Answer:
[tex]\frac{CL}{AC}=\frac{1}{3}[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠BAC+m∠ABC=90° -----> by complementary angles
m∠BAC=30° ----> given problem
so
m∠ABC=60°
If BL is an angle bisector of m∠ABC
then
m∠ABL=m∠LBC=30°
In the right triangle LBC
[tex]tan(30\°)=CL/BC[/tex]
Solve for BC
[tex]BC=CL/tan(30\°)[/tex] ----> equation A
In the right triangle ABC
[tex]tan(60\°)=AC/BC[/tex]
Solve for BC
[tex]BC=AC/tan(60\°)[/tex] ----> equation B
Equate equation A and equation B
[tex]\frac{CL}{tan(30\°)}=\frac{AC}{tan(60\°)}[/tex]
[tex]\frac{CL}{AC}=\frac{tan(30\°)}{tan(60\°)}[/tex]
Remember that
[tex]tan(30\°)=\frac{\sqrt{3}}{3}[/tex]
[tex]tan(60\°)=\sqrt{3}[/tex]
substitute
[tex]\frac{CL}{AC}=\frac{1}{3}[/tex]
