Question Number 192988 by Mingma last updated on 01/Jun/23
Answered by Subhi last updated on 01/Jun/23
$${suppose}\:{line}\:=\:{l} \\ $$$${line}\:=\:\sqrt{{l}^{\mathrm{2}} +{l}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{2}\:}\:{l} \\ $$$$\frac{\sqrt{\mathrm{2}}\:{l}}{{sin}\left(\mathrm{180}−\left(\mathrm{45}−{x}+{x}\right)\right)}=\frac{{l}}{{sin}\left(\mathrm{45}−{x}\right)} \\ $$$${sin}\left(\mathrm{45}−{x}\right)=\frac{{sin}\left(\mathrm{135}\right)}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{45}−{x}=\mathrm{30} \\ $$$${x}=\mathrm{15} \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��
Answered by HeferH last updated on 01/Jun/23
Commented by HeferH last updated on 01/Jun/23
$$\mathrm{45}°+{x}\:=\:\mathrm{60}° \\ $$$${x}\:=\:\mathrm{15}° \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��