Question

in the circle below, suppose m\\(\overarc{wxu}\\) = 114° and m\\(\angle xwv\\) = 77°. find the following. (a) m\\(\angle wxu\\) = \\(\square\\)° (b) m\\(\angle xuv\\) = \\(\square\\)°

Explanation:

Step1: Recall Inscribed Angle Theorem

The measure of an inscribed angle is half the measure of its intercepted arc. For \(\angle WXV\) (wait, let's correct: the angle \(\angle WXV\) intercepts arc \(WV\)? Wait, no, the problem has \(\angle XWV = 77^\circ\) and arc \(WXU = 114^\circ\). Wait, first, let's handle part (a): \(m\angle WXV\). Wait, \(\angle WXV\) is an inscribed angle intercepting arc \(WV\)? Wait, no, arc \(WXU\) is \(114^\circ\), so the remaining arc \(WVU\)? Wait, the total circumference is \(360^\circ\), but maybe we need to find the arc intercepted by the angle. Wait, actually, \(\angle XWV\) is an inscribed angle. Wait, let's re-express:

Wait, the problem says "in the circle below, suppose \(m\overarc{WXU} = 114^\circ\) and \(m\angle XWV = 77^\circ\). Find the following: (a) \(m\angle WXV\), (b) \(m\angle XVU\)".

First, for part (a): \(\angle XWV\) is an inscribed angle intercepting arc \(XV\)? Wait, no, \(\angle XWV\) is at point \(W\), with sides \(WX\) and \(WV\), so it intercepts arc \(XV\). Wait, the measure of an inscribed angle is half the measure of its intercepted arc. So \(m\angle XWV=\frac{1}{2}m\overarc{XV}\). So \(77^\circ=\frac{1}{2}m\overarc{XV}\), so \(m\overarc{XV}=154^\circ\). But arc \(WXU\) is \(114^\circ\), so arc \(WX\) + arc \(XU\) = \(114^\circ\). Wait, maybe arc \(WXU\) is a minor arc? Wait, no, maybe we need to find arc \(WV\). Wait, total circle is \(360^\circ\), but maybe the quadrilateral \(WXUV\) is cyclic, so opposite angles sum to \(180^\circ\)? Wait, no, let's think again.

Wait, \(\angle WXV\) intercepts arc \(WV\). The arc \(WXU\) is \(114^\circ\), so the arc \(WV\) can be found? Wait, no, \(\angle XWV = 77^\circ\) is an inscribed angle intercepting arc \(XV\), so \(m\overarc{XV}=2\times77^\circ = 154^\circ\). Then arc \(WXU = 114^\circ\), so arc \(WX\) + arc \(XU = 114^\circ\). Then arc \(WV\) is \(360^\circ - arc\ WXU - arc\ XV\)? Wait, no, that can't be. Wait, maybe the circle is divided into arcs: \(WXU\) is \(114^\circ\), so the arc from \(W\) to \(X\) to \(U\) is \(114^\circ\). Then the arc from \(U\) to \(V\) to \(W\) would be \(360 - 114 = 246^\circ\), but that's too big. Wait, maybe \(WXU\) is a minor arc, so \(114^\circ\), and \(\angle XWV = 77^\circ\) is an inscribed angle. Wait, perhaps \(\angle WXV\) is an inscribed angle intercepting arc \(WV\), and arc \(WXU\) is \(114^\circ\), so arc \(WV\) is \(180^\circ - 114^\circ\)? No, that doesn't make sense. Wait, maybe the key is that \(\angle WXV\) and \(\angle WVU\) are related, but let's do part (a) first.

Wait, the inscribed angle theorem: \(m\angle WXV=\frac{1}{2}m\overarc{WV}\). And we know that arc \(WXU = 114^\circ\), so the arc \(WV\) can be found as \(360^\circ - arc\ WXU - arc\ XV\). But we found arc \(XV = 2\times77^\circ = 154^\circ\) (from \(\angle XWV = 77^\circ\) intercepting arc \(XV\)). So arc \(WV = 360^\circ - 114^\circ - 154^\circ = 92^\circ\)? No, that can't be. Wait, maybe I made a mistake. Let's start over.

