Hello,
[tex]m\ \widehat{ABC}=x\\m\ \widehat{BAC}=2*x\\\\So:\\ x+2x=90^o\\x=30^o\\[/tex]
[tex]cos(30^o)=\dfrac{\sqrt{3} }{2} \\[/tex]
In the triangle ABC,
[tex]cos(30^o)=\frac{BC}{BA} \\\\BA=\dfrac{cos(30^o)}{BC} \\\\BA=\frac{\dfrac{\sqrt{3} }{2} }{24} =16*\sqrt{3} \\\\[/tex]
[tex]sin(30^o)=\dfrac{1 }{2} =\dfrac{AC}{AB} \\\\AC=\dfrac{1}{2} *16\sqrt{3} =8\sqrt{3}[/tex]
In the triangle ACB,
[tex]cos(30^o)=\dfrac{AC}{AL} \\\\AL=\dfrac{8\sqrt{3} *2}{\sqrt{3} } =16\\[/tex]
