<s>[INST] 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
 [/INST] 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}
</s>