A kite

ABCD is a kite. Angle OBC = 20° and angle OCD = 35°. O is the intersection of diagonals. Find angle ABC, angle ADC and angle BAD.

Correct answer:

x =  40 °
y =  110 °
z =  55 °

Step-by-step explanation:

$$\begin{gathered} \mathrm{\angle }OBC=20^{\circ } \\ \mathrm{\angle }OCD=35^{\circ } \\ OBA=\mathrm{\angle }OBC=20=20^{\circ } \\ ABC=OBA+\mathrm{\angle }OBC=20+20=40^{\circ } \\ x=ABC=40=40^{\circ } \end{gathered}$$

$$\begin{gathered} OAD=\mathrm{\angle }OCD=35=35^{\circ } \\ \mathrm{\angle }ADO=90-OAD=90-35=55^{\circ } \\ ODC=\mathrm{\angle }ADO=55=55^{\circ } \\ ADC=\mathrm{\angle }ADO+ODC=55+55=110^{\circ } \\ y=ADC=110=110^{\circ } \end{gathered}$$

$$\begin{gathered} BAD=OBA+OAD=20+35=55^{\circ } \\ z=BAD=55=55^{\circ } \end{gathered}$$

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