In kite WXYZ , m∠XWY=47° and m∠ZYW=18° .What is m∠WZY ?Enter your answer in the box.

Answer: [tex]\angle WZY=115^{\circ}[/tex]Explanation: Since, here WXYZ is a kite where [tex]\angle XWY=47^{\circ}[/tex] and [tex]\angle ZYW=18^{\circ}[/tex]Thus according to the property of a kite ,Exactly one pair of opposite angles are equal and The main diagonal bisects a pair of opposite angles.Therefore, In kite  WXYZ ,[tex]\angle WZY=\angle WXY[/tex]   -------(1)And, WY is the angle bisector of  kite WXYZ.So,  [tex]\angle ZWX=2\angle XWY=2\times 47^{\circ}=94^{\circ}[/tex] ( because WY bisects [tex]\angle ZWX[/tex] into two equal angles [tex]\angle XWY[/tex] and [tex]\angle ZWY[/tex])⇒[tex]\angle ZWX=94^{\circ}[/tex] -----(2)Similarly, [tex]\angle XYZ=2\angle ZYW=2\times 18^{\circ}=36^{\circ}[/tex]⇒[tex]\angle XYZ=36^{\circ}[/tex] -----(3)Since, WXYZ is a kite ⇒[tex]\angle WXY+\angle XYZ+\angle WZY+\angle ZWX=360^{\circ}[/tex] -------(4)Therefore,  From equation (1), (2), (3) and (4),We get, [tex]\angle WZY=115^{\circ}[/tex]