Question

Value of x.. help PLEASE!

Answer 1

Let M be the intersection with BC

Here we have to apply the law of sin in order to solve the problem, but prior to that let me refresh your memory:

In any triangle we can write the proportion of sin over opposite side:

sin A/a = sin B/b =sin C/c ,   Also sin(α) = sin(180° α)
Moreover Angle A1 = A₂ (given)

In Δ ABM==>(sin A₁)/(2x-1) = (sin M₁)/9 ==> let's calculate sin M₁==>
sin M₁ =(9.sin A₁) / (2x-1)

In Δ AMC==>(sin A₂)/(3x) = (sin M₂)/(15), but sin M₂ = sin(180°-M1) = sin M₁

Then let's replace sin M₂ = sin M₁ ===>or sin M₁ =(15sin A₂ / 3x)

Notice that the 2 equation of M₁ in bold are equal, then
=(9.sin A₁) /(2x-1)=(15sin A₂ / 3x). Also remember that A₁=A₂. Now solve for x
after simplifying A₁ from both sides:

9/(2x-1) = 15/3x .Cross multiplication  ==> 30x-15=27x==>3x =15 And X=5

Answer 2

Any arbitrary bisector from a vertex to an opposite side divides any given triangle into proportional ratios.  In this case

9/(2x-1)=15/(3x)  cross multiply

27x=30x-15  subtract 30x from both sides

-3x=-15 divide both sides by -3

x=5