# 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 result:

x =  40 °
y =  110 °
z =  55 °

#### Solution:

$\angle OBC=20 \ ^\circ \ \\ \angle OCD=35 \ ^\circ \ \\ \ \\ OBA=\angle OBC=20=20 \ ^\circ \ \\ ABC=OBA + \angle OBC=20 + 20=40 \ ^\circ \ \\ \ \\ x=ABC=40=40 ^\circ$
$OAD=\angle OCD=35=35 \ ^\circ \ \\ \angle ADO=90 - OAD=90 - 35=55 \ ^\circ \ \\ ODC=\angle ADO=55=55 \ ^\circ \ \\ \ \\ ADC=\angle ADO + ODC=55 + 55=110 \ ^\circ \ \\ \ \\ y=ADC=110=110 ^\circ$
$BAD=OBA + OAD=20 + 35=55 \ ^\circ \ \\ \ \\ z=BAD=55=55 ^\circ$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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