If $3 \mathrm{p}$ and $4 \mathrm{p}$ are resultant of a force $5 \mathrm{p}$, then the angle between $3 \mathrm{p}$ and $5 \mathrm{p}$ is

• A

  $\sin ^{-1}\left(\frac{3}{5}\right)$
• Correct answer
  B

  $\sin ^{-1}\left(\frac{4}{5}\right)$
• C

  $90^{\circ}$
• D

  None of these

Solution

$$ \quad \mathrm{Q}=\frac{\mathrm{R} \sin \alpha}{\sin (\alpha+\beta)} $$ $$ \begin{array}{l} \text { Also, }(5 \mathrm{P})^{2}=(4 \mathrm{P})^{2}+(3 \mathrm{P})^{2} \\ \quad+2(4 \mathrm{P})(3 \mathrm{P}) \cos (\alpha+\beta) \\ \Rightarrow 25 \mathrm{P}^{2}=16 \mathrm{p}^{2}+9 \mathrm{P}^{2}+24 \mathrm{P}^{2} \cos (\alpha+\beta) \\ \Rightarrow 24 \mathrm{P}^{2} \cos (\alpha+\beta)=0 \\ \Rightarrow \cos (\alpha+\beta)=0=\cos 90^{\circ} \\ \Rightarrow \alpha+\beta=90^{\circ} \end{array} $$
$$ \begin{array}{l} \text { Now, } 4 \mathrm{P}=\frac{5 \mathrm{P} \sin \alpha}{\sin 90^{\circ}} \\ \Rightarrow \sin \alpha=\frac{4}{5} \\ \Rightarrow \alpha=\sin ^{-1}\left(\frac{4}{5}\right) \end{array} $$

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