Question

find ( moverarc{ad} ).
( moverarc{ad} = square^circ )

Explanation:

Step1: Recall inscribed quadrilateral rule

In a cyclic quadrilateral, an inscribed angle is half the measure of its intercepted arc. Also, opposite angles sum to $180^\circ$, and the total of all arcs in a circle is $360^\circ$. First, find arc $\widehat{BC}$ using $\angle A$.
$\text{m}\widehat{BC} = 2 \times \text{m}\angle A = 2 \times 99^\circ = 198^\circ$

Step2: Find arc $\widehat{AB}$ using $\angle D$

$\text{m}\widehat{AB} = 2 \times \text{m}\angle D = 2 \times 92^\circ = 184^\circ$

Step3: Calculate sum of known arcs

We know $\text{m}\widehat{CD}=124^\circ$, $\text{m}\widehat{BC}=198^\circ$, $\text{m}\widehat{AB}=184^\circ$. But note that $\text{m}\widehat{AB} + \text{m}\widehat{BC} + \text{m}\widehat{CD} + \text{m}\widehat{AD} = 360^\circ$. Rearrange to solve for $\text{m}\widehat{AD}$:
$\text{m}\widehat{AD} = 360^\circ - (\text{m}\widehat{AB} + \text{m}\widehat{BC} + \text{m}\widehat{CD})$
Substitute values:
$\text{m}\widehat{AD} = 360^\circ - (184^\circ + 198^\circ + 124^\circ)$
First compute the sum inside parentheses: $184+198+124=506$
$\text{m}\widehat{AD} = 360^\circ - 506^\circ$
This is incorrect, so use the correct rule: the measure of an inscribed angle is half the measure of its intercepted arc, so the arc opposite the angle is $2\times$ the angle, but the arc adjacent is $180^\circ - 2\times$ the angle? No, correct: For cyclic quadrilateral, the arc intercepted by $\angle A$ is $\widehat{BCD}$, so $\text{m}\angle A = \frac{1}{2}\text{m}\widehat{BCD}$. $\text{m}\widehat{BCD} = \text{m}\widehat{BC} + \text{m}\widehat{CD} = \text{m}\widehat{BC} + 124^\circ$. So $99^\circ = \frac{1}{2}(\text{m}\widehat{BC} + 124^\circ)$
Solve for $\text{m}\widehat{BC}$:
$198^\circ = \text{m}\widehat{BC} + 124^\circ$
$\text{m}\widehat{BC} = 198^\circ - 124^\circ = 74^\circ$

Step4: Find $\text{m}\widehat{AB}$ using $\angle D$

$\text{m}\angle D = \frac{1}{2}\text{m}\widehat{ABC} = \frac{1}{2}(\text{m}\widehat{AB} + \text{m}\widehat{BC})$
$92^\circ = \frac{1}{2}(\text{m}\widehat{AB} + 74^\circ)$
$184^\circ = \text{m}\widehat{AB} + 74^\circ$
$\text{m}\widehat{AB} = 184^\circ - 74^\circ = 110^\circ$

Step5: Calculate $\text{m}\widehat{AD}$

Total circle is $360^\circ$, so:
$\text{m}\widehat{AD} = 360^\circ - (\text{m}\widehat{AB} + \text{m}\widehat{BC} + \text{m}\widehat{CD})$
$\text{m}\widehat{AD} = 360^\circ - (110^\circ + 74^\circ + 124^\circ)$
$\text{m}\widehat{AD} = 360^\circ - 308^\circ = 52^\circ$

Answer:

$52$