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
