One of the bisectors of the triangle is 10 cm and is extended by the point of intersection

One of the bisectors of the triangle is 10 cm and is extended by the point of intersection of the bissetrices in the ratio 3: 2, counting from the top. Find the length of the side of the triangle to which this bisector is drawn.

Given:

Triangle ABC. ВO / OM = 3/2. ВM = 10.

Solution:

In the triangle of the IUD, the bisector is OC, then BC / MC = VO / OM = 3/2. In the triangle ABC, the bisector BM, then MC / BC = AM / AB = 2/3.

AB / BC = AM / MS = 2/3.

MC ^ 2 = ВС ^ 2 + ВМ ^ 2 – 2 * ВС * ВМ * cosА (By t. Cosines)

X ^ 2 = (3/2 * X) ^ 2 + 100 – 2 * (3/2 * X) * 10 * cosA;

cosA = (5/4 * X ^ 2 + 100) / 30 * X.

AM ^ 2 = AB ^ 2 + BM ^ 2 – 2 * AB * BM * cosA (By t. Cosines)

(2/3 * X) ^ 2 = X ^ 2 + 100 – 2 * X * 10 * cosA;

cosA = (5/9 * X ^ 2 + 100) / 20 * X.

Equating the expressions for the cosines.

X = 2√30.

AB = 2√30.

AM = 2/3 * AB = (4/3) √30.

BC = 3/2 * AB = 3√30,

MC = 2/3 * BC = 2√30.

AC = AM + MC = (10/3) * √30. (cosA is the cosine of half of angle B)