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Two forces are such that the sum of their magnitudes is $18 \mathrm{~N}$ and their resultant is $12 \mathrm{~N}$ which is perpendicular to the smaller force. Then the magnitudes of the forces are
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$13 \mathrm{~N}, 5 \mathrm{~N}$
$13 \mathrm{~N}, 5 \mathrm{~N}$
Use $\tan \alpha=\frac{P \sin \theta}{Q+P \cos \theta} \Rightarrow \tan 90^{\circ}=\frac{P \sin \theta}{Q+P \cos \theta}=\infty$
$\therefore Q+P \cos \theta=0 \Rightarrow P \cos \theta=-Q$
$\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta} \mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{Q}^2}$ or $\mathrm{R}=\sqrt{\mathrm{P}^2-\mathrm{Q}^2}=12$
$144=(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})$ or $\mathrm{P}-\mathrm{Q}=144 / 18=8 \quad \therefore \mathrm{P}=13 \mathrm{~N}$ and $\mathrm{Q}=5 \mathrm{~N}$
$\therefore Q+P \cos \theta=0 \Rightarrow P \cos \theta=-Q$
$\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta} \mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{Q}^2}$ or $\mathrm{R}=\sqrt{\mathrm{P}^2-\mathrm{Q}^2}=12$
$144=(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})$ or $\mathrm{P}-\mathrm{Q}=144 / 18=8 \quad \therefore \mathrm{P}=13 \mathrm{~N}$ and $\mathrm{Q}=5 \mathrm{~N}$
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