Question

in the triangles, xy = mp and yz = pn. if m∠p = 85°, which is a possible measure for ∠y?
○ m∠y = 80°
○ m∠y = 85°
○ m∠y = 100°
○ m∠y = 180°

Explanation:

Step1: Analyze side lengths

In triangle \( XYZ \), \( XZ = 23 \) in. In triangle \( MPN \), \( MN = 19 \) in. Given \( XY = MP \) and \( YZ = PN \). So, \( XZ>MN \).

Step2: Apply Hinge Theorem

By Hinge Theorem, if two sides of one triangle are congruent to two sides of another triangle, but the third side of the first triangle is longer, then the included angle of the first triangle is larger. Here, included angle for \( XZ \) is \( \angle Y \), included angle for \( MN \) is \( \angle P = 85^\circ \). So \( m\angle Y>m\angle P = 85^\circ \) is not possible? Wait, no, wait: Wait, \( XZ \) is opposite? Wait, no, \( XY = MP \), \( YZ = PN \), so sides \( XY=MP \), \( YZ = PN \), and \( XZ = 23 \), \( MN = 19 \). So the included angle for \( XY \) and \( YZ \) is \( \angle Y \), and included angle for \( MP \) and \( PN \) is \( \angle P \). Since \( XZ>MN \), by Hinge Theorem, \( m\angle Y>m\angle P \)? Wait, no, Hinge Theorem: If \( AB = DE \), \( BC = EF \), and \( AC>DF \), then \( \angle B>\angle E \). So here, \( XY = MP \), \( YZ = PN \), \( XZ>MN \), so \( \angle Y>\angle P = 85^\circ \)? Wait, but \( 100^\circ>85^\circ \), \( 180^\circ \) is impossible (triangle angle sum). Wait, triangle angle sum is \( 180^\circ \), so \( \angle Y \) must be less than \( 180^\circ \). Now check options: \( 80^\circ<85^\circ \), \( 85^\circ = 85^\circ \), \( 100^\circ>85^\circ \), \( 180^\circ \) invalid. Wait, maybe I mixed up. Wait, \( XZ = 23 \), \( MN = 19 \). So \( XZ>MN \), so the angle opposite? No, Hinge Theorem is about included angles. Wait, \( XY = MP \), \( YZ = PN \), so the two sides are \( XY, YZ \) with included angle \( \angle Y \), and \( MP, PN \) with included angle \( \angle P \). So if \( XZ>MN \), then \( \angle Y>\angle P \). But \( \angle P = 85^\circ \), so \( \angle Y \) must be greater than \( 85^\circ \)? But \( 100^\circ \) is greater, \( 180^\circ \) is impossible. Wait, but wait, maybe I got the Hinge Theorem reversed. Wait, no: If the third side is longer, the included angle is larger. So \( XZ>MN \), so \( \angle Y>\angle P \). So \( \angle Y \) must be greater than \( 85^\circ \). But \( 100^\circ \) is greater, \( 180^\circ \) is invalid. But wait, triangle angle sum: \( \angle Y \) must be less than \( 180^\circ \), and also, in a triangle, each angle is less than \( 180^\circ \), and sum to \( 180^\circ \). Now check options: \( 80^\circ \) is less than \( 85^\circ \), \( 85^\circ \) equal, \( 100^\circ \) greater, \( 180^\circ \) invalid. Wait, maybe I made a mistake. Wait, let's check the side lengths again. \( XZ = 23 \), \( MN = 19 \). So \( XZ>MN \), so the angle at \( Y \) (included angle for \( XY, YZ \)) should be greater than angle at \( P \) (included angle for \( MP, PN \)). So \( \angle Y>\angle P = 85^\circ \). So \( 100^\circ \) is possible (since \( 100^\circ>85^\circ \)), \( 80^\circ \) is less, \( 85^\circ \) equal, \( 180^\circ \) invalid. So \( m\angle Y = 100^\circ \) is possible? Wait, but wait, maybe the Hinge Theorem is applied the other way. Wait, maybe \( XZ \) is shorter? No, 23 is longer than 19. Wait, maybe I messed up the included angle. Wait, \( XY = MP \), \( YZ = PN \), so the sides are \( XY, YZ \) (lengths equal to \( MP, PN \)), and the third sides \( XZ = 23 \), \( MN = 19 \). So by Hinge Theorem, since \( XZ>MN \), then \( \angle Y>\angle P \). So \( \angle Y \) must be greater than \( 85^\circ \). Among the options, \( 100^\circ \) is greater than \( 85^\circ \), and \( 180^\circ \) is impossible. So \( m\angle Y = 100^\circ \) is a possible measure.

Answer:

\( m\angle Y = 100^\circ \) (the option with \( m\angle Y = 100^\circ \))