Question

deriving the trigonometric area formula from a different vertex

the altitude of the triangle abc, segment da, has been drawn. derive a formula for the area of triangle bca that depends on the measure of angle b along with side lengths $a$ and $c$.

area = $\frac{1}{2}$(base)(height)
following this setup:
area = $\frac{1}{2}ah$
$sin(b) = $ , so
$h = $ , and
area =

Explanation:

Step1: Define sin(B) from right triangle

In right triangle ABD, $\sin(B) = \frac{h}{c}$

Step2: Solve for height h

Rearrange to isolate $h$: $h = c\sin(B)$

Step3: Substitute h into area formula

Replace $h$ in $\text{Area} = \frac{1}{2}ah$: $\text{Area} = \frac{1}{2}ac\sin(B)$

Answer:

$\sin(B) = \boldsymbol{\frac{h}{c}}$
$h = \boldsymbol{c\sin(B)}$
$\text{Area} = \boldsymbol{\frac{1}{2}ac\sin(B)}$