Let AB be the uniform bar of weight W suspended at rest by the two strings OA and O′B which make angles 36.9° and 53.1° respectively with the vertical. $\therefore \angle OAA' = 90^{\circ} - 36.9^{\circ} = 53.1^{\circ}$ Similarly $\angle O'BB' = 36.9^{\circ}$ $AB = 2\,\text{m}, AC = d\,\text{m}.$ Let $T_1 \text{ and } T_2$ be the tensions in the strings OA and O′B respectively and their rectangular components are shown in the figure.1As the rod is at rest, so the vector sum of the forces acting along A′B′ axis and ⊥ to it are zero i.e. $- T_1 \cos 53.1^{\circ} + T_2 \cos 36.9^{\circ} = 0 \ldots (i)$and $T_1 \sin 53.1^{\circ} + T_2 \sin 36.9^{\circ} - W = 0 \ldots (ii)$Taking the torques about A and equating the sum of torques to zero, we get$- (T_2 \sin 36.9^{\circ}) \times 2 + Wd = 0$or $T_2 = \frac{W d}{2 \sin 36.9^{\circ}} \ldots (iii)$∴ From (ii) and (iii), we get $\therefore T_1 = \frac{W}{\sin 53.1^{\circ}} \left(1 - \frac{d}{2}\right) \ldots (iv)$∴ From (i), (iii) and (iv), we get $T_1 \cos 53.1^{\circ} = T_2 \cos 36.9^{\circ}$or $\frac{W \left(1 - \frac{d}{2}\right)}{\tan 53.1^{\circ}} = \frac{W d}{2 \tan 36.9^{\circ}}$or $\frac{1 - \frac{d}{2}}{1.3319} = \frac{\frac{d}{2}}{0.7508}$or $1 - \frac{d}{2} = \frac{d}{2} \times \frac{1.3319}{0.7508}$$= \frac{d}{2} \times 1.7740 = 0.8870 d$or $0.5 d + 0.8870 d = 1$ or $d = \frac{1}{1.3870} = 0.721 \text{ m} = 72.1 \text{ cm}$