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 = 1/2 (base)(height)
following this setup:
area = 1/2 ah
sin(b) =
so
h =
and
area =

Explanation:

Step1: Recall sine - definition in right - triangle

In right - triangle $BDA$, $\sin(B)=\frac{h}{c}$, because $\sin$ of an angle in a right - triangle is the ratio of the opposite side to the hypotenuse.

Step2: Solve for height $h$

From $\sin(B)=\frac{h}{c}$, we can cross - multiply to get $h = c\sin(B)$.

Step3: Substitute $h$ into area formula

The area of triangle $BCA$ is $A=\frac{1}{2}ah$. Substituting $h = c\sin(B)$ into the area formula, we get $A=\frac{1}{2}ac\sin(B)$.

Answer:

The first blank should be $\frac{h}{c}$, the second blank should be $c\sin(B)$, and the third blank should be $\frac{1}{2}ac\sin(B)$.