Two forces of magnitude 8 , N and 15 , N resp | vedclass

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(C) The resultant force $R$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos

Vedclass · Accepted Answer

$90$

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(C) The resultant force $R$ of two vectors $A$ and $B$ with an angle $	heta$ between them is given by the formula: $R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$.Given: $A = 8 \, N$,$B = 15 \, N$,and $R = 17 \, N$.Substituting these values into the formula:$17^2 = 8^2 + 15^2 + 2(8)(15) \cos 	heta$$289 = 64 + 225 + 240 \cos 	heta$$289 = 289 + 240 \cos 	heta$Subtracting $289$ from both sides:$0 = 240 \cos 	heta$$\cos 	heta = 0$Since $\cos 	heta = 0$,the angle $	heta = 90^{\circ}$.

Two forces of magnitude $8 \, N$ and $15 \, N$ respectively act at a point. If the resultant force is $17 \, N$,the angle between the forces is: (in $^{\circ}$)

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