Question Number 127382 by I want to learn more last updated on 29/Dec/20 | ||
Answered by mr W last updated on 29/Dec/20 | ||
Commented by mr W last updated on 29/Dec/20 | ||
$${CD}=\frac{{DE}}{\mathrm{sin}\:\mathrm{45}°} \\ $$$${BD}=\frac{{DE}}{\mathrm{sin}\:\mathrm{30}°} \\ $$$$\mathrm{cos}\:\alpha=\frac{{CD}}{{BD}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{45}°}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow\alpha=\mathrm{45}°=\angle{ABD} \\ $$ | ||
Commented by I want to learn more last updated on 29/Dec/20 | ||
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$ | ||