Math Challenges

Monday, August 21, 2006

PROBLEM OF THE WEEK 2

In the triangle ABC, the foot of the perpendicular from A divides the opposite side into parts length 3 and 17, and tan A = 22/7. Find the area of triangle ABC.

posted by zarcsman at 10:02 PM

2 Comments:

  • Sir Zarco. . .
    nakuha ko na p0h ung answer. . .
    tanA = 22/7
    let x: be the base angle of the side whose length is 17.
    let y: be the base angle of the side whose length is 3.
    Let s be the perpendicular side :
    therefore: x+y=A
    tanx=17/s ;tany=3/s
    s=17/tanx;s=3/tany
    17/tanx=3/tany
    3tanx=17tany
    tanx=17tany/3
    tanA=tan(x+y)
    22/7=tanx+tany/1-tanxtany
    22(1-tanxtany)=7(tanx+tany)
    22-22tanxtany=7tanx+7tany
    Since tanx =17tany/3,
    22-22(17tany/3)tany=7(17tany/3)+ 7tany
    22-(374tan²y/3)=(119tany/3)+7tany
    66-374tan²y=119tany+21tany
    -374tan²y-140tany+66=0
    374tan²y+140tany-66=0
    Using quadratic formula :
    tany =3/11 ; -11/17
    Using 3/11=tany,y=15.2551187...deg
    tanx=17tany/3
    tanx=17(3/11)/3
    tanx=17/11
    x=57.09475708
    s=17/tanx
    s=17/(17/11) ; s= 11
    Area = ½bh
    = ½(20)(11)
    =10(11)
    =110sq. units.
    hehhe answer by: Libres Emmanuel
    From:PUP Taguig Campus

    By Anonymous Anonymous, at 10:37 PM  

  • just click the link below
    for my answer
    http://img60.imageshack.us/my.php?image=week2500210uw2.jpg

    By Anonymous Anonymous, at 7:39 AM  

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