If the resultant R, of two forces P and Q acting at an angle θ makes an angle α with P, then

\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\sin {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\sin {\rm{\theta }}}}} \right]{\rm}\)

\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\sin {\rm{\theta }}}}{{{\rm{Q}}~ -~ {\rm{P}}\sin {\rm{\theta }}}}} \right]{\rm}\)

\({\tan~\alpha} =\left[ {\frac{{{\rm{Q}}\sin {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\cos {\rm{\theta }}}}} \right]{\rm}\)

\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\cos {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\sin {\rm{\theta }}}}} \right]{\rm}\)

If the resultant R, of two forces P and Q acting at an angle θ makes

1.	If the resultant R, of two forces P and Q acting at an angle θ makes an angle α with P, then
A.	\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\sin {\rm{\theta }}}}{{{\rm{Q}}~ -~ {\rm{P}}\sin {\rm{\theta }}}}} \right]{\rm}\)
B.	\({\tan~\alpha} =\left[ {\frac{{{\rm{Q}}\sin {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\cos {\rm{\theta }}}}} \right]{\rm}\)
C.	\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\sin {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\sin {\rm{\theta }}}}} \right]{\rm}\)
D.	\({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\cos {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\sin {\rm{\theta }}}}} \right]{\rm}\)
Answer» C. \({\tan~\alpha} =\left[ {\frac{{{\rm{P}}\sin {\rm{\theta }}}}{{{\rm{P}} ~+~ {\rm{Q}}\sin {\rm{\theta }}}}} \right]{\rm}\)

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