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Prove: $\displaystyle \text{Area of a triangle} \:=\:\tfrac{1}{2}bc \sin A$ Code:
B
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*| *
c * | * a
* |h *
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A * * * * * * * C
b
We know that the area is: .$\displaystyle \text{Area} \:=\:\tfrac{1}{2}\text{(base)(height)}$ .[1]
The base of the triangle is side $\displaystyle b.$ .[2]
In the right triangle at the left: .$\displaystyle \sin A \,=\,\frac{h}{c} \quad\Rightarrow\quad h \,=\,c\sin A$ .[3]
Substitute [2] and [3] into [1]:
. . . $\displaystyle \text{Area} \:=\:\tfrac{1}{2}\underbrace{\text{(base)}}_{b} \underbrace{\text{(height)}}_{c\sin A} \;=\;\tfrac{1}{2}bc\sin A$