+0 In triangle ABC, the angle bisector of angle BAC meets BC at D. If angle BAC=60 degrees, angle ABC=60 degrees and AD=24 then find the area of triangle ABC.
+1632 In triangle ABC, angle BAC and angle ABC both measure 60 degrees. Because the interior angle sum of a triangle adds up to 180 degrees, our third angle BCA must equal 60 degrees... and then since all three angles are the same, we have an equilateral triangle ABC.
Because ABC is equilateral, then the angle bisector of BAC also acts as the perpendicular height of the triangle = 24.
Angle ABC = 60 degrees, angle BAD = 60/2 = 30 degrees, and angle ADB = 90 degrees, so we have a 30 - 60 - 90 triangle with long leg of 24 units, thus the shorter leg BD would have length 24/sqrt(3) = 8sqrt(3) after rationalizing the denominator. BC = 2*BD because the height of the isosceles triangle also bisects the base.
Base * Height / 2 = Area[ABC] = 24*16sqrt(3)/2 = \(192\sqrt{3}\) units\(^2\).
