Area of a Triangle Using Sine

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 $△BCD$ is a right triangle, the height of the triangle can be related to the sine of $\angle C$ using the sine ratio. Next, substitute the expression found for $h$ into the general formula for the area of a triangle.

The first formula was obtained. To obtain the second formula, notice that $△ABD$ is also a right triangle. Therefore, the sine ratio can be applied again, this time to connect $h$ and $\angle A.$ By substituting this expression into the general formula for the area of a triangle, the second formula can be obtained.

To deduce the third formula, the altitude from $C$ or $A$ should be drawn. In this case, the altitude from $C$ will be arbitrarily drawn and labeled $D$ with a length of $h.$ Because $\angle B$ is obtuse, the altitude will lie outside the triangle.

In this case, the length of the base is $c$ and the height is $h.$ Since $△BDC$ is a right triangle, the sine ratio can be used to connect $\angle CBD$ and $h.$ Since $\angle CBD$ and $\angle B$ form a linear pair, they are supplementary angles. Recall that the sine of an angle is equal to the sine of its supplementary angle. With this information, and using the Substitution Property of Equality, a formula connecting $\angle B$ and $h$ can be written. Multiplying both sides of the last equation by $a,$ it is obtained that $h=a\sin B.$ Finally, substitute this expression for $h$ into the formula for the area of $△ABC.$