The law of sines

In every triangle, a side divided by the sine of the angle opposite it equals the diameter of the circle through the triangle’s vertices.

• $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ – the sides; $\displaystyle \alpha$, $\displaystyle \beta$, $\displaystyle \gamma$ – the angles opposite them.
• $\displaystyle R$ – the radius of the circumscribed circle.
• It links sides and angles in every triangle, not only in a right-angled one.

Given: the triangle $\displaystyle ABC$ is inscribed in a circle with centre $\displaystyle O$ and radius $\displaystyle R$, and the angle $\displaystyle {\alpha}={\angle BAC}$ is acute.
To prove: $\displaystyle {\frac{a}{\sin\alpha}}={2R}$.

The proof below draws the diameter $\displaystyle BD$ and shows that the triangle $\displaystyle BCD$ is right-angled and that its angle at $\displaystyle D$ equals $\displaystyle \alpha$:

From here it is trigonometry in the right-angled triangle $\displaystyle BCD$: its hypotenuse is $\displaystyle {BD}={2R}$, and the side opposite the angle at $\displaystyle D$ is $\displaystyle {BC}={a}$.

In the same way, with a diameter through another vertex, $\displaystyle {\frac{b}{\sin\beta}}={2R}$ and $\displaystyle {\frac{c}{\sin\gamma}}={2R}$.

When two angles and a side are known, any other side can be found: take two of the ratios – one with everything known, one with the wanted side.

In the triangle $\displaystyle ABC$, $\displaystyle {\angle B}={40^{\circ}}$, $\displaystyle {\angle C}={75^{\circ}}$ and $\displaystyle {AB}={10}$.
We find $\displaystyle AC$.

• $\displaystyle AC$ is opposite $\displaystyle \angle B$, and $\displaystyle AB$ is opposite $\displaystyle \angle C$:

One side and the angle opposite it are enough to find the radius of the circumscribed circle.

In the triangle $\displaystyle ABC$, $\displaystyle {BC}={12}$ and $\displaystyle {\angle A}={50^{\circ}}$.
We find $\displaystyle R$:

When two sides are known, and the angle opposite one of them, the law of sines gives the sine of the angle opposite the other.

In the triangle $\displaystyle ABC$, $\displaystyle {BC}={8}$, $\displaystyle {AC}={6}$ and $\displaystyle {\angle A}={70^{\circ}}$.
We find $\displaystyle \angle B$.

• $\displaystyle \angle B$ is opposite $\displaystyle {AC}={6}$, shorter than $\displaystyle {BC}={8}$, so it is smaller than $\displaystyle \angle A$ – an acute angle.
• $\displaystyle {\angle B}={\sin^{-1}0.7048\approx44.81^{\circ}}$.

In the triangle $\displaystyle ABC$, $\displaystyle {\angle A}={50^{\circ}}$, $\displaystyle {\angle B}={65^{\circ}}$ and $\displaystyle {BC}={12}$.
Round to two decimal places.

The triangle $\displaystyle ABC$ is inscribed in a circle of radius $\displaystyle 5$.
$\displaystyle {\angle C}={36^{\circ}}$.
Find $\displaystyle AB$, to two decimal places.

In the triangle $\displaystyle ABC$, $\displaystyle {AB}={9}$, $\displaystyle {BC}={7}$ and $\displaystyle {\angle C}={80^{\circ}}$.
Find $\displaystyle \angle A$, to one decimal place.

The triangle $\displaystyle ABC$ is inscribed in a circle with centre $\displaystyle O$.
$\displaystyle OM$ is perpendicular to the side $\displaystyle AB$.
Prove: $\displaystyle {\angle AOM}={\angle ACB}$.

In the triangle of Exercise 4, also:
The distance of the centre from the side $\displaystyle AB$ is $\displaystyle 8$.
$\displaystyle {AC}={18}$ and $\displaystyle {\angle ACB}={72^{\circ}}$.
Round to two decimal places.