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