Question-85447

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

∠ABC + ∠ADC = 180^o ⇒ ∠ADC = 180^o −60^o =120^o ∠ADC + x = 180^o x = 180^o −120^o = 60^o

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

Commented by TawaTawa1 last updated on 22/Mar/20