Question

I have no idea how to do this, it is due in two days. Hopefully someone sees this before then.

Answer 1

Hello,

`m\ \widehat{ABC}=x\\m\ \widehat{BAC}=2*x\\\\So:\\ x+2x=90^o\\x=30^o\\`

`cos(30^o)=(√(3) )/(2) \\`

In the triangle ABC,

`cos(30^o)=(BC)/(BA) \\\\BA=(cos(30^o))/(BC) \\\\BA=((√(3) )/(2) )/(24) =16*√(3) \\\\`

`sin(30^o)=(1 )/(2) =(AC)/(AB) \\\\AC=(1)/(2) *16√(3) =8√(3)`

In the triangle ACB,

`cos(30^o)=(AC)/(AL) \\\\AL=(8√(3) *2)/(√(3) ) =16\\`