Answer:
KL = [tex]\frac{20\sqrt{6}}{1+\sqrt{3}}[/tex] = 17.93
MK = [tex]\frac{40\sqrt{3}}{1+\sqrt{3}}[/tex] = 25.36
Explanation:
According to the Law of Sines:
[tex]\frac{a}{sinA}=\frac{b}{sinB}= \frac{c}{sinC}[/tex]
where:
A, B, and C are angles
a, b, and c are the sides opposite to the angles
First of all, let's find m∠L: the sum of the angles of a triangle is 180°, therefore
m∠K + m∠L + m∠M = 180°
m∠L = 180° - m∠K - m∠M
m∠L = 180° - 105° - 30°
m∠L = 45°
Now, we can apply the Law of Sines to our case (see picture attached):
[tex]\frac{LM}{sinK}=\frac{MK}{sinL}=\frac{KL}{sinM}[/tex]
Let's solve one side at the time:
[tex]\frac{LM}{sinK}=\frac{MK}{sinL}[/tex]
[tex]\frac{20\sqrt{3}}{sin(105)}=\frac{MK}{sin(45)}[/tex]
[tex]MK = \frac{20\sqrt{3} }{sin(105)} \cdot sin(45)[/tex]
MK = [tex]\frac{40\sqrt{3} }{1+\sqrt{3} }[/tex] = 25.36
Similarily:
[tex]\frac{LM}{sinK}=\frac{KL}{sinM}[/tex]
[tex]\frac{20\sqrt{3}}{sin(105)}=\frac{KL}{sin(30)}[/tex]
[tex]KL = \frac{20\sqrt{3} }{sin(105)} \cdot sin(30)[/tex]
KL = [tex]\frac{20\sqrt{6}}{1+\sqrt{3}}[/tex] = 17.93
