in the circle below, ( overline{uw} ) is a diameter. suppose ( moverarc{vw} = 136^circ ) and ( mangle wvx = 62^circ ). find the following. (a) ( mangle uvx = square^circ ) (b) ( mangle uwv = square^circ )

Question

Accepted Answer

# Explanation:
## Step1: Recall Inscribed Angle Theorem
The measure of an inscribed angle is half the measure of its intercepted arc. Also, a diameter divides the circle into two semicircles (180° arcs).

## Step2: Find $ m\angle UVX $
$\angle UVX$ intercepts arc $UX$. First, find arc $UX$. Since $UW$ is a diameter, $m\overset{\frown}{UW} = 180^\circ$. We know $m\overset{\frown}{VW}=136^\circ$, so $m\overset{\frown}{UV}=m\overset{\frown}{UW}-m\overset{\frown}{VW}=180 - 136 = 44^\circ$. Wait, no, $\angle WVX = 62^\circ$, which is an inscribed angle intercepting arc $WX$? Wait, maybe better: $\angle UVX$ and $\angle WVX$ – wait, let's re-examine. Wait, $\angle WVX = 62^\circ$, which is an inscribed angle. Wait, maybe part (a): $\angle UVX$ – let's see, $UW$ is diameter, so $\angle UVW = 90^\circ$ (since angle inscribed in a semicircle is right angle). Wait, no, maybe I messed up. Wait, the problem says $m\angle WVX = 62^\circ$. Let's do part (b) first maybe.

## Step3: Find $ m\angle UWV $
$\angle UWV$ is an inscribed angle intercepting arc $UV$. We found $m\overset{\frown}{UV}=180 - 136 = 44^\circ$? Wait, no, $UW$ is diameter, so arc $UW = 180^\circ$. Arc $VW = 136^\circ$, so arc $UV = 180 - 136 = 44^\circ$. Then $\angle UWV$ intercepts arc $UV$, so $m\angle UWV=\frac{1}{2}m\overset{\frown}{UV}=\frac{1}{2}\times44 = 22^\circ$? Wait, no, maybe not. Wait, $\angle UWV$: vertex at $W$, sides $WU$ and $WV$. So it intercepts arc $UV$. Yes, so $m\angle UWV=\frac{1}{2}m\overset{\frown}{UV}$. Since $m\overset{\frown}{UV}=180 - 136 = 44^\circ$, so $m\angle UWV = 22^\circ$? Wait, but let's check part (a).

Wait, maybe I made a mistake. Let's start over.

### For part (a): $ m\angle UVX $
$\angle UVX$ intercepts arc $UX$. Wait, $\angle WVX = 62^\circ$ intercepts arc $WX$, so $m\overset{\frown}{WX}=2\times62 = 124^\circ$? No, that can't be, since arc $VW = 136^\circ$, and arc $VW + arc WX + arc XU + arc UV = 360^\circ$, but $UW$ is diameter, so arc $UW = 180^\circ$, so arc $VW + arc WU$? No, $UW$ is diameter, so arc $UW = 180^\circ$, so arc $VW$ (136°) and arc $WU$? No, $UW$ is diameter, so from $U$ to $W$ is 180°, so arc $UV + arc VW = 180^\circ$, so arc $UV = 180 - 136 = 44^\circ$. Then, $\angle WVX = 62^\circ$ – maybe $\angle WVX$ intercepts arc $WX$, so $m\overset{\frown}{WX}=2\times62 = 124^\circ$? But arc $UW$ is 180°, so arc $UX + arc WX = 180^\circ$? Wait, $UW$ is diameter, so arc $UW = 180^\circ$, so arc $UX + arc XW = 180^\circ$? No, $U$ to $W$ is diameter, so the arc from $U$ to $W$ through $X$ is $UX + XW$, and through $V$ is $UV + VW$, both 180°. So $UX + XW = 180$, and $UV + VW = 180$. We know $VW = 136$, so $UV = 44$. We know $\angle WVX = 62$, which is inscribed angle over $WX$, so $WX = 2\times62 = 124$. Then $UX = 180 - 124 = 56$? Wait, now I'm confused. Maybe the problem is:

Wait, the diagram: points $U, V, W, X$ on the circle, $UW$ diameter. $\angle WVX = 62^\circ$, arc $VW = 136^\circ$.

For part (a): $\angle UVX$ – let's see, $\angle UVX$ and $\angle WVX$: maybe $\angle UVX = \angle UVW - \angle WVX$. Since $\angle UVW = 90^\circ$ (inscribed in semicircle), so $90 - 62 = 28^\circ$? No, that doesn't match. Wait, maybe I was wrong about $\angle UVW$. Wait, $UW$ is diameter, so triangle $UVW$ is right-angled at $V$, so $\angle UVW = 90^\circ$. Then $\angle WVX = 62^\circ$, so $\angle UVX = \angle UVW - \angle WVX = 90 - 62 = 28^\circ$? But that seems off. Wait, maybe part (b): $\angle UWV$ intercepts arc $UV$, which is $44^\circ$, so $\angle UWV = 22^\circ$. Wait, let's check with arc $VW = 136^\circ$, so the inscribed angle over arc $VW$ would be $68^\circ$, but $\angle UWV$ is over arc $UV$, which is $44^\circ$, so $22^\circ$. Maybe part (a) is $62^\circ$? No, the problem says $m\angle WVX = 62^\circ$, maybe $\angle UVX$ is equal to $\angle WVX$? No, that doesn't make sense. Wait, maybe I misread the problem.

