In triangle ABC, the angle bisector of angle BAC meets AC at D. If angle BAC = 60, angle ABC = 60, and AD = 24, then find the area of triangle ABC.
+34 AD is the angle bisector of angle BAC, we have:
angle BAD = angle CAD = 30
Therefore, triangle ABD is a 30-60-90 triangle, which means that:
AD = BD / sqrt(3) = 8 * sqrt(3)
Now, since angle ABC = 60, triangle ABC is equilateral, which means that all sides are equal. Therefore, we have:
BC = AB
Let x be the length of AB. Then, using the law of cosines in triangle ABC, we have:
x^2 = 24^2 + (8 * sqrt(3))^2 - 2(24)(8 * sqrt(3))(cos 60)
x^2 = 576 + 192 - 384
x^2 = 384
x = 8 * sqrt(6)
Therefore, the area of triangle ABC is:
= (sqrt(3) / 4) * (8 * sqrt(6))^2
= 96 * sqrt(3)
Therefore, the area of triangle ABC is 96 * sqrt(3).
