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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

$$\mathrm{\angle }ABC+\mathrm{\angle }ADC={180}^o$$$$\Rightarrow \mathrm{\angle }ADC={180}^o-{60}^o={120}^o$$$$\mathrm{\angle }ADC+x={180}^o$$$$x={180}^o-{120}^o={60}^o$$

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

$${60}^o$$

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