Wait, the problem says: "Suppose $m\overset{\frown}{VW} = 136^\circ$ and $m\angle WVX = 62^\circ$. Find the following."

(a) $m\angle UVX$

(b) $m\angle UWV$

Let's solve (b) first:

$\angle UWV$ is an inscribed angle intercepting arc $UV$. Since $UW$ is a diameter, $m\overset{\frown}{UW} = 180^\circ$. $m\overset{\frown}{VW} = 136^\circ$, so $m\overset{\frown}{UV} = m\overset{\frown}{UW} - m\overset{\frown}{VW} = 180 - 136 = 44^\circ$. Then, by the inscribed angle theorem, $m\angle UWV = \frac{1}{2}m\overset{\frown}{UV} = \frac{1}{2} \times 44 = 22^\circ$.

For (a): $\angle UVX$ and $\angle WVX$ – wait, $\angle WVX = 62^\circ$ intercepts arc $WX$, so $m\overset{\frown}{WX} = 2 \times 62 = 124^\circ$. But $m\overset{\frown}{UW} = 180^\circ$, so $m\overset{\frown}{UX} = m\overset{\frown}{UW} - m\overset{\frown}{WX} = 180 - 124 = 56^\circ$? No, that contradicts. Wait, maybe $\angle UVX$ intercepts arc $UX$, and $\angle WVX$ intercepts arc $WX$, and since $UW$ is diameter, $arc UX + arc WX = 180^\circ$. So $m\overset{\frown}{UX} = 180 - 124 = 56^\circ$, then $m\angle UVX = \frac{1}{2}m\overset{\frown}{UX} = 28^\circ$? But earlier we had $m\overset{\frown}{UV} = 44^\circ$, which would mean $arc UV + arc VW = 44 + 136 = 180$, correct (since $UW$ is diameter). Then $arc WX + arc XU = 180$, so if $arc WX = 124$, then $arc XU = 56$, and $arc UV = 44$, $arc VW = 136$, total $44 + 136 + 124 + 56 = 360$, correct. Then $\angle UVX$ intercepts $arc UX = 56^\circ$, so $m\angle UVX = 28^\circ$? But the problem has $\angle WVX = 62^\circ$ intercepting $arc WX = 124^\circ$, which is correct ($62 \times 2 = 124$). Then $\angle UVX$: let's see, $\angle UVW = 90^\circ$ (right angle), $\angle WVX = 62^\circ$, so $\angle UVX = 90 - 62 = 28^\circ$. Yes, that matches. So:

(a) $m\angle UVX = 28^\circ$? Wait, no, wait: $\angle UVW$ is 90°, $\angle WVX$ is 62°, so $\angle UVX = \angle UVW - \angle WVX = 90 - 62 = 28^\circ$.

(b) $m\angle UWV$: intercepts $arc UV = 44^\circ$, so $44/2 = 22^\circ$.

Wait, but maybe I made a mistake. Let's confirm:

- $UW$ is diameter, so $\angle UVW = 90^\circ$ (inscribed angle in semicircle). Correct.

- $m\overset{\frown}{VW} = 136^\circ$, so the inscribed angle over $VW$ would be $68^\circ$, but $\angle UWV$ is over $UV$, which is $180 - 136 = 44^\circ$, so $22^\circ$. Correct.

- $\angle WVX = 62^\circ$, which is over $WX$, so $WX = 124^\circ$, so $UX = 180 - 124 = 56^\circ$, but $\angle UVX$ is over $UX$? No, $\angle UVX$ has vertex at $V$, sides $UV$ and $VX$, so it intercepts $UX$. Wait, but $\angle UVW$ is 90°, $\angle WVX$ is 62°, so $\angle UVX = 90 - 62 = 28^\circ$, which would mean $UX = 56^\circ$ (since $28 \times 2 = 56$), which matches $UX = 180 - 124 = 56$. So that's correct.

So:

(a) $m\angle UVX = 28^\circ$? Wait, no, wait the problem's $\angle WVX = 62^\circ$, maybe $\angle UVX$ is equal to $\angle WVX$? No, that doesn't fit. Wait, maybe I messed up the arcs. Let's try again.

Alternative approach for (a):

$\angle UVX$ and $\angle UWV$ – no, $\angle UVX$ is an inscribed angle. Wait, maybe the problem is that $\angle WVX = 62^\circ$ is equal to $\angle UVX$? No, that doesn't make sense. Wait, maybe the diagram shows that $VX$ is parallel to something, but no, it's a circle.

Wait, maybe the answer for (a) is 62°? No, that can't be. Wait, let's check the problem again. The user provided the diagram: points $U, V, W, X$ on the circle, $UW$ diameter, $V$ connected to $U, W, X$, $X$ connected to $U$. So $\angle WVX = 62^\circ$, $\angle UVX$ – maybe they are equal? No, unless arcs are equal. Wait, maybe I made a mistake in arc $UV$. Wait, $UW$ is diameter, so arc $UW = 180^\circ$, arc $VW = 136^\circ$, so arc $UV = 180 - 136 = 44^\circ$, correct. Then $\angle UWV$ is 22°, correct. Then $\angle UVX$: since $\angle WVX = 62^\circ$, and $\angle UVW = 90^\circ$, then $\angle UVX = 90 - 62 = 28^\circ$, correct.

# Answer:
(a) $ \boldsymbol{28^\circ} $
(b) $ \boldsymbol{22^\circ} $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Inscribed Angle Theorem

Step2: Find \( m\angle UVX \)

Step3: Find \( m\angle UWV \)

For part (a): \( m\angle UVX \)

Answer: