question 4 (multiple choice worth 1 points)
(07.01 mc)
in circle a, the measure of $ \angle bad $ is $ 148 ^ { \circ } $.
if $ m \widehat { b c } $ is $ 102 ^ { \circ } $, what is $ m \widehat { c d } $?

Question

Accepted Answer

# Explanation:
## Step1: Recall the sum of arcs in a circle
The sum of the measures of the arcs in a circle is $360^{\circ}$. Let $m\overset{\frown}{BC} = 102^{\circ}$, $m\overset{\frown}{BAD}=148^{\circ}$. We know that $m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{BAD}=360^{\circ}$ (since the sum of arcs in a circle is $360^{\circ}$).

## Step2: Solve for $m\overset{\frown}{CD}$
Substitute the known values into the equation: $102^{\circ}+m\overset{\frown}{CD}+148^{\circ}=360^{\circ}$.
First, simplify the left - hand side: $102 + 148+m\overset{\frown}{CD}=360$, so $250^{\circ}+m\overset{\frown}{CD}=360^{\circ}$.
Then, subtract $250^{\circ}$ from both sides: $m\overset{\frown}{CD}=360^{\circ}-250^{\circ}$.

# Answer:
$m\overset{\frown}{CD}=110^{\circ}$

Wait, there is a mistake. Let's re - check.

# Explanation:
## Step1: Recall the sum of arcs in a circle
The sum of the measures of the arcs in a circle is $360^{\circ}$. Let $m\overset{\frown}{BAD} = 148^{\circ}$, and assume the circle has arcs $\overset{\frown}{AB}$, $\overset{\frown}{BC}$, $\overset{\frown}{CD}$, $\overset{\frown}{DA}$. But if $\angle BAD$ is a central angle, and we know that $m\overset{\frown}{BAD}$ (the arc from $B$ to $D$ passing through the major arc) is $148^{\circ}$, and $m\overset{\frown}{BC}=102^{\circ}$.

We know that the sum of arcs in a circle $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$. But if we consider the relationship between the given arcs.

Wait, another approach: The sum of arcs in a circle is $360^{\circ}$. If we assume that the arc $\overset{\frown}{BAD}$ is $148^{\circ}$ (a major arc), and we know that $m\overset{\frown}{BC} = 102^{\circ}$.

We use the formula $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$. But if $\angle BAD$ is a central angle, and $m\overset{\frown}{BAD}$ (the arc $B - D$) $=m\overset{\frown}{BC}+m\overset{\frown}{CD}$ (assuming $A$ is the center).

## Step1: Use the formula for the sum of arcs related to the central angle
Since the sum of arcs in a circle is $360^{\circ}$, and if we assume the central - angle - arc relationship. Let $x=m\overset{\frown}{CD}$.
We know that $m\overset{\frown}{BAD}=m\overset{\frown}{BC}+m\overset{\frown}{CD}$ (if $A$ is the center and we consider the arc from $B$ to $D$ passing through $C$). But wait, no, the sum of arcs: $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$. If $\angle BAD$ is a central angle, $m\overset{\frown}{BAD}$ (the arc $B - D$) $=360-(m\overset{\frown}{AB}+m\overset{\frown}{DA})$.

Wait, correct formula: The measure of an arc formed by two adjacent arcs is the sum of the measures of the two arcs. The sum of all arcs in a circle is $360^{\circ}$.

Let $m\overset{\frown}{CD}=x$. We know that $m\overset{\frown}{BAD} = 148^{\circ}$ (central - angle - arc measure) and $m\overset{\frown}{BC}=102^{\circ}$.

We use the fact that $m\overset{\frown}{BAD}+m\overset{\frown}{AB}+m\overset{\frown}{DA}=360^{\circ}$, but this is not helpful.

Another way: The sum of arcs: If we assume that the arcs are $\overset{\frown}{AB}$, $\overset{\frown}{BC}$, $\overset{\frown}{CD}$, $\overset{\frown}{DA}$ and $m\overset{\frown}{BAD}=m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}$ (no, wrong).

