Rule

Area of a Triangle Using Sine

The area of a triangle is half the product of the lengths of any two sides and the sine of the included angle.

Based on the diagram above, the area of $△ A B C$ can be found using any of the three formulas below.

$Area Area Area = \frac{1}{2} a b sin C = \frac{1}{2} b c sin A = \frac{1}{2} a c sin B$

Proof

To find the first formula, start by drawing the altitude from $B$ and let $h$ be its length. Since the altitude is perpendicular to the base, it generates two right triangles.

Because $△ B C D$ is a right triangle, the height of the triangle can be related to the sine of $∠ C$ using the sine ratio. $sin C = \frac{h}{a} \Rightarrow h = a sin C$ Finally, substitute the expression found for $h$ into the general formula for finding the area of a triangle.

$Area = \frac{1}{2} b h ⇓ Area = \frac{1}{2} a b sin C$

To obtain the second formula, notice that $△ A B D$ is also a right triangle and then, the sine ratio can be applied. $sin A = \frac{h}{c} \Rightarrow h = c sin A$ Substituting this expression into the general formula for finding the area of a triangle gives the first formula.

$Area = \frac{1}{2} b h ⇓ Area = \frac{1}{2} b c sin A$

To deduce the third formula, it is needed to draw the height from $C$ $($ or $A) .$

In this case, the length of the base is $c$ and the height is $h .$ Since $△ B D C$ is a right triangle, the sine ratio can be used. $sin B = \frac{h}{a} \Rightarrow h = a sin B$ Lastly, substitute the expression for $h$ into the formula to find the area of $△ A B C .$

$Area = \frac{1}{2} c h ⇓ Area = \frac{1}{2} a c sin B$