Question Number 226525 by wongb1506 last updated on 02/Dec/25
Answered by som(math1967) last updated on 02/Dec/25
$${AF}=\sqrt{\mathrm{160}}\:,{AE}=\sqrt{\mathrm{180}}\:,\:{EF}=\sqrt{\mathrm{100}} \\ $$$${Cosx}=\frac{\mathrm{160}+\mathrm{180}−\mathrm{100}}{\mathrm{2}×\sqrt{\mathrm{180}×\mathrm{160}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\:{x}=\mathrm{45}° \\ $$
Answered by mr W last updated on 02/Dec/25
Commented by mr W last updated on 02/Dec/25
$$\mathrm{2}{x}=\mathrm{90}°\:\Rightarrow{x}=\mathrm{45}° \\ $$