Wait, \(\angle XWV\) is an inscribed angle intercepting arc \(XV\), so \(m\angle XWV = \frac{1}{2}m\overarc{XV}\) ⇒ \(m\overarc{XV} = 2 \times 77^\circ = 154^\circ\). Arc \(WXU\) is \(114^\circ\), so arc \(WX + XU = 114^\circ\). Then arc \(WV\) is \(360^\circ - arc\ WXU - arc\ XV = 360 - 114 - 154 = 92^\circ\). Then \(\angle WXV\) intercepts arc \(WV\), so \(m\angle WXV = \frac{1}{2}m\overarc{WV} = \frac{1}{2} \times 92^\circ = 46^\circ\)? Wait, no, that doesn't match. Wait, maybe arc \(WXU\) is a major arc? No, \(114^\circ\) is minor. Wait, maybe the quadrilateral \(WXUV\) is cyclic, so \(\angle WXV + \angle WVU = 180^\circ\)? No, that's for cyclic quadrilaterals, but opposite angles. Wait, no, \(\angle WXV\) and \(\angle WVU\) are not opposite. Wait, maybe I messed up the intercepted arc.

Wait, let's look at \(\angle WXV\): vertex at \(X\), sides \(XW\) and \(XV\), so it intercepts arc \(WV\). So \(m\angle WXV = \frac{1}{2}m\overarc{WV}\). Now, arc \(WXU\) is \(114^\circ\), so arc \(WX + arc\ XU = 114^\circ\). Arc \(XV\) is intercepted by \(\angle XWV\), so \(m\overarc{XV} = 2 \times 77^\circ = 154^\circ\). Then arc \(WV = arc\ WX + arc\ XV + arc\ VU\)? No, that's not right. Wait, the total circle is \(360^\circ\), so arc \(WXU + arc\ UVW = 360^\circ\). So arc \(UVW = 360 - 114 = 246^\circ\). But arc \(UVW\) is arc \(UV + arc\ VW\). Wait, this is getting confusing. Maybe another approach: in a circle, the measure of an inscribed angle is half the measure of its intercepted arc. So \(\angle XWV\) intercepts arc \(XV\), so \(m\overarc{XV} = 2 \times 77 = 154^\circ\). Then arc \(WXU\) is \(114^\circ\), so the arc from \(W\) to \(X\) to \(U\) is \(114^\circ\), so the arc from \(U\) to \(V\) to \(W\) is \(360 - 114 = 246^\circ\). But arc \(XV\) is \(154^\circ\), so arc \(XU + arc\ UV = 154^\circ\)? No, arc \(XV\) is from \(X\) to \(V\), so arc \(XU + arc\ UV = arc\ XV\). Wait, arc \(WXU\) is from \(W\) to \(X\) to \(U\), so arc \(WX + arc\ XU = 114^\circ\). Then arc \(WV\) is arc \(WX + arc\ XV\)? No, arc \(WV\) is from \(W\) to \(V\), so arc \(WX + arc\ XV = arc\ WV\)? No, that would be if \(X\) is between \(W\) and \(V\), but \(X\) is another point. Wait, maybe the problem has a typo, but let's assume that arc \(WXU\) is \(114^\circ\), so the inscribed angle over arc \(WXU\) would be half, but \(\angle WXV\) is related to the remaining arc. Wait, maybe \(\angle WXV\) is an inscribed angle intercepting arc \(WV\), and arc \(WV\) is \(180^\circ - 114^\circ = 66^\circ\)? No, that's not. Wait, let's check the sum of angles in triangle \(WXV\): \(\angle XWV = 77^\circ\), \(\angle WXV =?\), \(\angle XVW =?\) No, it's a circle, so maybe \(\angle WXV\) is half of arc \(WV\), and arc \(WV\) is \(360 - 114 - 154 = 92\), so \(46\). But let's do part (b): \(m\angle XVU\). \(\angle XVU\) is an inscribed angle intercepting arc \(XW\). Wait, arc \(WXU\) is \(114^\circ\), so arc \(XW\) is part of that. Wait, \(\angle XVU\) intercepts arc \(XW\), so \(m\angle XVU = \frac{1}{2}m\overarc{XW}\). But arc \(WXU = arc\ WX + arc\ XU = 114^\circ\), so if we can find arc \(XW\), but maybe \(\angle XVU\) is related to arc \(WXU\). Wait, no, let's go back to part (a):

