\begin{aligned} & {[\text { Law of Cosines }]} \\ & {[\text { Law of Sines }]}\end{aligned}

Find the radius of the circle if an inscribed angle with sides of lengths 1 and 2 subtends an arc of $120^{\circ}$.
Apply the Law of Cosines.

## Solution

Let $A$ be the vertex of the given angle, $AC$ and $AB$ be the sides of the angle, and points $B$ and $C$ lie on the circle, with $AC=1$ and $AB=2$. By the inscribed angle theorem, $\angle BAC = \frac{1}{2} \cdot 120^{\circ} = 60^{\circ}$. By the Law of Cosines,

$$
BC = \sqrt{AC^2 + AB^2 - 2 \cdot AC \cdot AB \cdot \cos \angle BAC} = \sqrt{1 + 4 - 2 \cdot 1 \cdot 2 \cdot \cos 60^{\circ}} = \sqrt{5 - 2} = \sqrt{3}
$$

If $R$ is the desired radius, then

$$
R = \frac{BC}{2 \cdot \sin 60^{\circ}} = \frac{\sqrt{3}}{\sqrt{3}} = 1
$$

## Answer

$R=1$.

A obtuse angle with sides of lengths 3 and 6 is inscribed in a circle of radius $\sqrt{21}$. Determine the measure of the arc on which it subtends.

## Answer

$240^{\circ}$.