Find the area of a triangle if its two sides measure $6 in.$ and $9 in.$, and the bisector of the angle between the sides is $4\sqrt{3}$ in. Find the area of a triangle if its two sides measure $6$ in. and $9$ in., and the bisector of the angle between the sides is $4\sqrt{3}$ in. I'm thinking of using the formula $A$=$\frac{1}{2}bh$ I can't find the base or height, I used the Angle bisector formula which is $l=$$\frac{\sqrt{ab[(a+b)^2-c^2]}}{a+b}$ So i found out C which i think is the base should i multiply it by $2$? because I think it's the half. From here I'm lost

If you know the lengths of two sides $a$ and $b$, and the angle $\theta$ between them, then the area of the triangle is $\frac{1}{2}ab\sin \theta$. Let's label the sides $a = 6$, $b = 4\sqrt 3$, and $c = 9$, and the half-angle as $\theta$. The sum of each sub triangle area must equal the total triangle area:
$$A = \frac{1}{2}ab\sin\theta + \frac{1}{2}bc\sin\theta = \frac{1}{2}ac\sin2\theta $$
Solving this,
$$ ab\sin\theta + bc\sin\theta = ac\sin2\theta = 2ac\sin\theta\cos\theta$$
$$ \cos\theta = \frac{b(a+c)}{2ac} = \frac{5}{3\sqrt 3}$$
Then,
$$\sin\theta = \sqrt{1-\cos^2\theta} = \frac{\sqrt{2}}{3\sqrt{3}} \qquad \sin2\theta = \frac{10\sqrt 2}{27}$$
And so the area is
$$ A = \frac{1}{2}ac\sin 2\theta = 10\sqrt 2 \text{ sq. in.} $$
