1. ## Trig Question

Given 2 angels, and the length of one side, how do i determine the height of a triangle?

2. There are many possible answers to that. To have only one possible answer, you need to tell us the relationships of the two given angles and the lone given side. Like, the two angles are base angles while the given side is the base of the triangle. The height of the triangle here is the line segment from the 3rd angle perpendicular to the given side.

3. the two angles are base angles and the side given is the base.

sorrry

4. Have you considered "solving" the triangle?

The Law of Sines will lead you to the lengths of the other two sides. There isn't much work left after that.

5. 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: . $\angle A,\:\angle B,\:\text{ side c}$
Code:
            C
*
*: *
b  * :   *  a
*  :h    *
*   :       *
*    :         *
A * * * * * * * * * * B
c

We have: . $\sin A \:=\:\frac{h}{b}\quad\Rightarrow\quad h\:=\:b\!\cdot\!\sin A$ .[1]
. . and we need to determine $b.$

From the Law of Sines: . $\frac{b}{\sin B} \:=\:\frac{c}{\sin C}\quad\Rightarrow\quad b \:=\;\frac{c\!\cdot\!\sin B}{\sin C}$ .[2]

. . Since . $C \:=\:180^o - (A+B)$ .and . $\sin[180^o - (A + B)] \:=\:\sin(A+B)$

. . . . then: . $\sin C \:=\:\sin(A+B)$

Substitute into [2]: . $b \:=\:\frac{c\!\cdot\!\sin B}{\sin(A+B)}$

Substitute into [1]: . $h \:=\:\frac{c\!\cdot\sin B}{\sin(A+B)}\cdot \sin A$

Therefore: . $\boxed{h \;=\;\frac{c\!\cdot\!\sin A\!\cdot\!\sin B}{\sin(A+B)}}$