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The resultant of two forces $\mathrm{P~N}$ and $3 \mathrm{~N}$ is a force of $7 \mathrm{~N}$. If the direction of $3 \mathrm{~N}$ force were reversed, the resultant would be $\sqrt{19} \mathrm{~N}$. The value of $\mathrm{P}$ is
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The correct answer is:
$5 \mathrm{~N}$
$5 \mathrm{~N}$
$\begin{aligned}
& 7^2=P^2+3^2+2 \times 3 \times P \cos \theta \quad \dots(1)\\
& (\sqrt{19})^2=P^2+(-3)^2+2 \times(-3) \times P \cos \theta \quad \dots (2)
\end{aligned}$
adding we get
$68=2 P^2+18 \Rightarrow P=5$.
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