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

• Correct answer
  A

  $5 \mathrm{~N}$
• B

  $6 \mathrm{~N}$
• C

  $3 \mathrm{~N}$
• D

  $4 \mathrm{~N}$

Solution

$\begin{aligned} & 7^2=P^2+3^2+2 \times 3 \times P \cos \theta \quad \dots(1)\\ & (\sqrt{19})^2=P^2+(-3)^2+2 \times(-3) \times P \cos \theta \quad \dots (2) \end{aligned}$ adding we get $68=2 P^2+18 \Rightarrow P=5$.

Hence, the correct answer is option: 'A'.