At what angle should the two forces 2P and sq | vedclass

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Question

(A) The resultant force $R$ of two vectors $A$ and $B$ acting at an angle $	heta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = 2P$

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) The resultant force $R$ of two vectors $A$ and $B$ acting at an angle $	heta$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given $A = 2P$,$B = \sqrt{2}P$,and $R = P\sqrt{10}$.Substituting these values into the formula:$P\sqrt{10} = \sqrt{(2P)^2 + (\sqrt{2}P)^2 + 2(2P)(\sqrt{2}P) \cos 	heta}$.Squaring both sides:$10P^2 = 4P^2 + 2P^2 + 4\sqrt{2}P^2 \cos 	heta$.$10P^2 = 6P^2 + 4\sqrt{2}P^2 \cos 	heta$.$4P^2 = 4\sqrt{2}P^2 \cos 	heta$.$1 = \sqrt{2} \cos 	heta$.$\cos 	heta = \frac{1}{\sqrt{2}}$.Therefore,$	heta = 45^{\circ}$.

At what angle should the two forces $2P$ and $\sqrt{2}P$ act so that the resultant force is $P\sqrt{10}$ (in $^{\circ}$)?

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