Hello, nikkaloo!
There are a number approaches to this problem.
. . Here's one of them.
Given two base angles and the length of the base,
determine the height of the triangle.
Suppose we are given: .$\displaystyle \angle A,\:\angle B,\:\text{ side c}$ Code:
C
*
*: *
b * : * a
* :h *
* : *
* : *
A * * * * * * * * * * B
c
We have: .$\displaystyle \sin A \:=\:\frac{h}{b}\quad\Rightarrow\quad h\:=\:b\!\cdot\!\sin A$ .[1]
. . and we need to determine $\displaystyle b.$
From the Law of Sines: .$\displaystyle \frac{b}{\sin B} \:=\:\frac{c}{\sin C}\quad\Rightarrow\quad b \:=\;\frac{c\!\cdot\!\sin B}{\sin C} $ .[2]
. . Since .$\displaystyle C \:=\:180^o - (A+B)$ .and .$\displaystyle \sin[180^o - (A + B)] \:=\:\sin(A+B)$
. . . . then: .$\displaystyle \sin C \:=\:\sin(A+B)$
Substitute into [2]: .$\displaystyle b \:=\:\frac{c\!\cdot\!\sin B}{\sin(A+B)} $
Substitute into [1]: .$\displaystyle h \:=\:\frac{c\!\cdot\sin B}{\sin(A+B)}\cdot \sin A$
Therefore: .$\displaystyle \boxed{h \;=\;\frac{c\!\cdot\!\sin A\!\cdot\!\sin B}{\sin(A+B)}} $