Question Number 226796 by sonukgindia last updated on 15/Dec/25
Answered by mr W last updated on 15/Dec/25
Commented by mr W last updated on 15/Dec/25
$$\frac{\mathrm{sin}\:\left(\mathrm{180}°−\alpha\right)}{\mathrm{sin}\:\varphi}=\frac{{AD}}{{ED}}=\frac{{AD}}{{DC}}=\frac{\mathrm{sin}\:\beta}{\mathrm{sin}\:\varphi} \\ $$$$\mathrm{sin}\:\alpha=\mathrm{sin}\:\beta \\ $$$$\Rightarrow\alpha=\beta \\ $$$$\angle{A}+\angle{D}+\beta+\mathrm{180}°−\alpha=\mathrm{360}° \\ $$$$\Rightarrow\angle{A}=\mathrm{180}°−\angle{D}=\mathrm{180}°−\mathrm{90}°=\mathrm{90}° \\ $$