Wait, correct: The measure of a full - circle arc is $360^{\circ}$. If we know that $m\overset{\frown}{BAD}$ (the arc from $B$ to $D$ passing through the major arc) $=148^{\circ}$, then the minor arc $m\overset{\frown}{BD}=360 - 148=212^{\circ}$. But if $m\overset{\frown}{BC} = 102^{\circ}$, then $m\overset{\frown}{CD}=m\overset{\frown}{BD}-m\overset{\frown}{BC}$

## Step1: Find the measure of the minor arc $\overset{\frown}{BD}$
Since the sum of the major arc $\overset{\frown}{BAD}$ and the minor arc $\overset{\frown}{BD}$ is $360^{\circ}$. If $m\overset{\frown}{BAD} = 148^{\circ}$, then $m\overset{\frown}{BD}=360 - 148=212^{\circ}$ (this is wrong, major arc $m\overset{\frown}{BAD}$ and minor arc $m\overset{\frown}{BD}$, no, actually if $A$ is the center, $m\overset{\frown}{BAD}$ (central - angle - arc) is $148^{\circ}$, then the remaining arc $m\overset{\frown}{AB}+m\overset{\frown}{DA}=360 - 148=212^{\circ}$.

Wait, no, correct formula: If $A$ is the center of the circle, then $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$. And if $m\overset{\frown}{BAD}=m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}$ (assuming $A$ is the center and we go from $B$ to $D$ passing through $C$) is wrong.

Correct approach: The sum of arcs in a circle is $360^{\circ}$. If we assume that the central angle $\angle BAD = 148^{\circ}$ (so the arc $\overset{\frown}{BAD}$ (the arc from $B$ to $D$ not containing the other part) is $148^{\circ}$). Then $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$, and $m\overset{\frown}{BAD}=m\overset{\frown}{AB}+m\overset{\frown}{DA}$ (if $A$ is the center).

No, another way: The measure of an arc is equal to the measure of its central angle. Let $A$ be the center. Then $m\overset{\frown}{BAD}$ (central - angle - arc) $=148^{\circ}$, $m\overset{\frown}{BC}=102^{\circ}$.

We know that $m\overset{\frown}{AB}+m\overset{\frown}{BC}+m\overset{\frown}{CD}+m\overset{\frown}{DA}=360^{\circ}$. And if we assume that $m\overset{\frown}{BAD}=m\overset{\frown}{AB}+m\overset{\frown}{DA}$ (central - angle - arc).

Let $m\overset{\frown}{CD}=x$. Then $m\overset{\frown}{AB}+m\overset{\frown}{DA}=360-(102 + x)$. But $m\overset{\frown}{BAD}=m\overset{\frown}{AB}+m\overset{\frown}{DA}$ (central - angle - arc). So $148=360-(102 + x)$

## Step1: Set up the equation
$148=360-(102 + x)$

## Step2: Solve the equation
First, simplify the right - hand side: $148=360 - 102-x$.
$148=258 - x$.
Then, add $x$ to both sides: $148+x=258$.
Subtract $148$ from both sides: $x=258 - 148$.

# Answer:
$m\overset{\frown}{CD}=110^{\circ}$

Wait, there is a mistake in the problem's options. Maybe the problem was mis - transcribed. If we assume that the formula is $m\overset{\frown}{CD}=360-(m\overset{\frown}{BAD}+m\overset{\frown}{BC})$ (wrong, but if we follow the wrong logic of adding two arcs and subtracting from $360$): $360-(148 + 102)=110$ (not in the options). But if we assume that the problem has a typo and $\angle BAD$ is an inscribed angle (but the problem says "in circle $A$" implying $A$ is the center, so central angle).

Another approach: If we consider that the sum of arcs: Let the circle have center $A$. The sum of arcs $\overset{\frown}{AB}+\overset{\frown}{BC}+\overset{\frown}{CD}+\overset{\frown}{DA}=360^{\circ}$. If $\angle BAD$ (central angle) $=148^{\circ}$, then $\overset{\frown}{BD}$ (the arc $B - D$) $=148^{\circ}$ (if $A$ is the center, central angle equals arc measure). Then $m\overset{\frown}{CD}=m\overset{\frown}{BD}-m\overset{\frown}{BC}$

## Step1: Use the arc - subtraction formula
Since $m\overset{\frown}{BD}$ (central - angle - arc) $=148^{\circ}$ and $m\overset{\frown}{BC}=102^{\circ}$

## Step2: Calculate $m\overset{\frown}{CD}$
$m\overset{\frown}{CD}=148 - 102$

# Answer:
$m\overset{\frown}{CD}=46^{\circ}$ (Option C)

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the sum of arcs in a circle

Step2: Solve for \(m\overset{\frown}{CD}\)

Step1: Recall the sum of arcs in a circle

Step1: Use the formula for the sum of arcs related to the central angle

Step1: Find the measure of the minor arc \(\overset{\frown}{BD}\)

Answer: