Question

what is m∠uvw?
diagram: right angles at w (xw ⊥ wv) and u (xu ⊥ uv), segments xw and xu marked congruent, angle ∠xvw = 33°, m∠uvw = ☐°

Explanation:

Step1: Identify the figure type

The figure has two right angles (at \( W \) and \( U \)) and two equal - length segments (implies some congruency or symmetry). The triangles \( XVW \) and \( XVU \) seem to be related. Also, we know that \( \angle WVX = 33^{\circ} \).

Step2: Analyze the angle \( \angle UVW \)

Since the figure has right angles at \( W \) and \( U \), and the segments \( XW \) and \( XU \) are equal (marked with the same tick), the triangles \( XWV \) and \( XUV \) are congruent (by HL - Hypotenuse - Leg theorem, as \( XV \) is common hypotenuse and \( XW = XU \), \( \angle W=\angle U = 90^{\circ} \)). So the angle \( \angle UVX=\angle WVX = 33^{\circ} \).

To find \( m\angle UVW \), we note that \( \angle UVW=\angle UVX+\angle WVX \) (since \( \angle UVW \) is composed of \( \angle UVX \) and \( \angle WVX \)). Wait, actually, since \( \angle W \) and \( \angle U \) are right angles, and the triangles are congruent, \( \angle UVW = 90^{\circ}+ 33^{\circ}\)? No, wait, let's re - examine.

Wait, the right angles are at \( W \) (between \( XW \) and \( WV \)) and at \( U \) (between \( XU \) and \( UV \)). The segment \( XV \) is the angle - bisector? Wait, no. Let's think about the sum of angles.

Actually, since \( XW\perp WV \) and \( XU\perp UV \), and \( XW = XU \), \( XV = XV \), so \( \triangle XWV\cong\triangle XUV \) (HL). Then \( \angle VXW=\angle VXU \), and \( \angle UVX=\angle WVX = 33^{\circ} \).

Now, \( \angle UVW \) is a right angle plus \( 33^{\circ} \)? No, wait, \( \angle W \) is \( 90^{\circ} \), and we have \( \angle WVX = 33^{\circ} \), and since the two right - angled triangles are congruent, \( \angle UVX = 33^{\circ} \). So \( \angle UVW=\angle U VW=90^{\circ}+ 33^{\circ} \)? No, that's not right. Wait, actually, \( \angle UVW \) is made up of \( \angle UVX \) and \( \angle XVW \)? No, let's look at the diagram again.

Wait, the angle at \( V \), \( \angle UVW \): we know that \( \angle WVX = 33^{\circ} \), and since \( \triangle XWV\) and \( \triangle XUV \) are congruent, \( \angle UVX=\angle WVX = 33^{\circ} \). So \( \angle UVW=\angle UVX+\angle WVX=33^{\circ}+33^{\circ}+ 90^{\circ}- 90^{\circ} \)? No, I think I made a mistake.

Wait, another approach: The figure is a quadrilateral? No, it's two right - angled triangles sharing the hypotenuse \( XV \). The angle at \( V \), \( \angle UVW \): since \( \angle W = 90^{\circ} \), \( \angle U = 90^{\circ} \), and \( XW = XU \), \( XV \) is common. So the triangles are congruent, so \( \angle UVX=\angle WVX = 33^{\circ} \). Then \( \angle UVW=\angle UVX+\angle WVX + 90^{\circ}- 90^{\circ} \)? No, wait, \( \angle UVW \) is equal to \( 90^{\circ}+ 33^{\circ} \)? No, let's calculate the angle.

Wait, the sum of angles in a right - angled triangle: in \( \triangle XWV \), \( \angle XWV = 90^{\circ} \), \( \angle WVX = 33^{\circ} \), so \( \angle VXW=90 - 33=57^{\circ} \). Since \( \triangle XWV\cong\triangle XUV \), \( \angle VXU = 57^{\circ} \). Then in quadrilateral \( XW V U \), the sum of interior angles is \( 360^{\circ} \). We have \( \angle XWV = 90^{\circ} \), \( \angle XUV = 90^{\circ} \), \( \angle VXW = 57^{\circ} \), \( \angle VXU = 57^{\circ} \). So \( \angle UVW=360-(90 + 90+57 + 57)=360 - 294 = 66^{\circ} \)? No, that's not right.

Wait, I think the correct way is: Since \( XW\perp WV \) and \( XU\perp UV \), and \( XW = XU \), \( XV \) is the angle - bisector of \( \angle UVW \)? No, wait, the angle \( \angle UVW \): we know that \( \angle WVX = 33^{\circ} \), and because the two right triangles are congruent, \( \angle UVX=\angle WVX = 33^{\circ} \), so \( \angle UVW=\angle UVX+\angle WVX=33 + 33=66^{\circ} \)? No, that can't be. Wait, no, the right angle is at \( W \) and \( U \), so \( \angle UVW \) is a right angle plus \( 33^{\circ} \)? Wait, no, let's look at the diagram again. The angle at \( V \), between \( UV \) and \( WV \). We have a right angle at \( W \) (so \( XW\perp WV \)) and a right angle at \( U \) (so \( XU\perp UV \)). The segment \( XV \) splits the angle at \( V \) into two equal parts? Wait, no, the given angle is \( 33^{\circ} \) at \( V \) between \( WV \) and \( XV \). So since \( XW = XU \), \( \triangle XWV\) and \( \triangle XUV \) are congruent, so \( \angle UVX=\angle WVX = 33^{\circ} \). Therefore, \( \angle UVW=\angle UVX+\angle WVX = 33^{\circ}+33^{\circ}=66^{\circ} \)? No, that's not correct. Wait, I think I messed up.

Wait, the correct approach: In the right - angled triangle \( XWV \), \( \angle XWV = 90^{\circ} \), \( \angle WVX = 33^{\circ} \), so \( \angle VXW=90 - 33 = 57^{\circ} \). Since \( XW = XU \) and \( \angle XWV=\angle XUV = 90^{\circ} \), \( XV = XV \), so \( \triangle XWV\cong\triangle XUV \) (HL). Then \( \angle VXU=\angle VXW = 57^{\circ} \). Now, the angle \( \angle UVW \): we know that the sum of angles around point \( V \) is not the case, but in the quadrilateral \( XW V U \), the sum of interior angles is \( 360^{\circ} \). So \( \angle XWV+\angle WVU+\angle VUX+\angle UXW = 360^{\circ} \). We know \( \angle XWV = 90^{\circ} \), \( \angle VUX = 90^{\circ} \), \( \angle UXW=\angle VXW+\angle VXU=57 + 57 = 114^{\circ} \). So \( 90+\angle WVU + 90+114 = 360 \), so \( \angle WVU=360-(90 + 90+114)=66^{\circ} \). Wait, but that's the same as before. But actually, the angle \( \angle UVW \) is \( 90^{\circ}+ 33^{\circ} \)? No, I think the correct answer is \( 66^{\circ} \). Wait, no, let's think again.

Wait, the two right - angled triangles: \( XW\perp WV \), \( XU\perp UV \), \( XW = XU \), \( XV \) is common. So \( \triangle XWV\cong\triangle XUV \), so \( \angle UVX=\angle WVX = 33^{\circ} \). Then \( \angle UVW=\angle UVX+\angle WVX=33 + 33 = 66^{\circ} \). Yes, that makes sense. Because the angle between \( UV \) and \( WV \) is composed of two angles each equal to \( 33^{\circ} \) (since the triangles are congruent, the angles at \( V \) with respect to \( XV \) are equal).

Answer:

\( 66 \)