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Question Number 188965 by normans last updated on 09/Mar/23

Answered by mr W last updated on 09/Mar/23

$$sayedgelengthofcubeis1.$$$$baseofisoscelestriangleis\sqrt{2}.$$$$lengthofitslegsis\sqrt{1^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}=\frac{\sqrt{6}}{2}.$$$$\sin \frac{\alpha }{2}=\frac{\sqrt{2}}{2}\times \frac{2}{\sqrt{6}}=\frac{1}{\sqrt{3}}$$$$\cos \alpha =1-2\times {\left(\frac{1}{\sqrt{3}}\right)}^2=\frac{1}{3}$$$$\Rightarrow \alpha ={\cos }^{-1}\frac{1}{3}\approx 70.529°$$

Answered by manxsol last updated on 10/Mar/23

$$A=(l,0,l)$$$$B=(l,l,0)$$$$M=(0,\frac{l}{2},\frac{l}{2})$$$$MA=(l,-\frac{l}{2},\frac{l}{2})\mid MA\mid =\sqrt{\frac{3}{2}}l$$$$MB=(l,\frac{l}{2},-\frac{l}{2})\mid MB\mid =\sqrt{\frac{3}{2}l}\mid$$$$MA\bullet MB=\mid MA\mid \mid MB\mid cos\theta$$$$(l,-\frac{l}{2},\frac{l}{2})\bullet (l,\frac{l}{2},-\frac{l}{2})=\sqrt{\frac{3}{2}}\sqrt{\frac{3}{2}}{l}^{2}cos\theta$$$$\frac{l^2}{2}=\sqrt{\frac{3}{2}}\sqrt{\frac{3}{2}}{l}^{2}cos\theta$$$$\frac{1}{3}=cos\theta$$$${\cos }^{-1}\left(\frac{1}{3}\right)$$$$70.528{}^o$$$$$$

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