Question

$\square$ABCD is a kite in which diagonal AC and diagonal BD intersect at point O. $\angle$OBC = 20$°$ and $\angle$OCD = 40$°$ find i) $\angle$ABC (ii) ADC (iii) BAD

Answer 1

Consider the given figure.

$$\begin{gathered} \angle \mathrm{DAO}=\angle \mathrm{DCO}=40°(\mathrm{AD}=\mathrm{CD},\mathrm{as}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{kite}) \\ \mathrm{Consider}△\mathrm{AOD}. \\ \angle \mathrm{DAO}+\angle \mathrm{AOD}+\angle \mathrm{ODA}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 40°+90°+\angle \mathrm{ODA}=180° \\ \Rightarrow \angle \mathrm{ODA}=50° \\ \mathrm{Similarly}, \\ \mathrm{Consider}△\mathrm{COD}, \\ \mathrm{Here}, \\ \angle \mathrm{ODC}=50° \\ \therefore \angle \mathrm{ADC}=\angle \mathrm{ADO}+\angle \mathrm{ODC} \\ =100°...\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{considering}△\mathrm{COB}. \\ \angle \mathrm{COB}+\angle \mathrm{OBC}+\angle \mathrm{BCO}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow 90°+20°+\angle \mathrm{BCO}=180° \\ \Rightarrow \angle \mathrm{BCO}=70° \\ \Rightarrow \angle \mathrm{BCO}=\hspace{0.17em}\angle \mathrm{BAO}=70°(\because \mathrm{AB}=\mathrm{AC},\mathrm{as}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{kite}) \\ \mathrm{Thus},\mathrm{in}△\mathrm{AOB}, \\ \angle A\mathrm{OB}+\angle \mathrm{OBA}+\angle \mathrm{BAO}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \angle \mathrm{OBA}=180°-\left(90°+70°\right) \\ \therefore \angle \mathrm{OBA}=20° \\ \mathrm{Now}, \\ \angle \mathrm{BAD}=\angle \mathrm{BAO}+\angle \mathrm{DAO} \\ =110° \\ \angle \mathrm{ABC}=\angle \mathrm{OBA}+\angle \mathrm{OBC} \\ =40° \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \left(\mathrm{i}\right)\angle \mathrm{ABC}=40° \\ \left(\mathrm{ii}\right)\angle \mathrm{ADC}=100° \\ \left(\mathrm{iii}\right)\angle \mathrm{BAD}=110° \end{gathered}$$