Question

at what angle do the forces p+q and p-q act so that their resultant is under root 3p square +q square

Answer 1

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Answer 2

Answer:

Angle will be 60^{\circ}  

Explanation:

$3 p^{2}+q^{2}=(p+q)_{2}^{2}+(p-q)^{2}+2(p+q)(p-q) \cos \theta$

According to the formula of $(a+b)^{2}$ and $(a-b)^{2}$ we can write it as following-

$3\mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{p}^{2}+\mathrm{q}^{2}+2 \mathrm{pq}+\mathrm{p}^{2}+\mathrm{q}^{2}-2 \mathrm{pq}+2\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right) \cos \theta$

$3 p^{2}+q^{2}=2 p^{2}+2 q^{2}+2\left(p^{2}-q^{2}\right) \cos \theta$ [+2pq and -2pq will be cancel]

$\begin{array}{l}{\mathrm{p}^{2}-\mathrm{q}^{2}=2\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right) \cos \theta} \\ {\cos \theta=\mathrm{p}^{2}-\mathrm{q}^{2} / 2\left(\mathrm{p}^{2}-\mathrm{q}^{2}\right)} \\ {\cos \theta=1 / 2}\end{array}$

Theta$(\theta)=60^{\circ}$