Question

ABCD is a kite in which diagonals AC and BD meet at O. If angle OBC = 20° and

angle OCD= 35°. Find angle ABC, angle ADC, angle BAD.​

Answer 1

Answer:

Step-by-step explanation:

Here,

=> ABCD is a kite in which diagonals AC and BD meet at O.

Thus, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90⁰

=> In ΔBOC,

∠BOC + ∠OBC + ∠OCB = 180°

90°+20°+∠OCB = 180°

110°+∠OCB = 180°

∠OCB = 180° - 110° = 70°

Also, adjacent sides have equal length. thus, AB = BC

The angles opposite to equal sides are also equal.

∴ ∠ACB = ∠BAC = 70°

=> Now, In ΔAOB

∠ABO + ∠AOB + ∠OAB = 180°

∠ABO = 180° -∠AOB - ∠OAB

= 180° - 90° - 70°

= 20°

=> AD = DC

∠DAC = ∠DCA = 35°

=> In ΔCOD,

∠COD + ∠DCO + ∠CDO = 180°

90° + 35° + ∠CDO = 180°

∠CDO = 55°

Similarly, ∠ADO = 55°

Thus,

=>  ∠ADC=∠CDO+∠ADO

=55°+55°

=110°

=> ∠BAD=∠DAO+∠BAO

=35°+70°

=105°

=> ∠ABC=∠ABO+∠OBC

=20°+20°

=40°