Solving Triangles Using the Law of Sines

Acute Triangles

Consider an acute triangle with height $h$ drawn from one of its vertices. Because $h$ is perpendicular to the base, the original triangle is split into two right triangles.

In these two right triangles, $\angle A$ and $\angle B$ are both opposite angles to $h.$ Therefore, by applying the definition of the sine ratio to $\angle A$ and $\angle B,$ it is possible to relate the sine of these angles with the side lengths $a$ and $b.$ Next, $h$ can be isolated and written in terms of the corresponding side length and angle, for both right triangles. By the Transitive Property of Equality, it can be stated that $b\sin A$ and $a\sin B$ are equal. Finally, the obtained equation can be rearranged to obtain the desired result. By following the same procedure but drawing the height from vertex $A,$ it can be shown that $\frac{\sin B}{b}=\frac{\sin C}{c}.$ Putting these two results together, the Law of Sines is obtained.

Obtuse Triangles

An obtuse triangle will now be considered.

This proof is very similar to the proof for acute triangles, but it uses an interior and an exterior height. First, the height $h_1$ from the vertex where the obtuse angle is located will be drawn. Just as before, this generates two right triangles.

The sine ratio will be written for these right triangles. In the equations, $h_1$ can be isolated. By the Transitive Property of Equality, it can be stated that $c\sin A$ and $a\sin C$ are equal. The obtained equation can be rearranged to obtain the desired result. Next, the exterior height $h_2$ from vertex $C$ will be drawn. Let $E$ be the point of intersection of this height and the extension of $\overline{AB}.$ Here, $\angle ABC$ and $\angle CBE$ form a linear pair and are therefore supplementary angles. Because the sine of supplementary angles is the same, the sine of $\angle ABC$ equals the sine of $\angle CBE.$ Also, using the sine ratio on $△BCE,$ it can be stated that the sine of $\angle CBE$ is the ratio of $h_2$ to $a.$ By the Transitive Property of Equality, the sine of $\angle ABC$ can be written in terms of $h_2$ and $a.$ Now $△ACE$ will be considered. By using the sine ratio, it follows that the sine of $\angle A$ is the ratio of $h_2$ to $b.$ Now, $h_2$ can be written in terms of $\sin ABC$ and $a,$ and in terms of $\sin A$ and $b.$ The Transitive Property of Equality can be used one more time. If only $△ABC$ is considered, then $\angle ABC$ can be named as $\angle B.$ This allows $\sin ABC$ to be written as $\sin B.$ Therefore, $a\sin B=b\sin A,$ and this equation can be rearranged to obtain the desired formula. Finally, this result can be combined with the previous determination $\frac{\sin A}{a}=\frac{\sin C}{c}$ to derive the Law of Sines.