Question

what is the measure of $angle wzx$ in rhombus $wxyz$?

a. 38.5
b. 51.5
c. 77
d. 103

Explanation:

Step1: Recall properties of a rhombus

In a rhombus, the diagonals bisect the angles. Also, adjacent angles in a rhombus are supplementary, but here we use the fact that the diagonal bisects the angle at vertex \( W \). The angle \( \angle XWY = 77^\circ \)? Wait, no, looking at the diagram, \( \angle XWZ \) is part of the angle at \( W \), and the diagonal \( WY \) and \( XZ \) intersect. Wait, actually, in a rhombus, the diagonals bisect the internal angles. Also, the triangle \( WXZ \) – wait, no, the angle at \( W \), \( \angle XWZ \): Wait, the given angle is \( 77^\circ \) at \( W \) between \( WX \) and \( WY \)? Wait, no, the diagram shows \( \angle XWW \)? No, the label is \( 77^\circ \) at \( W \) between \( WX \) and the diagonal? Wait, maybe the angle \( \angle XWZ \) is such that the diagonal bisects the angle, but actually, in a rhombus, the diagonals are perpendicular bisectors, and they bisect the angles. Wait, the angle at \( W \): let's consider triangle \( WXZ \). Wait, no, the key property: in a rhombus, the diagonals bisect the angles. So if we have angle at \( W \), say \( \angle XWZ \), but wait, the given angle is \( 77^\circ \), and we need to find \( \angle WZX \)? Wait, no, the question is \( \angle WZX \). Wait, maybe I misread. Wait, the question is \( \angle WZX \). Let's re-examine.

Wait, in rhombus \( WXYZ \), sides \( WX = WZ \) (since all sides of a rhombus are equal). The diagonal \( XZ \) divides the rhombus into two isosceles triangles? No, \( WX = WZ \), so triangle \( WXZ \) is isosceles? Wait, no, all sides of the rhombus are equal, so \( WX = WZ = ZY = YX \). Wait, the angle at \( W \): let's say \( \angle XWZ \) – no, the given angle is \( 77^\circ \) at \( W \) between \( WX \) and \( WY \). Wait, maybe the angle \( \angle XWY = 77^\circ \), but actually, the diagonal \( WY \) and \( XZ \) intersect. Wait, perhaps the angle at \( W \) is \( 180^\circ - 2 \times \) something, but no. Wait, the correct approach: in a rhombus, the diagonals bisect the angles. So if we have angle \( \angle XWZ \), but wait, the given angle is \( 77^\circ \), and we need to find \( \angle WZX \). Wait, maybe the triangle \( WXZ \): since \( WX = WZ \) (sides of rhombus), triangle \( WXZ \) is isosceles with \( WX = WZ \). Then, the angle at \( W \) is \( 77^\circ \)? No, wait, the angle between \( WX \) and \( WY \) is \( 77^\circ \), but the diagonal \( WY \) and \( XZ \) are perpendicular? No, diagonals of a rhombus are not necessarily perpendicular? Wait, no, diagonals of a rhombus are perpendicular bisectors of each other. Oh right! Diagonals of a rhombus are perpendicular, so they form right angles where they intersect. Wait, but the angle at \( W \): let's consider angle \( \angle XWZ \). Wait, maybe the angle given is \( \angle XWY = 77^\circ \), but actually, the angle at \( W \) is \( 180^\circ - 2 \times \) (angle we need)? No, let's start over.

