Results 1 to 5 of 5

Math Help - angle trisector?

  1. #1
    Junior Member Kaloda's Avatar
    Joined
    Sep 2010
    Posts
    74

    angle trisector?

    In triangle ABC, one of the trisector of angle A is a median, and the other is an altitude. If BC = 24, what is the area of triangle ABC?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Kaloda View Post
    In triangle ABC, one of the trisector of angle A is a median, and the other is an altitude. If BC = 24, what is the area of triangle ABC?
    1. I've drawn a sketch (see attachment)

    2. \Delta(AMC) is an isosceles triangle and therefore the height h bisects half of \overline{BC}.

    3. In the right triangle \Delta(AMF) you know about the angle \alpha:

    \tan(\alpha)=\dfrac6h

    4. According to the question \angle(BAM)=\alpha.
    In the right triangle \Delta(ABF) you know:

    \tan(2\alpha)=\dfrac{18}h

    5. Since \tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-(\tan(\alpha))^2} you'll get:

    \dfrac{18}h = \dfrac{2\cdot \frac6h}{1-\left(\frac6h\right)^2}

    Solve for h and subsequently calculate the area of the triangle \Delta(ABC)
    Spoiler:
    I've gotten: Area(\Delta(ABC)) = 72\sqrt{3}
    Attached Thumbnails Attached Thumbnails angle trisector?-winkl3teilung.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2010
    Posts
    16
    I was looking over this, and am somewhat confused about step 5 the solution. I am not sure how you got the second half of that fraction. If possible, could you elaborate on that? Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by fireballs619 View Post
    I was looking over this, and am somewhat confused about step 5 the solution. I am not sure how you got the second half of that fraction. If possible, could you elaborate on that? Thanks!
    I'm not certain that I understand your question correctly ...

    1. This is a formula: \tan(2\alpha)=\dfrac{2\tan(\alpha)}{1-\left(\tan(\alpha) \right)^2}

    2. In step #3 you find \tan(\alpha)=\dfrac6h

    3. I then replaced \tan(\alpha) in the formula by \dfrac6h

    ... or do you want me to derive the formula?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2010
    Posts
    16
    Oh, my apologies, as I did not realize that it was a formula. I thought you had somehow derived it from the diagram, and I could not for the life of me figure out how. Thanks for the time, though!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: March 1st 2010, 09:53 AM
  2. Replies: 1
    Last Post: September 11th 2009, 06:46 AM
  3. Replies: 2
    Last Post: March 28th 2009, 02:18 PM
  4. Replies: 5
    Last Post: November 10th 2008, 05:42 AM
  5. Replies: 2
    Last Post: February 15th 2007, 10:16 PM

Search Tags


/mathhelpforum @mathhelpforum