Results 1 to 2 of 2

Math Help - intruiging geometry problem

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    1

    intruiging geometry problem

    okay so here's the problem that i found on the net...

    "in the picture, AD is angle bisector if AB + AD = CD and AD + AC = BC ,
    angle ACB = 20 degree,"
    find angle ABC

    here's the link to the picture (http://img694.imageshack.us/img694/6562/pb3.png)

    and there are hints onto solving this problem as well... but i still can't doooo it.

    "AD is a bisector implies
    AB : AC = area(ABD) : area(ADC) = BD : DC
    Use this to express BC/AB and hence cos(angle ACB) in
    terms of k = AC/AB. Now, what is sin(angle ABC) in terms
    of k and sin(angle ACB)?"

    TIA
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: intriguing geometry problem

    Quote Originally Posted by avitos View Post
    okay so here's the problem that i found on the net...

    "in the picture, AD is angle bisector if AB + AD = CD and AD + AC = BC ,
    angle ACB = 20 degree,"
    find angle ABC

    here's the link to the picture (http://img694.imageshack.us/img694/6562/pb3.png)

    and there are hints onto solving this problem as well... but i still can't doooo it.

    "AD is a bisector implies
    AB : AC = area(ABD) : area(ADC) = BD : DC
    Use this to express BC/AB and hence cos(angle ACB) in
    terms of k = AC/AB. Now, what is sin(angle ABC) in terms
    of k and sin(angle ACB)?"
    Following the hint, let k = \frac{AC}{AB} = \frac{CD}{BD}, so that AC=k.AB and CD=k.BD. Then

    \frac{BC}{AB} = \frac{BD+CD}{AB} = (1+k)\frac{BD}{AB}.

    But also \frac{BC}{AB} = \frac{AC+AD}{AB} = k + \frac{CD-AB}{AB} = k-1+\frac{CD}{AB} = k-1+k\frac{BD}{AB}.

    Comparing those two equations, you see that k-1 = \frac{BD}{AB} and hence \frac{BC}{AB} = k^2-1.

    Thus AC=k.AB and BC = (k^2-1)AB. Now the cosine formula in triangle ABC tells you that

    \cos20^\circ = \frac{AC^2+BC^2-AB^2}{2AC.BC}. Divide top and bottom by AB^2 to get

    \cos20^\circ = \frac{k^2 + (k^2-1)^2-1}{2k(k^2-1)} = \frac k2 (after a bit of simplifying). Therefore k=2\cos20^\circ.

    Finally, the sine rule tells you that \frac{\sin(ABC)}{AC} = \frac{\sin20^\circ}{AB}, from which

    \sin(\angle ABC) = k\sin20^\circ = 2\sin20^\circ \cos20^\circ = \sin40^\circ, and so angle ABC = 40^\circ.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 10th 2009, 05:49 AM
  2. geometry problem
    Posted in the Geometry Forum
    Replies: 2
    Last Post: November 19th 2008, 08:22 AM
  3. Geometry problem
    Posted in the Geometry Forum
    Replies: 4
    Last Post: April 15th 2008, 09:45 AM
  4. Help this geometry problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: March 29th 2008, 03:25 AM
  5. A Geometry problem, please help me
    Posted in the Geometry Forum
    Replies: 3
    Last Post: February 9th 2008, 08:54 PM

Search Tags


/mathhelpforum @mathhelpforum