what is the measure of $\widehat{au}$? $\bigcirc$ $44^{\circ}$ $\bigcirc$ $50^{\circ}$ $\bigcirc$ $64^{\circ}$ $\bigcirc$ $92^{\circ}$ (with a circle diagram having points q, u, a, d; arc qu is 88°, angle at u is 111°)

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Accepted Answer

# Explanation:
## Step1: Recall the property of a cyclic quadrilateral (or the total degrees in a circle)
The sum of the measures of the arcs in a circle is $360^\circ$. Also, in a cyclic quadrilateral, the sum of opposite angles is $180^\circ$, but here we can use the fact that the sum of all arcs should be $360^\circ$. Wait, actually, the inscribed angle or the arc measures: Wait, the given angle at $U$ is $111^\circ$, and the arc $QU$ is $88^\circ$. Wait, maybe we can use the property that the sum of the measures of the arcs corresponding to the angles in a cyclic quadrilateral? Wait, no, let's think again.

Wait, the total circumference (in terms of degrees) is $360^\circ$. Let's denote the arc $AU$ as $x$, arc $AD$ as something, but maybe we can use the fact that in a cyclic quadrilateral, the sum of the measures of the arcs between the vertices should add up to $360^\circ$. Wait, the arc $QU$ is $88^\circ$, the angle at $U$ is $111^\circ$, but maybe we need to find the arc $AU$. Wait, another approach: The sum of the measures of the arcs in a circle is $360^\circ$. Let's assume that the quadrilateral $QUAD$ is cyclic (since all points are on the circle). Then, the measure of an inscribed angle is half the measure of its intercepted arc. Wait, the angle at $U$ is $111^\circ$, which is an inscribed angle? Wait, no, the angle at $U$ is formed by chords $UQ$ and $UA$, so the inscribed angle at $U$ intercepts arc $QA$. Wait, maybe I made a mistake. Let's try to find the arc $AU$.

Wait, the arc $QU$ is $88^\circ$. Let's denote arc $AU = x$, arc $AD = y$, arc $DQ = z$. Then $88 + x + y + z = 360$. Also, the angle at $U$ is $111^\circ$, which is an inscribed angle intercepting arc $QD$ (since in a cyclic quadrilateral, the angle is half the measure of the opposite arc). Wait, in a cyclic quadrilateral, the measure of an angle is half the measure of the arc opposite to it. So angle at $U$ ($111^\circ$) intercepts arc $QD$, so arc $QD = 2 \times 111^\circ = 222^\circ$? Wait, no, that can't be, because $222 + 88 = 310$, and then $x + y = 50$, but that doesn't match. Wait, maybe the angle at $U$ is not an inscribed angle but a central angle? No, the center is marked, but the angle at $U$ is a vertex of the quadrilateral on the circle. Wait, maybe I messed up. Let's try again.

Wait, the arc $QU$ is $88^\circ$. Let's find the arc opposite to the angle at $U$. In a cyclic quadrilateral, the sum of a pair of opposite angles is $180^\circ$. So angle at $U$ is $111^\circ$, so the angle at $D$ should be $180 - 111 = 69^\circ$. But maybe that's not helpful. Wait, the total of all arcs is $360^\circ$. Let's assume that the arc $QU$ is $88^\circ$, and we need to find arc $AU$. Wait, maybe the angle at $U$ is related to the arc $QD$. Wait, the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. So if angle at $U$ is $111^\circ$, then the arc $QD$ (which is opposite to angle $U$) is $2 \times 111 = 222^\circ$. Then, the remaining arcs $QU$ and $AU + AD$? Wait, no, arc $QU$ is $88^\circ$, arc $QD$ is $222^\circ$, so the remaining arcs $AU + AD$ would be $360 - 88 - 222 = 50^\circ$? No, that doesn't make sense. Wait, maybe the arc $AU$ is $64^\circ$? Wait, no, let's check the answer options. The options are $44^\circ$, $50^\circ$, $64^\circ$, $92^\circ$. Wait, maybe I made a mistake in the approach. Let's try another way.

Wait, the sum of the measures of the arcs in a circle is $360^\circ$. Let's consider that the angle at $U$ is $111^\circ$, which is an inscribed angle, so the arc it intercepts is $2 \times 111 = 222^\circ$ (arc $QD$). Then, arc $QU$ is $88^\circ$, so the arc $AU$ and arc $AD$ plus arc $QU$ plus arc $QD$ should be $360^\circ$. Wait, arc $QD = 222^\circ$, arc $QU = 88^\circ$, so arc $AU + arc AD = 360 - 222 - 88 = 50^\circ$? No, that's not right. Wait, maybe the angle at $U$ is not intercepting arc $QD$ but another arc. Wait, maybe the angle at $U$ is formed by chords $UQ$ and $UA$, so the inscribed angle at $U$ intercepts arc $QA$. Then, the measure of angle $U$ is half the measure of arc $QA$. So $111^\circ = \frac{1}{2} \times arc QA$, so arc $QA = 222^\circ$. But arc $QA$ is arc $QU + arc UA = 88^\circ + x$ (where $x$ is arc $AU$). So $88 + x = 222$, so $x = 222 - 88 = 134^\circ$, which is not an option. So that's wrong.

Wait, maybe the angle at $U$ is a central angle? No, the center is a dot, not at $U$. Wait, maybe the problem is using the fact that the sum of the arcs is $360^\circ$ and the given angle is related to the arc. Wait, the arc $QU$ is $88^\circ$, so the inscribed angle over arc $QU$ would be $44^\circ$, but that's not helpful. Wait, the answer options include $64^\circ$, $50^\circ$, etc. Wait, maybe I made a mistake in the initial assumption. Let's try to calculate the arc $AU$ as follows:

The total degrees in a circle: $360^\circ$. Let's assume that the quadrilateral has arcs: $QU = 88^\circ$, $AU = x$, $AD = y$, $DQ = z$. Then $88 + x + y + z = 360$. Also, in a cyclic quadrilateral, the sum of opposite angles is $180^\circ$. The angle at $U$ is $111^\circ$, so the angle at $D$ is $180 - 111 = 69^\circ$. The angle at $D$ is an inscribed angle intercepting arc $QUA$ (arc $QU + arc UA = 88 + x$), so the measure of angle $D$ is half the measure of arc $QUA$. So $69^\circ = \frac{1}{2} \times (88 + x)$. Then, $88 + x = 138$, so $x = 138 - 88 = 50^\circ$. Ah, that works!

## Step2: Solve for arc $AU$
We know that in a cyclic quadrilateral, the sum of opposite angles is $180^\circ$. So angle at $U$ ($111^\circ$) and angle at $D$ are supplementary:
$ \text{Angle at } D = 180^\circ - 111^\circ = 69^\circ $.

The angle at $D$ is an inscribed angle intercepting arc $QUA$ (arc $QU + arc AU$). By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc:
$ \text{Angle at } D = \frac{1}{2} \times (\text{arc } QU + \text{arc } AU) $.

Substitute the known values:
$ 69^\circ = \frac{1}{2} \times (88^\circ + \text{arc } AU) $.

Multiply both sides by 2:
$ 138^\circ = 88^\circ + \text{arc } AU $.

Subtract $88^\circ$ from both sides:
$ \text{arc } AU = 138^\circ - 88^\circ = 50^\circ $.

# Answer:
$50^\circ$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the property of a cyclic quadrilateral (or the total degrees in a circle)

Answer: