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Q.
The maximum and minimum of the resultant of two forces are $F$ and $G$ (the angle between these
forces is $2\alpha$), Then, the resultant of $F$ and $G$ is
Let two forces be $F _{1}$ and $F _{2}$.
Given, $F$ and $G$ are maximum and minimum
resultant of two forces $F _{1}$ and $F _{2}$ respectively and angle between two forces $F _{1}$ and $F _{2}$ is $2 \alpha$.
$\therefore F= F_{1}+ F_{2}$ and $G=F_{1}-F_{2}$
$\Rightarrow F_{1}=\frac{F+G}{2}$
and $F_{2}=\frac{F-G}{2}$
Resultant
$R =\sqrt{ F _{1}^{2}+ F _{2}^{2}+2 F _{1} F _{2} \cos 2 \alpha}$
$=\sqrt{\left(\frac{F+G}{2}\right)^{2}+\left(\frac{F-G}{2}\right)^{2}+2\left(\frac{F+G}{2}\right)\left(\frac{F-G}{2}\right) \cos 2 \alpha}$
$=\sqrt{\frac{1}{4}\left[2\left(F^{2}+G^{2}\right)+2\left(F^{2}-G^{2}\right) \cos 2 \alpha\right]}$
$=\sqrt{\frac{1}{2}\left[F^{2}(1+\cos 2 \alpha)+G^{2}(1-\cos 2 \alpha)\right]}$
$=\sqrt{\frac{1}{2}\left[F^{2}\left(2 \cos ^{2} \alpha\right)+G^{2}\left(2 \sin ^{2} \alpha\right)\right]}$
$=\sqrt{F^{2} \cos ^{2} \alpha+ G^{2} \sin ^{2} \alpha}$
$=\left(F^{2} \cos ^{2} \alpha+ G^{2} \sin ^{2} \alpha\right)^{1 / 2}$