Law of Sines

If $A$, $B$, and $C$ are the angles of a triangle and $a$, $b$, and $c$ are the corresponding opposing side lengths, then the Law of Sines states

\[\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\]

The easiest way to prove the Law of Sines is to break it up into 3 cases: acute, right, and obtuse triangles. An acute angle is an angle less than $90^{\circ}$. A right angle is an angle that is exactly $90^{\circ}$. An obtuse angle is an angle greater than $90^{\circ}$. An acute triangle is a triangle with all acute angles. A right triangle is a triangle with one right angle. An obtuse triangle is a triangle with one obtuse angle.


Acute Triangles

    


Using the definition of sine on the left triangle

\[\sin B = \frac{h_A}{c} \qquad\qquad \sin C = \frac{h_A}{b}\]

Therefore,

\[\frac{\sin B}{b} = \frac{\sin C}{c}\]

Likewise, using the definition of sine on the right triangle

\[\sin A = \frac{h_B}{c} \qquad\qquad \sin C = \frac{h_B}{a}\]

Therefore,

\[\frac{\sin A}{a} = \frac{\sin C}{c}\]

This completes the proof.


Right Triangles


Using the definition of sine on angles $A$, $B$, and $C$.

\[\sin A = \frac{a}{b} \qquad \sin B = \sin 90^{\circ} = 1 \qquad \sin C = \frac{c}{b}\]

Therefore

\[\frac{\sin A}{a} = \frac{\sin C}{c} = \frac{1}{b} = \frac{\sin B}{b}\]


Obtuse Triangles

    


This proof is the same as the acute triangle case. Using the definition of sine on the left triangle

\[\sin B = \frac{h_A}{c} \qquad\qquad \sin C = \frac{h_A}{b}\]

Therefore,

\[\frac{\sin B}{b} = \frac{\sin C}{c}\]

Likewise, using the definition of sine on the right triangle

\[\sin A = \frac{h_B}{c} \qquad\qquad \sin C = \frac{h_B}{a}\]

Therefore,

\[\frac{\sin A}{a} = \frac{\sin C}{c}\]

This completes the proof.

Trigonometry Series