Two forces of magnitude P & Q acting at a | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\cos 	heta=\frac{Q-P}{Q}$

$\cos 	heta=1-\frac{P}{Q}$

$1-2 \sin ^{2} \frac{	heta}{2}=1-\frac{P}{Q}$

$	herefore 2 \sin ^{2} \frac{	heta}{2}=\fr

Vedclass · Accepted Answer

$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$

Vedclass · Answer

$\cos 	heta=\frac{Q-P}{Q}$

$\cos 	heta=1-\frac{P}{Q}$

$1-2 \sin ^{2} \frac{	heta}{2}=1-\frac{P}{Q}$

$	herefore 2 \sin ^{2} \frac{	heta}{2}=\frac{P}{Q}$

$\sin ^{2} \frac{	heta}{2}=\frac{P}{2 Q}$

$\Rightarrow \sin \frac{	heta}{2}=\left(\frac{P}{2 Q}ight)^{1 / 2}$

$\Rightarrow 	heta=2 \sin ^{-1}\left(\frac{P}{2 Q}ight)^{1 / 2}$

Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :