if there is no triangle information, what... if necessary, you may know what the markings on a figure indicate. (then there are two figures with triangles and some angle/side markings, and options like cannot be determined etc.)

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# Explanation: ## Step1: Analyze the first triangle (WXZ) In triangle $ WXY $ and $ WZY $, we have $ XY = ZY = 16 $, $ WY $ is common. The angle at $ Y $ for $ \triangle WXY $ is $ 91^\circ $ and for $ \triangle WZY $ is $ 16^\circ $. Since $ 91^\circ>16^\circ $, by the Hinge Theorem (if two sides of one triangle are congruent to two sides of another triangle, but the included angle is larger, then the third side is longer), in $ \triangle WXY $ and $ \triangle WZY $, $ WX > WZ $ because the included angle $ \angle WYX = 91^\circ $ is larger than $ \angle WYZ = 16^\circ $ and $ XY = ZY $, $ WY = WY $. ## Step2: Analyze the second triangle (TVU and DEF) In $ \triangle TVU $, sides: $ TV $ (let's say the equal sides are $ TU = VU $? Wait, no, looking at $ \triangle DEF $, $ DE = EF $? Wait, $ \triangle TVU $ has sides: $ TU $ (marked equal), $ VU = 10 $, and $ \triangle DEF $ has $ DE $ (marked equal), $ EF $ (marked equal), $ DF = 16 $. Wait, actually, $ \triangle TVU $: sides $ TU = VU $? No, $ VU = 10 $, $ TU $ is equal to some side, and $ \triangle DEF $: $ DE = EF $ (marked equal), $ DF = 16 $. Wait, the sides: in $ \triangle TVU $, the two equal sides (marked) and the third side is... Wait, no, the Hinge Theorem: if two sides of one triangle are congruent to two sides of another triangle, then the larger included angle has the longer third side. In $ \triangle TVU $ and $ \triangle DEF $, $ TU = DE $, $ VU = EF $ (wait, $ VU = 10 $, $ EF $ is marked equal to $ DE $, and $ DF = 16 $, $ TV $ is... Wait, no, $ \triangle TVU $: sides $ TU $ (equal to $ DE $), $ VU = 10 $, and $ \triangle DEF $: $ DE = EF $, $ DF = 16 $. Wait, the included angles: $ \angle T $ in $ \triangle TVU $ and $ \angle E $ in $ \triangle DEF $. The third side of $ \triangle TVU $ is $ TV $, and $ \triangle DEF $ is $ DF = 16 $. Since $ VU = 10 < DF = 16 $? Wait, no, maybe I mixed up. Wait, the Hinge Theorem: if $ AB = DE $, $ BC = EF $, and $ \angle B > \angle E $, then $ AC > DF $. Here, in $ \triangle TVU $ and $ \triangle DEF $, $ TU = DE $, $ VU = EF $ (assuming the marked sides are equal), and the third side of $ \triangle TVU $ is $ TV $, $ \triangle DEF $ is $ DF = 16 $. Wait, but $ VU = 10 $, $ DF = 16 $? No, maybe the sides are $ TU = DE $, $ TV = DF $? Wait, no, the first triangle: $ WX > WZ $ (from Step1). The second triangle: let's see, $ \triangle TVU $ has two sides equal (marked) and $ VU = 10 $, $ \triangle DEF $ has two sides equal (marked) and $ DF = 16 $. Wait, no, the angle at $ T $ and $ E $: since $ DF = 16 $ is longer than $ VU = 10 $, but wait, no, the Hinge Theorem: if two sides are equal, the larger included angle gives the longer third side. Wait, maybe I messed up. Wait, the first question: $ WX $ vs $ WZ $: $ WX > WZ $ (from Hinge Theorem, included angle $ 91^\circ > 16^\circ $, sides $ XY = ZY $, $ WY $ common). The second question: $ m\angle T $ vs $ m\angle E $: in $ \triangle TVU $, the two equal sides (marked) and third side $ VU = 10 $; in $ \triangle DEF $, two equal sides (marked) and third side $ DF = 16 $. Since $ DF = 16 > VU = 10 $, by the converse of Hinge Theorem, the included angle $ \angle E $ (in $ \triangle DEF $) is larger than $ \angle T $ (in $ \triangle TVU $)? Wait, no: converse of Hinge Theorem: if two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer, then the included angle is larger. So in $ \triangle TVU $ and $ \triangle DEF $, $ TU = DE $, $ TV = EF $ (wait, no, $ VU = 10 $, $ DF = 16 $). Wait, maybe the sides are $ TU = DE $, $ VU = EF $, and $ TV $ vs $ DF $. If $ DF = 16 $ is longer than $ TV $ (assuming $ TV $ is the third side), then $ \angle E > \angle T $, so $ m\angle T < m\angle E $? Wait, no, maybe I need to re - check. But for the first part: $ WX $ and $ WZ $: since $ XY = ZY = 16 $, $ WY = WY $, and $ \angle WYX = 91^\circ $, $ \angle WYZ = 16^\circ $, by Hinge Theorem, $ WX > WZ $. For the second part: in $ \triangle TVU $ and $ \triangle DEF $, $ TU = DE $, $ VU = EF $ (marked equal), and $ DF = 16 $, $ TV $ is the third side. Wait, no, $ DF = 16 $ is longer than $ VU = 10 $? No, $ VU = 10 $ is a side, $ DF = 16 $ is a side. Wait, maybe the two triangles have two sides congruent: $ TU = DE $, $ TV = DF $? No, I think the first answer is $ WX > WZ $, and the second: since $ DF = 16 $ is longer than the third side of $ \triangle TVU $ (which is $ VU = 10 $? No, that can't be. Wait, maybe the second triangle: $ \triangle TVU $ has sides $ TU $ (equal to $ DE $), $ VU = 10 $, and $ \triangle DEF $ has $ DE = TU $, $ EF = VU $, and $ DF = 16 $. So by Hinge Theorem, since $ DF = 16 > VU = 10 $, the included angle $ \angle E $ is larger than $ \angle T $, so $ m\angle T < m\angle E $. But the question is probably about the first triangle: $ WX $ and $ WZ $, so $ WX > WZ $. # Answer: For the first comparison (WX vs WZ): $ WX > WZ $ (by Hinge Theorem, included angle $ 91^\circ > 16^\circ $, $ XY = ZY $, $ WY $ common). For the second comparison (m∠T vs m∠E): $ m\angle T < m\angle E $ (by converse of Hinge Theorem, since $ DF = 16 > VU = 10 $, so included angle $ \angle E > \angle T $). But if the question is about the first option (WX vs WZ), the answer is $ WX > WZ $.

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the first triangle (WXZ)

Step2: Analyze the second triangle (TVU and DEF)

Answer: