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Q.
At what angle the two vectors of magnitudes $( A + B )$ and $(A-B)$ must act, so that resultant is $\sqrt{A^{2}+B^{2}} ?$
Motion in a Plane
Solution:
$R =\sqrt{ P ^{2}+ Q ^{2}+2 PQ \cos \theta}$
$R =\sqrt{ A ^{2}+ B ^{2}}, P =( A + B ), Q =( A - B ) \\
\sqrt{ A ^{2}+ B ^{2}}=\sqrt{( A + B )^{2}+\left( A ^{2}- B ^{2}\right)+2( A + B )( A - B ) \cos \theta}$
Squaring both sides
$A^{2}+B^{2}=(A+B)^{2}+(A-B)^{2}+2\left(A^{2}-B^{2}\right) \cos \theta$
$A^{2}+B^{2}=A^{2}+B^{2}+2 A B+A^{2}+B^{2}-2 A B+2\left(A^{2}-B^{2}\right) \cos \theta$
$-A^{2}-B^{2}=2\left(A^{2}-B^{2}\right) \cos \theta$
$-\left(A^{2}+B^{2}\right)=2\left(A^{2}-B^{2}\right) \cos \theta$
Multiply by (-)
$A^{2}+B^{2}=2\left(B^{2}-A^{2}\right) \cos \theta$
$\cos ^{-1}\left(\frac{A^{2}+B^{2}}{2\left(B^{2}-A^{2}\right)}\right)=\theta$