Question

c) In the given figure , ABCD is a kite whose diagonals intersect at O.
If 2 DAB = 64° and ABC = 110°. Calculate (1)ODA (ii) BCD
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Answer 1

Given : ABCD is a kite whose diagonals intersect at O

∠DAB = 64° and ABC = 110°

To Find : Calculate ∠ODA (ii) ∠BCD

Solution:

Figure is missing

Case 1 : AB = AD  and BC = CD

∠DAB = 64°

∠ODA = ∠OBA

∠DAB + ∠BDA +  ∠DBA = 180°

=> ∠DAB + ∠ODA +  ∠OBA = 180°

=> 64° + 2∠ODA   = 180°

=> 2∠ODA   = 116°

=> ∠ODA   = 58°

∠DAB = 64°

=> ∠CAB = (1/2)64° = 32°

∠CAB + ∠ABC + ∠ACB = 180°

=> 32° + 110° + ∠ACB = 180°

=> ∠ACB =38°

∠BCD  = 2∠ACB =2 * 38° = 76°

∠BCD  =  76°

Case 2

Case 1 : AB = BC  and CD = AD

∠DAB = 64°

∠ABC = 110°

=>∠CBD =  ∠ABD = (1/2)110° = 55°

∠ABD  + ∠DAB + ∠BDA = 180°

=> 55° + 64° + ∠BDA = 180°

=> ∠BDA = 61°

=> ∠ODA = 61°

∠BDA = 61° =>  ∠BDC = 61°

∠CBD =   55°

∠BDC + ∠CBD  + ∠BCD = 180°

=> 61°  + 55°  + ∠BCD = 180°

=>  ∠BCD = 64°

∠ODA = 61° and ∠BCD = 64°

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