Area of a Triangle (SAS)
In triangle $ABC$, side $a = BC$ and side $b = CA$ meet at vertex $C$, where the included
angle is $C$. Write the area of the triangle in terms of $a$, $b$ and $C$.
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Answer
$ \tfrac{1}{2}\,a\,b\,\sin C $
Solution
Drop a perpendicular from $B$ to line $AC$; its length is $a\sin C$. Taking $b$ as the
base, the area is
$$\text{Area} = \tfrac{1}{2}\,b\,(a\sin C) = \tfrac{1}{2}\,a\,b\,\sin C.$$