Find the area of triangle ABC if AB = BC = 12 and angle ABC = 120 degrees.

Area=1/2absinC

Area=1/2*12*12*sin120

Area=1/2*144*sin120

Area=72*sin120

Area=62.4cm²(1dp)

sin 30 = h/12     h= 6

6^2 + (1/2b)^2) = 12^2

(1/2 b)^2 = 144-36 = 108

b=20.78

Area = 1/2 bh = 1/2  20.78  x 6 = 62.35 units^2

Thanks so much, both of you!