Question 1039827
Thank you all beforehand for helping me out. I really appreciate your time and effort.

Find the area of triangle ABC if AB = BC = 12 and angle ABC = 120.
<pre>You can use: Area of triangle = {{{highlight((1/2) * ac * sin (120) = (1/2) * ac * (sqrt(3)/2) = (1/2) * 12^2 * (sqrt(3)/2) = 144sqrt(3)/4 = highlight_green(36sqrt(3)))}}}

OR

The altitude that's drawn from &#8736ABC to the base (AC) is the SHORTER side of one of the two 30-60-90 special right-triangles that're formed, since it's opposite the smaller of the
2 acute angles ({{{30^o}}}). With the hypotenuse being 12, the shorter leg/height = {{{matrix(1,3, (1/2) * 12, or, 6)}}}, and the longer leg = {{{matrix(1,6, shorter, leg, "*", sqrt(3), or, 6 * sqrt(3))}}}. This means that the base = {{{matrix(1,3, 2 * 6sqrt(3), or, 12sqrt(3))}}}. 
Therefore, area = {{{highlight(matrix(1,7, (1/2) * bh, or, (1/2) * 12sqrt(3) * 6, or, (1/2) * 72sqrt(3), or, highlight_green(36sqrt(3))))}}}