SOLUTION: Hello all, could you please help me with this problem? Would greatly appreciate the help. <a rel=nofollow HREF="https://imgbb.com/"><img src="https://i.ibb.co/SKYyCNx/IMG-6263.j

Algebra ->  Triangles -> SOLUTION: Hello all, could you please help me with this problem? Would greatly appreciate the help. <a rel=nofollow HREF="https://imgbb.com/"><img src="https://i.ibb.co/SKYyCNx/IMG-6263.j      Log On


   



Question 1189026: Hello all, could you please help me with this problem? Would greatly appreciate the help.
IMG-6263

Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.

Since the shown angles are equal, the shown segments inside triangle are bisectors of respective angles.


    +----------------------------------------------------------------------+
    |    In any triangle, bisector of an angle divides the opposite side   |
    |        in the ratio equal to the ratio of adjacent sides.            |
    +----------------------------------------------------------------------+


THEREFORE, in the given triangle  y%2Fz = 6%2F4  and  y + z = 5.

    So, the side of the length 5 is divided in the ratio 6:4.

    It means that y = 5%2A%286%2F10%29 = 6%2F2 = 3;  z = 5-3 = 2.



Similarly, side of the length 6 is divided in propoprtion  x%2Fw = 5%2F4.

    So, the side of the length 6 is divided in the ratio 5:4.

    It means that  x = 6%2A%285%2F9%29 = 30%2F9 = 10%2F3 = 3 1%2F3,  w = 6 - 3 1%2F3 = 2 2%2F3.


ANSWER.  x = 3 1%2F3;  y = 3;  z = 2;  w = 2 2%2F3.

Solved.