Law of Sines

Law of Sines

Suppose we try to find the circumradius of a triangle \triangle{ABC}, as shown below:

How can we find it? Note that the triangle \triangle{AOC} is isosceles, with the equal sides being radii of the circumcircle. To find these, we only need to know the angle \angle{AOC} and side length AC. Angle \angle{AOC} intercepts the arc \stackrel{\frown}{AC}, with O at the center of the circle. Angle \angle{B} also intercepts the same arc, but with point B on the circumference of the circle. This means

(1)   \begin{equation*} m\angle{AOC} = 2m\angle{B} \end{equation*}

where symbol ‘m’ denotes magnitude of an angle. We are now ready to solve it. Construct point D as a mid-point of segment \overline{AC}. Segment \overline{DO} is an angle bisector, meaning

(2)   \begin{equation*} m\angle{DOC} = \frac{1}{2}m\angle{AOC} =  \frac{1}{2} (2m\angle{B}) = m\angle{B}  \end{equation*}

Therefore, m\angle{DOC} = m\angle{B}

Note that \triangle{DOC} is a right-angled triangle with hypotenuse OC. The length of segment \overline{OC} can be found by noting that:

(3)   \begin{equation*} sin DOC = \frac{\overline{DC}}{\overline{OC}} = \frac{\frac{1}{2}\overline{AC}}{\overline{OC} }\end{equation*}

Solving for OC, we find that \overline{OC} = \frac{\overline{AC}}{2 sin DOC}

However, since m\angle{DOC} = m\angle{B}, this can be written as

\frac{\overline{AC}}{2 sin B} = \overline{OC}, or \frac{\overline{AC}}{sin B} = 2R, where R is the circumradius

An important discovery here is that we could have done this for any side and its opposite angle and we should have gotten the same result for the circumradius. This means that

(4)   \begin{equation*} \frac{\overline{AB}}{sin C} = \frac{\overline{AC}}{sin B} =  \frac{\overline{BC}}{sin A} = 2R \end{equation*}

This is an important relationship, known as Law of Sines, that holds true for any triangle.

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