Q4 of 100 Page 17

In Fig. 16.179 ABCD is a cyclic quadrilateral. If ∠BCD=100° and ∠ABD=70°, find ∠ADB.

We have,

∠BCD = 100o


∠ABD = 70o


Therefore,


∠DAB + ∠BCD = 180o (Opposite angles of cyclic quadrilateral)


∠DAB + 100o = 180o


∠DAB = 180o – 100o


= 80o


In triangle DAB, by angle sum property


∠ADB + ∠DAB + ∠ABD = 180o


∠ABD + 80o + 70o = 180o


∠ABD = 30o


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