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Thread: area of triangle given angle bisector

  1. #16
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Vinod View Post
    Hello,
    Your answer is correct. Then why did WolframAlpha computaional intelligence give wrong answer?
    no wrong answers.

    Mathematica gives a list of possible solutions, among them some with negative values.

    After we eliminate those extra solutions we get two triangles satisfying the conditions given in the problem,
    both with area = 60
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  2. #17
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Idea View Post
    no wrong answers.

    Mathematica gives a list of possible solutions, among them some with negative values.

    After we eliminate those extra solutions we get two triangles satisfying the conditions given in the problem,
    both with area = 60
    Hello,
    If we put all the four equations given by SlipEternal in the Wolfram computational intelligence It gave the answer for the four unknown variables as follows $z(v)=26.4979,w=3.50207,x=5.31243,y=22.2655$. So the area of triangle assuming $base=y+4=26.2655,height=w=3.50207,$ we get the area of triangle 45.99 using the formula of the area of triangle $[\frac12*base*height]$ which is wrong and different from the correct answer 60
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  3. #18
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    Re: area of triangle given angle bisector

    Interesting!

    Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

    The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
    so AN is not the bisector
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  4. #19
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    Re: area of triangle given angle bisector

    Can someone confirm that there are two possible triangles that fit the requirements, both having an area of 60 ?

    (All dimensions approximate, 4dp)

    (1) AB =13.7262, BC = 8.7423, CA = 16.2738, /_BAC = 32.4933 deg.

    (2) AB = 4.8666, BC = 24.6577, CA = 25.1334, /_BAC = 78.8351 deg.
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  5. #20
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Idea View Post
    Interesting!

    Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

    The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
    so AN is not the bisector
    Hello,

    The WolframAlpha computational intelligence gave these four solutions 1)v≈26.4979, w≈3.50207, x≈-5.31643, y≈22.2655;2)v≈26.4979, w≈3.50207, x≈5.31643, y≈22.2655;
    3)v≈34.5687, w≈-4.56873, x≈-6.07234, y≈-38.2655; 4)v≈34.5687, w≈-4.56873, x≈-6.07234, y≈-38.2655.

    Now tell me which two solutions are acceptable?
    Secondly, You are telling $\overline{AN}$ is not the angle bisector. But your answer confirms that $\overline{AN}$ is the angle bisector and its length is 14.2971.

    So, WolframAlpha gave wrong answers.
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  6. #21
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by BobP View Post
    Can someone confirm that there are two possible triangles that fit the requirements, both having an area of 60 ?

    (All dimensions approximate, 4dp)

    (1) AB =13.7262, BC = 8.7423, CA = 16.2738, /_BAC = 32.4933 deg.

    (2) AB = 4.8666, BC = 24.6577, CA = 25.1334, /_BAC = 78.8351 deg.

    Hello,
    Both the triangles have the area of ${60^\circ}$.
    Last edited by Vinod; Nov 21st 2018 at 05:00 AM.
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  7. #22
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Idea View Post
    Interesting!

    Wolfram gives three solutions. Two solutions are acceptable and give an area = 60

    The third solution w=3.50207 gives an angle BAN = 48.8 and angle NAC = 33.6 degrees
    so AN is not the bisector
    Hello,

    There is yet another equation to add in inputs to get correct solution i-e $\frac{y}{4}=\frac{z}{w}$.So WolframAlpha computational intelligence is correct, we omitted one input to add.
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  8. #23
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by SlipEternal View Post
    Let's use letters to represent the different lengths:

    $$\overline{AB} = w, \overline{AN} = x, \overline{NC}=y, \overline{AC} = z$$

    Now, we have the following information:

    By the Pythagorean Theorem:
    $$w^2+4^2=x^2$$

    By the Pythagorean Theorem:
    $$(4+y)^2+w^2 = z^2$$

    By the double angle formula for sine:
    $$\dfrac{4+y}{z} = 2\dfrac{4}{x}\cdot \dfrac{w}{x}$$

    Given:
    $$w+z=30$$

    That is four equations in four variables. So, this system may have a solution. You would need to solve it.
    Hello,

    There is one more equation to add which is $\frac{y}{4}=\frac{z}{w}$.
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