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
$${DB}=\mathrm{1} \\ $$$${DO}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${OB}=\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${AO}={DA}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×\mathrm{2}=\sqrt{\mathrm{2}}\:\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$$${DC}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\beta=\frac{\sqrt{\mathrm{2}}×\mathrm{2}}{\mathrm{2}×\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\:\Rightarrow\beta=\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{6}}}{\mathrm{3}}=\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}}=\alpha \\ $$
Commented by mr W last updated on 11/Aug/23
Commented by mr W last updated on 11/Aug/23
$${FG}={HG}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\gamma=\frac{\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)×\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\gamma=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{3}}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}\alpha+\gamma=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\sqrt{\mathrm{2}}=\mathrm{180}° \\ $$
Commented by mr W last updated on 12/Aug/23
$$\mathrm{12}\:{edges}: \\ $$$${AE},\:{BE},\:{CE},\:{DE},\:{CB},\:{CD},\:{DA},\:{DE},\:{CF},\:{GA},\:{GB},\:{GF} \\ $$$$\mathrm{7}\:{faces}: \\ $$$${CBE},\:{CED},\:{CDF},\:{ADE},\:{AEBG},\:{BCFG},\:{FDAG} \\ $$
Commented by mr W last updated on 11/Aug/23
