Question Number 189557 by normans last updated on 18/Mar/23 | ||
Answered by mr W last updated on 18/Mar/23 | ||
Commented by mr W last updated on 18/Mar/23 | ||
$$\theta=\mathrm{180}°−\mathrm{2}\alpha−\mathrm{2}\beta \\ $$$$\:\:=\mathrm{180}°−\mathrm{2}\left(\alpha+\beta\right) \\ $$$$\:\:=\mathrm{180}°−\mathrm{2}×\mathrm{45}° \\ $$$$\:\:=\mathrm{90}° \\ $$ | ||
Answered by HeferH last updated on 18/Mar/23 | ||
$$\:\alpha+\beta\:=\:\mathrm{135}° \\ $$$$\:\mathrm{180}°−\alpha\:+\:\mathrm{180}°−\beta\:+\:\mathrm{45}°\:+\:\mathrm{x}\:=\:\mathrm{360}° \\ $$$$\:\mathrm{x}\:=\:\left(\alpha\:+\beta\right)−\:\mathrm{45}°\:=\:\mathrm{135}°\:−\:\mathrm{45}°\:=\:\mathrm{90}° \\ $$$$\:\mathrm{x}\:=\:\mathrm{90}° \\ $$ | ||
Commented by HeferH last updated on 18/Mar/23 | ||