Law of Sines

A chord of a circle equals the product of the diameter and the sine of any inscribed angle subtending that chord.

If chord $MN$ subtends an inscribed angle of measure $\alpha$, and the circle has radius $R$, then:

$MN = 2R\sin\alpha$

The sides of a triangle are proportional to the sines of the opposite angles:

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

where $R$ is the circumradius (radius of the circumscribed circle).

In triangle $ABC$: $AC = \sqrt{2}$ cm, $BC = 1$ cm, $\angle A = 30°$. Find angle $B$.

Solution. By the Law of Sines: $\frac{BC}{\sin A} = \frac{AC}{\sin B} \implies \sin B = \frac{AC \cdot \sin A}{BC} = \frac{\sqrt{2} \cdot \sin 30°}{1} = \frac{\sqrt{2}}{2}$

So $\angle B = 45°$ or $\angle B = 135°$.

Since $BC < AC$, we have $\angle A < \angle B$, so both values are possible: $\angle B = 45°$ or $\angle B = 135°$. $\blacktriangleleft$

In an isosceles trapezoid, the bases are $21$ cm and $9$ cm, and the height is $8$ cm. Find the circumradius.

Solution. The leg $AB = \sqrt{6^2 + 8^2} = 10$ cm and $\sin A = 8/10 = 4/5$.

From the Law of Sines applied to triangle $ABD$: $R = \frac{BD}{2\sin A} = \frac{17}{2 \cdot \frac{4}{5}} = \frac{85}{8} \text{ cm} \qquad \blacktriangleleft$

The distance $d$ between the centers of the inscribed and circumscribed circles of a triangle satisfies:

$d = \sqrt{R^2 - 2Rr}$

where $R$ and $r$ are the circumradius and inradius respectively. In particular, $R \geq 2r$ for any triangle, with equality only for equilateral triangles.