# Question regarding resultant of force vectors

This problem is being asked in the regards of the following question

The resultant of two forces has magnitude $$20\hspace{1mm}N$$. One of the forces is of magnitude $$20\sqrt{3}\hspace{1mm}N$$ and makes an angle of $$30^\circ$$ with the resultant. Then what is the magnitude of the other force?

The solution for the above question in my textbook is given as follows

Let $$P$$ be the unknown force. $$P^2 = (20\sqrt{2})^2 + 20^2 - 2(20)(20\sqrt{3})\cos(30^\circ)$$ $$1600 - 1200 = 400$$

Therefore, $$P = \sqrt{400} = 20\hspace{1mm}N$$

My issue with the above solution is that first why is the term containing the cosine subtracted not added doesn't the formula add the cosine term.

Another question since the angle is given between the vector and the resultant and not between the 2 vectors how can it be used in the formula to give us the answer.

Any help would be highly appreciated!

• Have you tried drawing a diagram of the problem? The law of cosines is $\mathbf{A}\cdot\mathbf{B} = A^{2} + B^{2} - 2AB\cos{(C)}$, so the minus sign does belong there. Commented Apr 3 at 10:50
• This does not use the law of cosines rather it uses the formula for the resultant of two vectores i.e. $R = \sqrt{A^2 + B^2 + 2AB\cos(\theta)}$ Commented Apr 3 at 10:53
• @kandb Yeah thanks now i understood it thanks for the help, could you post your comment as an answer so that i could accept it Commented Apr 3 at 11:06
• I jsut did :) Good luck! Commented Apr 3 at 11:06
• Thanks for the help! Commented Apr 3 at 11:07

Suppose the two forces are $$\mathbf{F}_{1}$$ and $$\mathbf{F}_{2}$$ so that the resultant is $$\mathbf{F} = \mathbf{F}_{1} + \mathbf{F}_{2}$$. Then we have that $$\mathbf{F}_{2} = \mathbf{F} - \mathbf{F}_{1}$$, so that
$$\begin{equation*} F_{2} = \sqrt{(\mathbf{F} - \mathbf{F}_{1})\cdot(\mathbf{F} - \mathbf{F}_{1})} = \sqrt{F_{1}^{2} + F^{2} - 2\mathbf{F}_{1}\cdot\mathbf{F}} = \sqrt{F_{1}^{2}+F^{2} - 2F_{1}F\cos{(\theta)}}. \end{equation*}$$
$$\begin{equation*} F = \sqrt{\left(\mathbf{F}_{1}+\mathbf{F}_{2}\right)\cdot \left(\mathbf{F}_{1}+\mathbf{F}_{2}\right)} = \sqrt{F_{1}^{2}+F_{2}^{2} + 2F_{1}F_{2}\cos{(\theta')}}. \end{equation*}$$