A) \[1\]
B) \[\sqrt{2}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\sqrt{3}\]
Correct Answer: C
Solution :
Let x be the length of the cube \[\therefore \] \[BQ=x\] and \[SQ=\sqrt{{{x}^{2}}+{{x}^{2}}}\] \[=\sqrt{2}\,x\] From right angled \[\Delta \,BSQ\] \[\tan \theta =\frac{BQ}{SE}\] \[=\frac{x}{\sqrt{2x}}=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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