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MHT CET 2019 2nd May Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $\sqrt{A^2+B^2}$ represents the magnitude of resultant of two vectors $(\mathbf{A}+\mathbf{B})$ and $(\mathbf{A}-\mathbf{B})$, then the angle between two vectors is

A
$\cos ^{-1}\left[-\frac{2\left(A^2-B^2\right)}{\left(A^2+B^2\right)}\right]$
B
$\cos ^{-1}\left[-\frac{A^2-B^2}{A^2 B^2}\right]$
C
$\cos ^{-1}\left[-\frac{\left(A^2+B^2\right)}{2\left(A^2-B^2\right)}\right]$
D
$\cos ^{-1}\left[-\frac{\left(A^2-B^2\right)}{A^2+B^2}\right]$
MHT CET Subjects
EXAM MAP