**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#