Answer:
[tex]m\angle CAB=102\dfrac{6}{7}^{\circ}\\ \\m\angle ABC=51\dfrac{3}{7}^{\circ}\\ \\m\angle ACB=25\dfrac{5}{7}^{\circ}[/tex]
Step-by-step explanation:
AK is angle A bisector, then
[tex]m\angle BAK=m\angle KAC=x^{\circ}[/tex]
BL is angle B bisector, then
[tex]m\angle ABL=m\angle CBL=y^{\circ}[/tex]
Consider triangle ABL. The sum of the measures of all interior angles in this triangle is [tex]180^{\circ},[/tex] then
[tex]m\angle BAL+m\angle ALB+m\angle LBA=180^{\circ}\\ \\2x+2y+y=180\\ \\2x+3y=180[/tex]
Consider triangle ABK. In this triangle,
[tex]m\angle AKB=180^{\circ}-2x^{\circ} \ [\text{Supplementary angles}][/tex]
The sum of the measures of all interior angles in this triangle is [tex]180^{\circ},[/tex] then
[tex]m\angle BAK+m\angle AKB+m\angle KBA=180^{\circ}\\ \\x+(180-2x)+2y=180\\ \\2y-x=0[/tex]
Hence,
[tex]x=2y\\ \\2(2y)+3y=180\\ \\4y+3y=180\\ \\7y=180\\ \\y=\dfrac{180}{7}=25\dfrac{5}{7}\\ \\x=\dfrac{360}{7}=51\dfrac{3}{7}[/tex]
Find the measures of the triangle ABC:
[tex]m\angle CAB=2x^{\circ}=102\dfrac{6}{7}^{\circ}\\ \\m\angle ABC=2y^{\circ}=51\dfrac{3}{7}^{\circ}\\ \\m\angle ACB=180^{\circ}-102\dfrac{6}{7}^{\circ}-51\dfrac{3}{7}^{\circ}=25\dfrac{5}{7}^{\circ}[/tex]
