Find the sum of vector A and B as shown in the figure ,also find the direction of sum vector. Given `A=4"unit"` and `B=3 "unit"` .

Options

A. `sqrt(35)"unit"` , `alpha=tan^(-1)(0.472)=25.3^(@)`
B. `sqrt(25)"unit"` , `alpha=tan^(-1)(0.472)=25.3^(@)`
C. `sqrt(37)"unit"` , `alpha=tan^(-1)(0.472)=25.3^(@)`
D. `sqrt(37)"unit"` , `alpha=tan^(-1)(1)=45^(@)`

Answer

C

Explanation

According to the question ,we draw the following figure .

Resultant of the vectors A and B
`R=sqrt(A^(2)+B^(2) +2AB costheta)`
`=sqrt(16+9+2xx4xx3cos60^(@))=sqrt(37)unit`
`therefore`Direction of the sum vector ,
`tanalpha=(Bsintheta)/(A+Bcostheta)= (3sin60^(@))/(4+3cos60^(@))=0.472`
`alpha=tan^(-1)(0.472)=25.3^(@)`
Thus resultant of `|A|and |B|` is `sqrt(37)` unit at angle `25.3^(@)` from A in the direction shown in figure .