There are several ways to compute the area of a triangle. For instance, there’s the basic formula that the area of a triangle is half the base times the height. This formula only works, of course, when you know what the height of the triangle is.

Another is Heron’s formula which gives the area in terms of the
three sides of the triangle, specifically, as the square root of the product
*s*(*s* – *a*)(*s* – *b*)(*s* – *c*)
where *s* is the semiperimeter of the triangle, that is,
*s* = (*a* + *b* + *c*)/2.

Here, we’ll consider a formula for the area of a triangle when you know
two sides and the included angle of the triangle. Suppose we know the values
of the two sides *a* and *b* of the triangle, and the included
angle *C.*

As in the proof of the law of sines in the previous section,
drop a perpendicular *AD* from the vertex *A* of the triangle to the
side *BC,* and label this height *h.* Then triangle *ACD* is a right
triangle, so sin *C* equals *h*/*b.* Therefore,
*h* = *b* sin *C.* Since the area of the
triangle is half the base *a* times the height *h,* therefore the
area also equals half of *ab* sin *C.*
Although the figure is an acute triangle, you can see from the discussion in the
previous section that *h* = *b* sin *C* holds
when the triangle is right or obtuse as well. Therefore, we get the general
formula

That is to say, the area of a triangle is half the product of two sides times the sine of the included angle.