Question

The resultant of two vectors of magnitudes 2A and √
2
A
acting at an angle θ
is √
10
A
. The correct value of θ
is

Answer 1

Given

$\sqrt{(2A)^2+(\sqrt{2}A )^2+2(2A)(\sqrt{2}A )cos\alpha } =\sqrt{10}\\ Sq.LHSandRHS\\4A^2+2A^2+4\sqrt{2}A^2cos\alpha =10\\ A^2(6+4\sqrt2cos\alpha)=10\\(6+4\sqrt2cos\alpha)=\frac{10}{A^2}\\Hence ,\alpha = cos^{-1}((\frac{10}{A^2}-6)/4\sqrt2)\\Simplify,\\\alpha=cos^{-1}((10-6A^2)/4\sqrt2A^2)$

Hope this was right