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(D) Let the two forces be $F_1$ and $F_2$. The angle between them is $	heta = 120^{\circ}$.Let the resultant $R = 10 \,kg$-wt be perpendicular to for

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$\frac{10}{\sqrt{3}} \,kg$-wt

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(D) Let the two forces be $F_1$ and $F_2$. The angle between them is $	heta = 120^{\circ}$.Let the resultant $R = 10 \,kg$-wt be perpendicular to force $F_1$.Using the formula for the angle $\alpha$ that the resultant makes with force $F_1$:$	an \alpha = \frac{F_2 \sin 	heta}{F_1 + F_2 \cos 	heta}$Since the resultant is perpendicular to $F_1$, $\alpha = 90^{\circ}$, so $	an 90^{\circ} = \infty$.This implies the denominator must be zero: $F_1 + F_2 \cos 120^{\circ} = 0$.$F_1 + F_2 (-0.5) = 0 \implies F_1 = 0.5 F_2 \implies F_2 = 2 F_1$.The magnitude of the resultant is given by $R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos 120^{\circ}$.$10^2 = F_1^2 + (2 F_1)^2 + 2 F_1 (2 F_1) (-0.5)$.$100 = F_1^2 + 4 F_1^2 - 2 F_1^2 = 3 F_1^2$.$F_1^2 = \frac{100}{3} \implies F_1 = \frac{10}{\sqrt{3}} \,kg$-wt.Thus, the force perpendicular to the resultant is $\frac{10}{\sqrt{3}} \,kg$-wt.

The resultant of two forces acting at an angle of $120^{\circ}$ is $10 \,kg$-wt and is perpendicular to one of the forces. That force is

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