Wait, the key is that \(\angle WXV\) and \(\angle XWV\) are angles in triangle \(WXV\), but it's inscribed in a circle. Wait, another approach: the measure of an inscribed angle is half the measure of its intercepted arc. So \(\angle WXV\) intercepts arc \(WV\), and \(\angle XWV\) intercepts arc \(XV\). We know \(m\overarc{WXU} = 114^\circ\), so the arc \(WV\) can be found as \(360^\circ - 114^\circ - m\overarc{XV}\). But \(m\overarc{XV} = 2 \times 77^\circ = 154^\circ\) (since \(\angle XWV = 77^\circ\) is inscribed angle over arc \(XV\)). So \(m\overarc{WV} = 360 - 114 - 154 = 92^\circ\). Then \(m\angle WXV = \frac{1}{2} \times 92^\circ = 46^\circ\). That makes sense.

For part (b): \(\angle XVU\) is an inscribed angle intercepting arc \(XW\). Arc \(WXU\) is \(114^\circ\), which is arc \(WX + arc\ XU\). Wait, \(\angle XVU\) intercepts arc \(XW\), so \(m\angle XVU = \frac{1}{2}m\overarc{XW}\). But arc \(WXU = 114^\circ\), so if arc \(XW\) is part of that, but maybe arc \(XW\) is \(114^\circ - arc\ XU\)? No, wait, \(\angle XVU\) is at point \(U\), with sides \(UX\) and \(UV\), so it intercepts arc \(XV\)? No, \(\angle XVU\) intercepts arc \(XW\). Wait, no, \(\angle XVU\) intercepts arc \(XW\), so \(m\angle XVU = \frac{1}{2}m\overarc{XW}\). But we know that arc \(WXU = 114^\circ\), so arc \(XW\) is part of that. Wait, maybe \(\angle XVU\) is equal to half of arc \(WXU\)? No, that would be \(57^\circ\), since \(114^\circ / 2 = 57^\circ\). Ah! Wait, \(\angle XVU\) intercepts arc \(WX\)? No, arc \(WXU\) is from \(W\) to \(X\) to \(U\), so the inscribed angle intercepting arc \(WXU\) would be at point \(V\), so \(\angle XVU\) intercepts arc \(WX\)? No, wait, \(\angle XVU\) is at \(U\), so it intercepts arc \(XV\)? No, I'm getting confused. Wait, let's use the cyclic quadrilateral property: in a cyclic quadrilateral, the measure of an angle is half the measure of its intercepted arc. Wait, \(\angle XVU\) intercepts arc \(XW\), and arc \(WXU = 114^\circ\), so arc \(XW\) is part of that. Wait, maybe \(\angle XVU\) is half of arc \(WXU\), so \(114^\circ / 2 = 57^\circ\). Let's check:

For part (a): \(m\angle WXV = \frac{1}{2}(360^\circ - 114^\circ - 2 \times 77^\circ) = \frac{1}{2}(360 - 114 - 154) = \frac{1}{2}(92) = 46^\circ\).

For part (b): \(m\angle XVU = \frac{1}{2}m\overarc{WXU} = \frac{1}{2} \times 114^\circ = 57^\circ\). Yes, that makes sense because \(\angle XVU\) intercepts arc \(WXU\).

Step1: Solve for \(m\angle WXV\)

- The inscribed angle theorem states \(m\angle = \frac{1}{2}m\overarc{\text{intercepted}}\).
- \(\angle XWV = 77^\circ\) intercepts \(\overarc{XV}\), so \(m\overarc{XV} = 2 \times 77^\circ = 154^\circ\).
- Total circle: \(360^\circ\), so \(m\overarc{WV} = 360^\circ - m\overarc{WXU} - m\overarc{XV} = 360^\circ - 114^\circ - 154^\circ = 92^\circ\).
- \(\angle WXV\) intercepts \(\overarc{WV}\), so \(m\angle WXV = \frac{1}{2} \times 92^\circ = 46^\circ\).

Step2: Solve for \(m\angle XVU\)

- \(\angle XVU\) intercepts \(\overarc{WXU}\) (inscribed angle theorem).
- \(m\angle XVU = \frac{1}{2} \times m\overarc{WXU} = \frac{1}{2} \times 114^\circ = 57^\circ\).

Answer:

(a) \(46^\circ\)
(b) \(57^\circ\)