum. When you do this, the result simplifies to:
    -x^2+8=0

    \Rightarrow x = 2\sqrt2
    You can check that this gives a maximum value of \tan\theta, and that value, from (4), is:
    \frac{2\sqrt2\sqrt3}{8-6\sqrt2+8}
    =\frac{\sqrt3}{4\sqrt2-3}
    Grandad
    Attached Thumbnails Attached Thumbnails finding largest value of tan angle-untitled.jpg  
    Last edited by Grandad; December 19th 2009 at 04:34 AM. Reason: Corrected typo
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  7. #7
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    Quote Originally Posted by Grandad View Post
    Hello Vicky1997

    This is quite tricky - I've been puzzling over it for some time. But I have a solution - and it agrees with the one you've been given.

    In the attached diagram, I have called the length of AC, x cm, and I've drawn the perpendiculars from C and M (the mid-point of BC) to AB, meeting it at E and D.

    Here's my solution.

    The Cosine Rule on \triangle ABC gives:
    AB^2 = x^2-4x+16 (1)
    The area of \triangle ABC can be written in two ways, using the formulae \tfrac12AC.BC\sin \angle ACB and \tfrac12AB.EC. This gives:
    \tfrac124x\sin60^o=\sqrt3x=\tfrac12AB.EC

    \Rightarrow EC = \frac{2\sqrt3x}{AB}

    \Rightarrow DM = \frac{\sqrt3x}{AB}, using the similar triangles BMD, BCE (2)
    Next, using Pythagoras on \triangle BMD:
    DB^2=MB^2-DM^2
    = 4 - \frac{3x^2}{AB^2}, from (2)

    = \frac{4AB^2-3x^2}{AB^2}

    =\frac{x^2-16x+64}{AB^2}, from (1)

    =\frac{(8-x)^2}{AB^2}
    \Rightarrow DB = \frac{8-x}{AB} (3)
    You'll see that I've taken this sign, rather than (x-8), since x>8 will clearly not maximise the value of \tan\theta.

    Now:
    AD = AB-DB
    = \frac{AB^2-(8-x)}{AB}, from (3)

    =\frac{x^2-3x+8}{AB}, from (1)
    and:
    \tan\theta = \frac{MD}{AD}
    = \frac{\sqrt3x}{x^2-3x+8} (4)
    All the hard work is done now. It's a simple matter to differentiate this function, and equate the result to zero for a maximum or minimum. When you do this, the result simplifies to:
    -x^2+8=0

    \Rightarrow x = 2\sqrt2
    You can check that this gives a maximum value of \tan\theta, and that value, from (4), is:
    \frac{2\sqrt2\sqrt3}{8-6\sqrt2+8}
    =\frac{\sqrt3}{4\sqrt2-3}
    Grandad

    Hello Grandad

    Thank you!!!!

    Honestly, I only understand certain parts of the solution. But I have printed it out and this is what I will be studying over the winter break.

    Vicky.
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