m(∡ABC)=90°
BC=6√3
AC=12√3
⇒AB=√[(12√3)²-(6√3)²]=√(432-108)=√324=18 cm
A) sin ∡BAC=BC/AC=6√3/12√3=1/2 ⇒ m(∡BAC)=30°
B) cos ∡ACB=BC/AC=6√3/12√3=1/2 ⇒ m(∡ACB)=60°
C) tg ∡BAC=BC/AB=6√3/18=√3/3=1/√3
D) ctg ∡BAC=AB/BC=18/6√3=3/√3=√3