In a rhombus, all sides are equal, and diagonals bisect the angles. So, the diagonal \( WY \) bisects \( \angle XWZ \)? Wait, no, the diagonal \( XZ \) bisects \( \angle WXY \) and \( \angle WZY \), and diagonal \( WY \) bisects \( \angle XWZ \) and \( \angle XYZ \). Wait, the angle at \( W \): \( \angle XWZ \) – if the diagonal \( WY \) bisects it, then \( \angle XWY = \angle ZWY \). But the given angle is \( 77^\circ \), maybe that's \( \angle XWY = 77^\circ \), but we need \( \angle WZX \). Wait, in triangle \( WZX \), we know \( WZ = WX \) (sides of rhombus), so it's isosceles. The angle at \( W \): let's say \( \angle XWZ = 180^\circ - 2 \times 77^\circ \)? No, that doesn't make sense. Wait, maybe the angle given is \( \angle XWZ = 77^\circ \), and we need \( \angle WZX \). Wait, no, the sum of angles in a triangle is \( 180^\circ \). In triangle \( WZX \), \( WX = WZ \) (rhombus sides), so it's isosceles with \( \angle WXZ = \angle WZX \). Wait, no, \( WX = WZ \), so angles opposite are equal: \( \angle WZX = \angle WXZ \). The angle at \( W \) is \( \angle XWZ \). Wait, maybe the given angle is \( \angle XWY = 77^\circ \), but actually, the diagonal \( WY \) is perpendicular to \( XZ \)? No, diagonals of a rhombus are perpendicular, so \( \angle AOB = 90^\circ \) where diagonals intersect, but that's not directly helpful here.

Wait, maybe I made a mistake. Let's recall: in a rhombus, the diagonals bisect the angles. So if we have angle at \( W \), say \( \angle XWZ = 180^\circ - 2 \times \) (the angle we need)? No, wait, the correct approach: the angle given is \( 77^\circ \), and we need to find \( \angle WZX \). Since the diagonals bisect the angles, and adjacent angles in a rhombus are supplementary. Wait, no, let's look at the answer choices. The options are 38.5, 51.5, 77, 103. 38.5 is half of 77? Wait, no, 77 + 51.5 = 128.5, no. Wait, maybe the angle at \( W \) is \( 77^\circ \), and the diagonal bisects it, but no, the angle \( \angle WZX \): let's consider triangle \( WZX \). Wait, \( WX = WZ \), so it's isosceles. The angle at \( W \): \( \angle XWZ = 180^\circ - 2 \times \angle WZX \). But we need another relation. Wait, maybe the angle given is \( \angle XWY = 77^\circ \), and since diagonals are perpendicular, \( \angle XWY + \angle WXZ = 90^\circ \)? No, that's for a square. Wait, no, diagonals of a rhombus are perpendicular, so if they intersect at point \( O \), then \( \angle XOW = 90^\circ \). But that's not helpful here.

Wait, maybe the angle \( \angle XWZ = 77^\circ \), and we need \( \angle WZX \). Wait, no, the sum of angles in triangle \( WZX \) is \( 180^\circ \). If \( WX = WZ \), then \( \angle WZX = \angle WXZ \). Let \( \angle WZX = x \), then \( \angle WXZ = x \), and \( \angle XWZ = 180 - 2x \). But we need to find \( x \). Wait, maybe the given angle is \( \angle XWZ = 77^\circ \), then \( 180 - 2x = 77 \), so \( 2x = 180 - 77 = 103 \), so \( x = 51.5 \). Ah! That makes sense. So \( \angle WZX = 51.5^\circ \), which is option B.

Step1: Identify triangle properties

In rhombus \( WXYZ \), \( WX = WZ \) (all sides of rhombus are equal), so \( \triangle WZX \) is isosceles with \( \angle WZX = \angle WXZ \).

Step2: Sum of angles in a triangle

The sum of angles in a triangle is \( 180^\circ \). Let \( \angle WZX = x \), then \( \angle WXZ = x \), and \( \angle XWZ = 180^\circ - 2x \).

Step3: Determine \( \angle XWZ \)

From the diagram, \( \angle XWZ = 77^\circ \)? Wait, no, wait, the given angle is \( 77^\circ \), but actually, the angle at \( W \) is \( 180^\circ - 2 \times 77^\circ \)? No, wait, I think I mixed up. Wait, the angle given is \( 77^\circ \), which is \( \angle XWZ \)? No, no, the correct approach: in a rhombus, adjacent angles are supplementary, but here we have triangle \( WZX \) with \( WX = WZ \), so angles at \( X \) and \( Z \) are equal. The angle at \( W \) is \( 77^\circ \), so \( 180 - 77 = 103 \), then divide by 2: \( 103 / 2 = 51.5 \). So \( \angle WZX = 51.5^\circ \).

Answer:

B. 51.5