Given
#->/_ACB=60^@#
#->/_BAC=30^@#
#->Rest/_ABC=90^@#
#-> and BC = 6 cm#
Construction
#CD = BC " is cut off from " CA " and B,D are joined"#
Analysis
#"In " Delta BDC#
#/_BCD=60^@,BC=CD->/_CDB=/_CBD=60^@#
#:. Delta BDC " is equilateral "->BC=CD=BD=6 cm#
#"In " Delta ABD#
#/_ABD=/_ABC-/_DBC=(90-60)=30^@ #
#and /_BAD=30^@=>Delta ABD-> "isosceles triangle" #
#=>AD=BD=6 cm#
#"Now " AC = AD +CD =(6+6) cm=12 cm#
# " Now applying Pythagoras theorem for " Delta ABC#
#AB= sqrt(AC^2-BC^2)=sqrt(12^2-6^2)=6sqrt3 cm#
# " Area of " Delta ABC=1/2xxABxxBC=1/2*6sqrt3*6=18sqrt3 cm^2#