Dear Student, In cyclic quadrilateral ABCD ∠BCD+∠BAD=180°⇒∠BCD+70°=180°⇒∠BCD=110°Also∠DBC+∠BCD+∠BDC=180°⇒35°+110°+∠BDC=180°⇒∠BDC=35°Since ∠ADB is angle in semicircle∠ADB=90°In △ABD∠ABD+∠ADB+∠BAD=1800⇒∠ABD+90°+70°=180°⇒∠ABD=20° Regards!

Please answer this

Dear Student,

In cyclic quadrilateral ABCD ∠ BCD + ∠ BAD = 180 ° \Rightarrow ∠ BCD + 70 ° = 180 ° \Rightarrow ∠ BCD = 110 ° Also ∠ DBC + ∠ BCD + ∠ BDC = 180 ° \Rightarrow 35 ° + 110 ° + ∠ BDC = 180 ° \Rightarrow ∠ BDC = 35 ° Since ∠ ADB is angle in semicircle ∠ ADB = 90 ° In △ ABD ∠ ABD + ∠ ADB + ∠ BAD = 180^{0} \Rightarrow ∠ ABD + 90 ° + 70 ° = 180 ° \Rightarrow ∠ ABD = 20 °

Regards!

View Full Answer