Question #9c959

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Question #9c959

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Answer 1 · 2016-10-30T07:37:05

Option- C

Explanation:

The vertices of a regular tetrahedron are at four corners of a cube as shown in above figure.

If we consider it as an unit cube then its diagonal of any square face will have length $d_{s} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ $d_s=sqrt(1^2+1^2)=sqrt2$ unit

Again half of the length of the diagonal of the cube will be

$d_{c} = \frac{1}{2} \sqrt{1^{2} + 1^{2} + 1^{2}} = \frac{\sqrt{3}}{2}$ $d_c=1/2sqrt(1^2+1^2+1^2)=sqrt3/2$ unit

If the two half diogonals make an angle $θ$ $theta$ at the center then we can write

$cos θ = \frac{d_{c}^{2} + d_{c}^{2} - d_{s}^{2}}{2 \cdot d_{c} \cdot d_{c}}$ $costheta=(d_c^2+d_c^2-d_s^2)/(2*d_c*d_c)$

$\Rightarrow cos θ = \frac{{(\frac{\sqrt{3}}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} - {(\sqrt{2})}^{2}}{2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}}$ $=>costheta=((sqrt3/2)^2+(sqrt3/2)^2-(sqrt2)^2)/(2*sqrt3/2*sqrt3/2)$

$\Rightarrow cos θ = \frac{\frac{3}{2} - 2}{\frac{3}{2}} = - \frac{1}{3}$ $=>costheta=(3/2-2)/(3/2)=-1/3$

$\Rightarrow θ = {cos}^{- 1} (- \frac{1}{3})$ $=>theta=cos^-1(-1/3)$

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Question #9c959

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