Question

see the figure of $\triangle abc$ with auxiliary lines added.
if $c$ is the base of $\triangle abc$, the height is
$\sin(a) = $
the previous statement is leading to the derivation of which area formula?
area $\triangle abc = $

Explanation:

Step1: Identify height for base $c$

In $\triangle ABC$, base is $AB = c$. The height from $C$ to line $AB$ (extended to $D$) is the segment $CD$. In right $\triangle ACD$, $CD = b\sin(A)$ (opposite side to angle $A$).

Step2: Define $\sin(A)$

From right $\triangle ACD$, $\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{CD}{b}$.

Step3: Derive area formula

Area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$. Substitute base $c$ and height $b\sin(A)$, or use the relationship to get the formula with two sides and included angle.

Answer:

If $c$ is the base of $\triangle ABC$, the height is $\boldsymbol{b\sin(A)}$
$\sin(A) = \boldsymbol{\frac{CD}{b}}$ (or equivalently $\boldsymbol{\frac{a\sin(C)}{b}}$ but the direct right triangle ratio is $\frac{CD}{b}$; alternatively $\boldsymbol{\text{height}/b}$)
The previous statement is leading to the derivation of which area formula?
Area $\triangle ABC = \boldsymbol{\frac{1}{2}bc\sin(A)}$