9. a) find the measure of arc eu if arc eq is 112 degrees and the measure of ∠e is 50 degrees.

b) find the measure of ∠q if arc eq is 112 degrees and the measure of ∠e is 50 degrees.

Question

9. a) find the measure of arc eu if arc eq is 112 degrees and the measure of ∠e is 50 degrees.

b) find the measure of ∠q if arc eq is 112 degrees and the measure of ∠e is 50 degrees.

Accepted Answer

### Part A
# Explanation:
## Step1: Recall Inscribed Angle Theorem
The measure of an inscribed angle is half the measure of its intercepted arc. Also, in a triangle formed by chords in a circle, we can use the relationship between inscribed angles and arcs. Let's denote the measure of arc $ EU $ as $ x $. The inscribed angle $ \angle E $ intercepts arc $ QU $, but we also know about the arc $ EQ = 112^\circ $ and the total around a circle is $ 360^\circ $, but maybe a better approach is using the triangle angle - arc relationship. Wait, actually, for an inscribed angle, if we have a triangle with vertices on the circle, the measure of an angle is half the difference of the intercepted arcs? No, wait, the inscribed angle theorem: the measure of an inscribed angle is half the measure of its intercepted arc. But also, in a triangle, the sum of angles is $ 180^\circ $, and each angle is an inscribed angle intercepting an arc.

Wait, let's re - express. Let's consider the circle with points $ E $, $ Q $, $ U $ on the circumference. Let $ \angle E = 50^\circ $, arc $ EQ=112^\circ $. The inscribed angle $ \angle E $ intercepts arc $ QU $. Wait, no, maybe we can use the fact that the measure of an inscribed angle is half the measure of its intercepted arc. Let's denote arc $ EU $ as $ x $, arc $ QU $ as $ y $. We know that the sum of arcs in a circle is $ 360^\circ $, but maybe we are dealing with a triangle where the angles are inscribed angles.

Wait, another approach: In a circle, the measure of an inscribed angle is half the measure of its intercepted arc. Also, in triangle $ E QU $, the sum of the inscribed angles is $ 180^\circ $. The measure of $ \angle E $ is an inscribed angle intercepting arc $ QU $, so $ m\angle E=\frac{1}{2}m\ arc\ QU $, so $ m\ arc\ QU = 2\times m\angle E=2\times50 = 100^\circ $.

We know that arc $ EQ + arc\ QU+arc\ EU = 360^\circ $? No, that can't be right because the three arcs $ EQ $, $ QU $, and $ EU $ would make up the whole circle, but maybe the triangle is inscribed in a semicircle? Wait, no, the diagram shows a triangle inside a circle. Wait, maybe the arcs $ EQ $, $ EU $, and $ QU $ are the arcs between the points. Wait, let's correct. The inscribed angle $ \angle E $ intercepts arc $ QU $, so $ m\angle E=\frac{1}{2}m(arc\ QU) $, so $ m(arc\ QU) = 2\times50=100^\circ $. The inscribed angle $ \angle Q $ intercepts arc $ EU $, and $ \angle U $ intercepts arc $ EQ $. Since the sum of arcs in a circle is $ 360^\circ $, but if we consider the triangle, the sum of the inscribed angles is $ 180^\circ $, and each angle is half the measure of its intercepted arc. So $ m\angle E + m\angle Q+m\angle U=180^\circ $, and $ m\angle E=\frac{1}{2}m(arc\ QU) $, $ m\angle Q=\frac{1}{2}m(arc\ EU) $, $ m\angle U=\frac{1}{2}m(arc\ EQ) $.

We know $ m(arc\ EQ) = 112^\circ $, so $ m\angle U=\frac{1}{2}\times112 = 56^\circ $. Then, since $ m\angle E + m\angle Q+m\angle U=180^\circ $, and $ m\angle E = 50^\circ $, $ m\angle U = 56^\circ $, then $ m\angle Q=180-(50 + 56)=74^\circ $. But we need to find arc $ EU $. Since $ m\angle Q=\frac{1}{2}m(arc\ EU) $, then $ m(arc\ EU)=2\times m\angle Q $. Wait, but we can also use the fact that the sum of arcs $ EQ + arc\ QU+arc\ EU = 360^\circ $? No, that would be for the whole circle, but maybe the three arcs $ EQ $, $ QU $, and $ EU $ are the only arcs between the three points, so $ arc\ EQ+arc\ QU + arc\ EU=360^\circ $. We know $ arc\ EQ = 112^\circ $, $ arc\ QU = 100^\circ $, so $ arc\ EU=360-(112 + 100)=148^\circ $? No, that can't be, because the sum of angles in the triangle is $ 180^\circ $, and if $ arc\ EU = 148^\circ $, then $ m\angle Q=\frac{1}{2}\times148 = 74^\circ $, and $ 50+74 + 56=180 $, which works. Wait, but maybe the circle is not the whole $ 360^\circ $ for these three arcs? No, the sum of arcs between three non - collinear points on a circle is $ 360^\circ $.

