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Question Number 195814 by sonukgindia last updated on 11/Aug/23

Answered by mr W last updated on 11/Aug/23

Commented by mr W last updated on 11/Aug/23

$$DB=1$$$$DO=\frac{\sqrt{2}}{2}$$$$OB=\sqrt{1^2-{\left(\frac{\sqrt{2}}{2}\right)}^2}=\frac{\sqrt{2}}{2}$$$$AO=DA=\frac{1}{2}$$$$\tan \alpha =\frac{\sqrt{2}}{2}\times 2=\sqrt{2}\Rightarrow \alpha ={\tan }^{-1}\sqrt{2}$$$$DC=\frac{\sqrt{3}}{2}$$$$\sin \beta =\frac{\sqrt{2}\times 2}{2\times \sqrt{3}}=\frac{\sqrt{6}}{3}\Rightarrow \beta ={\sin }^{-1}\frac{\sqrt{6}}{3}={\tan }^{-1}\sqrt{2}=\alpha$$

Commented by mr W last updated on 11/Aug/23

Commented by mr W last updated on 11/Aug/23

$$FG=HG=\frac{\sqrt{3}}{2}$$$$\cos \gamma =\frac{2{\left(\frac{\sqrt{3}}{2}\right)}^2-1^2}{2\times \left(\frac{\sqrt{3}}{2}\right)\times \left(\frac{\sqrt{3}}{2}\right)}=\frac{1}{3}$$$$\Rightarrow \gamma ={\cos }^{-1}\frac{1}{3}={\tan }^{-1}2\sqrt{2}$$$$2\alpha +\gamma =2{\tan }^{-1}\sqrt{2}+{\tan }^{-1}2\sqrt{2}=180°$$

Commented by mr W last updated on 12/Aug/23

$$12edges:$$$$AE,BE,CE,DE,CB,CD,DA,DE,CF,GA,GB,GF$$$$7faces:$$$$CBE,CED,CDF,ADE,AEBG,BCFG,FDAG$$

Commented by mr W last updated on 11/Aug/23

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