\left.\begin{array}{l}{[\text { Inscribed angle is half of the central angle }]} \\ {[\quad \text { Cyclic quadrilaterals }}\end{array}\right]

Points $A, B, C$ and $D$ are sequentially located on a circle. It is known that the degree measures of the smaller arcs $A B, B C, C D$ and $A D$ are in the ratio $1: 3: 5: 6$. Find the angles of the quadrilateral $A B C D$.
Let the degree measure of the smaller arc $AB$ be $x$. Then the degree measures of the consecutive arcs $BC, CD$, and $AD$ are $3x, 5x$, and $6x$ respectively. From the equation $x + 3x + 5x + 6x = 360^{\circ}$, we find that $x = 24^{\circ}$. Therefore, the degree measures of the consecutive arcs $AB, BC, CD$, and $AD$ are $24^{\circ}, 3 \cdot 24^{\circ} = 72^{\circ}, 5 \cdot 24^{\circ} = 120^{\circ}$, and $6 \cdot 24^{\circ} = 144^{\circ}$ respectively.

The degree measure of an inscribed angle is half the degree measure of the arc it subtends, so

$$
\begin{aligned}
& \angle BAD = \frac{1}{2} BCD = \frac{1}{2} (72^{\circ} + 120^{\circ}) = 96^{\circ}, \angle ABC = \frac{1}{2} ADC = \frac{1}{2} (144^{\circ} + 120^{\circ}) = 132^{\circ} \\
& \angle BCD = \frac{1}{2} BAD = \frac{1}{2} (24^{\circ} + 144^{\circ}) = 84^{\circ}, \angle ADC = \frac{1}{2} ABC = \frac{1}{2} (24^{\circ} + 72^{\circ}) = 48^{\circ}.
\end{aligned}
$$

## Answer

$96^{\circ}, 132^{\circ}, 84^{\circ}, 48^{\circ}$