Triangle ABC is isosceles. Angle B = 120 degrees. AC = 18 cm. Find the area of the triangle.

Let’s draw the height of the ВN (it is the medial and the bisector, since this is an isosceles triangle)
angle A = angle B = (180-120) / 2 = 30 degrees
AH = AC / 2 = 18/2 = 9
The leg lying opposite an angle of 30 degrees is equal to half the hypotenuse
VN = VA / 2
Consider the triangle ABN
By the Pythagorean theorem, we find VA
BH ^ 2 + AH ^ 2 = BA ^ 2
(ВA / 2) ^ 2 + 9 ^ 2 = BA ^ 2
(BA ^ 2) / 4 + 81 = BA ^ 2 | * 4
ВA ^ 2 + 324 = 4ВA ^ 2
3BA ^ 2 = 324
ВA ^ 2 = 108
ВA = 6 √3
Sabc = 1/2 * sinBAC * BA * AC
Sabc = 1/2 * sin 30 * 6 √ 3 * 18 = 1/2 * 1/2 * 6√3 * 18 = 27 √3
There is another way to find the area
Sabc = 1/2 * BH * AC
BH = BA / 2
BH = (6√3) / 2
ВN = 3 √ 3
Sabc = 1/2 * 3 √ 3 * 18 = 27 √ 3