Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle `theta` between them .

Explanation

Let OP and OQ represent two vectors A and B making an angle `theta` (Fig . 4.10) . Then , using the parallelogram method of vector addition , OS represents the resultant vector R:
R = A+B
SN is normal to OP and PM is normal to OS . From the geometry of the figure,
`OS^(2)=ON^(2)+SN^(2)`
but `ON=OP+PN=A+Bcostheta`
`SN=Bsintheta`
`OS^(2)=(A+Bcostheta)^(2)+(Bsintheta)^(2)`
or , `R^(2)=A^(2)+B^(2)+2ABcostheta`
`R=sqrt(A^(2)+B^(2)+2ABcostheta)`
In `DeltaOSN SN =OS `sinalpha=Rsinalpha,and`
in `DeltaPSN,SN=PSsintheta=Bsintheta`
Therefore , `R sinalpha=Bsintheta`
or `(R)/(sintheta)=(B)/(sinalpha)`
similarly
`PM=Asinalpha=Bsinbeta`
or `(A)/(sinbeta)=(B)/(sinalpha)`
or Combining Eqs . (4.24b) and (4.24c),we get
`(R)/(sintheta)=(A)/(sinbeta)=(B)/(sinalpha)`
Using Eq . (424a).
or , `tanalpha=(SN)/(OP+PN)=(Bsintheta)/(A+Bcostheta)`
Equation (4.24a) gives the magnitude of the resultant and Eqs . (4.24e) and (4.24f) its direction. Equantion (4.24a) is known as the Law of cosines and Eq . (4.24d) as the Law of sines.