The image contains multiple geometry problems involving circles and angles. It is not possible to extract a single question.

• $\angle CAD = 40^\circ$, so $\angle CBD = 40^\circ$ (angles subtended by arc CD).
• $\angle ACB = 60^\circ$, so $\angle ADB = 60^\circ$ (angles subtended by arc AB).
• $\angle ABD = 10^\circ$, so $\angle ACD = 10^\circ$ (angles subtended by arc AD).

Let P be the intersection of the diagonals. $q$ is $\angle APD$.
In $\triangle APD$:
$\angle PAD = \angle CAD = 40^\circ$.
$\angle PDA = \angle ADB = 60^\circ$.
The sum of angles in a triangle is $180^\circ$.

$r$ is $\angle BCD$.

$s$ is $\angle ADC$. In a cyclic quadrilateral, opposite angles sum to $180^\circ$.
First, find $\angle ABC$:

Now, find $s$:

The values are: $\boxed{q = 80^\circ, r = 70^\circ, s = 130^\circ}$

:
We are given a circle with center O and a cyclic quadrilateral. We need to find $g$ and $h$.

The angle $30^\circ$ is $\angle OQR$. Since O is the center, OQ and OR are radii, making $\triangle OQR$ an isosceles triangle.
Therefore, $\angle ORQ = \angle OQR = 30^\circ$.
The sum of angles in $\triangle OQR$ is $180^\circ$.

$g$ is $\angle QSR$. This is an angle at the circumference subtended by arc QR.
$\angle QOR$ is the central angle subtended by the same arc QR.
The angle at the circumference is half the angle at the center.

The $126^\circ$ angle is $\angle QPS$, which is consistent with the other angles but not needed for finding $g$ and $h$.
The values are: $\boxed{g = 60^\circ, h = 120^\circ}$

:
We are given a cyclic quadrilateral WXYZ with equal chords $WX = XY = YZ$. We need to find $\angle WYX$ and $\angle YWZ$.

Given $\angle WZY = 42^\circ$. This angle subtends arc WXY.
The measure of the arc is twice the angle at the circumference.

Since chords $WX = XY = YZ$ are equal, the arcs they subtend are also equal.
Let $\text{arc WX} = \text{arc XY} = \text{arc YZ} = x_{arc}$.
Then $\text{arc WXY} = \text{arc WX} + \text{arc XY} = x_{arc} + x_{arc} = 2x_{arc}$.

So, $\text{arc WX} = 42^\circ$, $\text{arc XY} = 42^\circ$, and $\text{arc YZ} = 42^\circ$.

$\angle WYX$ subtends arc WX.

$\angle YWZ$ subtends arc YZ.

The values are: $\boxed{\text{i) } \angle WYX = 21^\circ, \text{ii) } \angle YWZ = 21^\circ}$

:
We are given a cyclic quadrilateral ABCD with equal chords $AB = BC = CD$. We need to find $\angle ACD$ and $\angle ADC$.

Since chords $AB = BC = CD$ are equal, they subtend equal angles at the circumference.
Let $\angle ADB = \angle BDC = \angle CAD = \alpha$.
Also, $\angle BCA$ subtends arc AB, so $\angle BCA = \alpha$.
And $\angle CBD$ subtends arc CD, so $\angle CBD = \alpha$.

Given $\angle BCD = p^\circ$.
We know $\angle BCD = \angle BCA + \angle ACD$.
Also, consider $\triangle BCD$. The angles are $\angle CBD$, $\angle BDC$, and $\angle BCD$.

We know $\angle BCD = \angle BCA + \angle ACD$.

Substitute the expression for $\alpha$:

$\angle ADC$ subtends arc ABC.
Arc ABC = arc AB + arc BC.
The angle subtended by arc AB at the circumference is $\alpha$. So $\text{arc AB} = 2\alpha$.
The angle subtended by arc BC at the circumference is $\alpha$. So $\text{arc BC} = 2\alpha$.
Thus, $\text{arc ABC} = 2\alpha + 2\alpha = 4\alpha$.
The angle at the circumference subtended by arc ABC is $\angle ADC$.

Substitute the expression for $\alpha$:

The values are: $\boxed{\text{i) } \angle ACD = \frac{3p^\circ}{2} - 90^\circ, \text{ii) } \angle ADC = 180^\circ - p^\circ}$