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=$$\frac{\sqrt{ab[(a+b)^2-c^2]}}{a+b}$ So i found out C which i think is the base should i multiply it by $2$? because I think it's the half. From here I'm lost
 A: If you know the lengths of two sides $a$ and $b$, and the angle $\theta$ between them, then the area of the triangle is $\frac{1}{2}ab\sin \theta$. Let's label the sides $a = 6$, $b = 4\sqrt 3$, and $c = 9$, and the half-angle as $\theta$. The sum of each sub triangle area must equal the total triangle area:
$$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.} $$
A: Hint
Use the SAS formula for solving this.
Note that the bisector divides the triangle in two triangles, so you can use the formula on both of them, and on the original triangle.
A: Let $|BC|=a$, $|AC|=b$, $|AB|=c$,
angle bisector $|AD|=\beta_a$ in $\triangle ABC$.
Given $b=6,\ c=9,\ \beta_a=4\sqrt3$, find the area $S_{ABC}$.
Using known expression for the length of the angle bisector,
\begin{align} 
\beta_a&=bc\left(
1-\frac{a^2}{(b+c)^2}
\right)
\tag{1}\label{1}
,
\end{align}
the missing side length is found as
\begin{align} 
a&=(b+c)\sqrt{1-\frac{\beta_a^2}{bc}}
=5
\tag{2}\label{2}
,
\end{align}
and the area is found as usual by Heron’s formula
\begin{align} 
S_{ABC}&=
\tfrac14\,\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
=10\sqrt2
\tag{3}\label{3}
.
\end{align}
Or, in terms of the given lengths,
\begin{align} 
S_{ABC}&=
\tfrac14\,\beta_a\,(b+c)\,\sqrt{4-\beta_a^2\,(\tfrac1b+\tfrac1c)^2}
=10\sqrt2
\tag{4}\label{4}
\end{align}