Wait, let's start over.

1. Recall the inscribed angle theorem: The measure of an inscribed angle is half the measure of its intercepted arc.
2. Let $ \angle E = 50^\circ $. $ \angle E $ is an inscribed angle intercepting arc $ QU $, so $ m(arc\ QU)=2\times m(\angle E)=2\times50 = 100^\circ $.
3. We know that $ m(arc\ EQ) = 112^\circ $. Let $ m(arc\ EU)=x $.
4. Since the sum of the measures of arcs $ EQ $, $ QU $, and $ EU $ is equal to the measure of the whole circle ($ 360^\circ $), we have $ m(arc\ EQ)+m(arc\ QU)+m(arc\ EU)=360^\circ $.
5. Substitute the known values: $ 112+100 + x=360 $.
6. Solve for $ x $: $ x=360-(112 + 100)=360 - 212 = 148^\circ $? Wait, that seems too big. Wait, maybe the triangle is inscribed in a semicircle? No, the diagram doesn't suggest that. Wait, maybe I made a mistake in identifying the intercepted arc.

Wait, maybe $ \angle E $ is not intercepting arc $ QU $. Let's look at the diagram (a triangle inside a circle with vertices $ E $, $ Q $, $ U $). The angle at $ E $ is formed by chords $ EQ $ and $ EU $, so it intercepts arc $ QU $. The angle at $ Q $ is formed by chords $ QE $ and $ QU $, so it intercepts arc $ EU $. The angle at $ U $ is formed by chords $ UE $ and $ UQ $, so it intercepts arc $ EQ $.

We know that the sum of the measures of the inscribed angles of a triangle is $ 180^\circ $. So $ m\angle E+m\angle Q + m\angle U=180^\circ $.

We know $ m\angle E = 50^\circ $, and $ m\angle U=\frac{1}{2}m(arc\ EQ)=\frac{1}{2}\times112 = 56^\circ $.

Then $ m\angle Q=180-(50 + 56)=74^\circ $.

Since $ m\angle Q=\frac{1}{2}m(arc\ EU) $, then $ m(arc\ EU)=2\times m\angle Q = 2\times74 = 148^\circ $. Wait, but that's the same as before.

### Part B
# Explanation:
## Step1: Recall Triangle Angle Sum
The sum of the interior angles of a triangle is $ 180^\circ $. Also, we can use the inscribed angle theorem. The measure of an inscribed angle is half the measure of its intercepted arc. The angle at $ U $ ( $ \angle U $) intercepts arc $ EQ $, so $ m\angle U=\frac{1}{2}m(arc\ EQ) $.

## Step2: Calculate $ m\angle U $
Given $ m(arc\ EQ) = 112^\circ $, then $ m\angle U=\frac{1}{2}\times112 = 56^\circ $.

## Step3: Calculate $ m\angle Q $
In triangle $ E QU $, $ m\angle E+m\angle Q + m\angle U=180^\circ $. We know $ m\angle E = 50^\circ $ and $ m\angle U = 56^\circ $. So $ m\angle Q=180-(50 + 56)=74^\circ $.

### Part A Answer:
The measure of arc $ EU $ is $ \boldsymbol{148^\circ} $

### Part B Answer:
The measure of $ \angle Q $ is $ \boldsymbol{74^\circ} $

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QUESTION IMAGE

Question

Explanation:

Part A

Step1: Recall Inscribed Angle Theorem

Step1: Recall Triangle Angle Sum

Step2: Calculate \( m\angle U \)

Step3: Calculate \( m\angle Q \)

Part A Answer:

Part B Answer:

Answer:

Step1: Recall Triangle Angle Sum

Step2: Calculate \( m\angle U \)

Step3: Calculate \( m\angle Q \)

Part A Answer:

Part B Answer: