1. **State the problem:** A body of mass 2 kg is held in equilibrium by two strings AP and AR. AP is inclined at 56° to the upward vertical, and AR is horizontal. We need to find the tensions $T_1$ (in AP) and $T_2$ (in AR). 2. **Given data:** - Mass $m = 2$ kg - Gravitational acceleration $g = 10$ m/s$^2$ - Angle of AP with vertical $\theta = 56^\circ$ - AR is horizontal 3. **Calculate the weight of the body:** $$ W = mg = 2 \times 10 = 20 \text{ N} $$ 4. **Set up equilibrium conditions:** Since the body is in equilibrium, the sum of forces in both vertical and horizontal directions must be zero. - Vertical forces: $$ T_1 \cos 56^\circ = W = 20 $$ - Horizontal forces: $$ T_1 \sin 56^\circ = T_2 $$ 5. **Calculate $T_1$:** $$ T_1 = \frac{20}{\cos 56^\circ} $$ Using $\cos 56^\circ \approx 0.5592$, $$ T_1 = \frac{20}{0.5592} \approx 35.77 \text{ N} $$ 6. **Calculate $T_2$:** $$ T_2 = T_1 \sin 56^\circ $$ Using $\sin 56^\circ \approx 0.8290$, $$ T_2 = 35.77 \times 0.8290 \approx 29.65 \text{ N} $$ **Final answer:** $$ T_1 \approx 35.77 \text{ N}, \quad T_2 \approx 29.65 \text{ N} $$