Question Number 85447 by TawaTawa1 last updated on 22/Mar/20
Commented by TawaTawa1 last updated on 22/Mar/20
$$\mathrm{Please}\:\mathrm{workings} \\ $$
Commented by john santu last updated on 22/Mar/20
$$\angle{ABC}\:+\:\angle{ADC}\:=\:\mathrm{180}^{{o}} \\ $$$$\Rightarrow\:\angle{ADC}\:=\:\mathrm{180}^{{o}} −\mathrm{60}^{{o}} \:=\mathrm{120}^{{o}} \\ $$$$\angle{ADC}\:+\:{x}\:=\:\mathrm{180}^{{o}} \\ $$$${x}\:=\:\mathrm{180}^{{o}} −\mathrm{120}^{{o}} \:=\:\mathrm{60}^{{o}} \\ $$
Commented by TawaTawa1 last updated on 22/Mar/20
$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{it}\:\mathrm{cyclic}\:\mathrm{quadrillateral}? \\ $$
Commented by john santu last updated on 22/Mar/20
$${yes}. \\ $$
Commented by TawaTawa1 last updated on 22/Mar/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by john santu last updated on 22/Mar/20
$$\mathrm{60}^{{o}} \\ $$