Question #9c959

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Answer 1 · 2016-10-30T07:37:05

Option- C

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)=sqrt2#unit

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

#d_c=1/2sqrt(1^2+1^2+1^2)=sqrt3/2# unit

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

#costheta=(d_c^2+d_c^2-d_s^2)/(2*d_c*d_c)#

#=>costheta=((sqrt3/2)^2+(sqrt3/2)^2-(sqrt2)^2)/(2*sqrt3/2*sqrt3/2)#

#=>costheta=(3/2-2)/(3/2)=-1/3#

#=>theta=cos^-1(-1/3)#