If CL is a bisector, how do i solve this?

Answer:
Step-by-step explanation:
First equation: Use angle bisector theorem
[tex]\frac{x}{y}[/tex] = [tex]\frac{35}{21}[/tex] , [tex]\frac{x}{y}[/tex] = [tex]\frac{5}{3}[/tex] ⇒ x = [tex]\frac{5}{3}[/tex] y ............ (1)
Second equation: Use cosine law
x² + y² - 2xycos 120° = ( 35 + 21 )² , cos 120° = - 0.5
x² + y² + xy = 56² ................. (2)
(1) ----> (2)
( [tex]\frac{5}{3}[/tex] y )² + y² + [tex]\frac{5}{3}[/tex] y² = 56² ............. (3)
9 × (3)
25y² + 9y² + 15y² =( 9 ) × 56²
7²y² = (168)²
y = 24
x = [tex]\frac{5}{3}[/tex] × 24 = 40
AC = 40
BC = 24
m∠A ≈ 22 ° ( see the attachment )
m∠B ≈ 38 °