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

Explanation

Let OP and OQ represent the two vectors A and B making an angle `theta`. Then, using the parallelogram method of vector addition, OS represents the resultant vector R.
R= A+B
SN is normal to OP and OP and PM is normal to OS.
From the geometry of the figure.
`(OS)^(2)= (ON)^(2) + (SN)^(2)`
but ON = OP + PN = A + Bcos`theta`
SN = B`sintheta`
`(OS)^(2)= (A+Bcostheta)^(2)+(Bsintheta)^(2)`
or, `R^(2) = sqrt(A^(2)+B^(2)+2ABcostheta)`.............(4.24a)
In `Delta` OSN, SN= OS`sinalpha = Rsinalpha`, and in `Delta`PSN, SN= PS `sintheta`
Therefore, `Rsinalpha = Bsintheta`
or, `R/(sintheta) = B/(sinalpha)`................(4.24b)
Similarly,
PM= A`sinalpha= Bsinbeta`
or `A/(sinbeta)= B/(sinalpha)` ............(4.24c)
Combining Eqs. we get,
`R/(sintheta) = A/(sinbeta)=B/(sinalpha)`................(4.24d)
Using Eqs, we get
`sin alph= B/Rsintheta` ...............(4.24 e)
Where R is given by Eq.
or `tanalpha = (SN)/(OP+PN) = (Bsintheta)/(A+Bcostheta)`...............(4.24f)
Eqs. (4.24a) givestjhe magnitude of the resultant and Eqs. (4.24e) and (4.24f) its direction. Equation (4.24a) is known as the law of cosines and Eq. (4.24d) as the law of sines.