Given forces: F1 = 7 N and F2 = 10 N.

The resultant of two vectors is a single vector that produces the same effect as the original two vectors combined. It is the vector sum of the two vectors. Given forces: $F_1 = 7$ N and $F_2 = 10$ N. The magnitude of the resultant force $R$ of two forces $F_1$ and $F_2$ with an angle $\theta$ between them is given by the formula: $$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta}$$ a) The forces are parallel and act in the same direction. Step 1: Determine the angle between the forces. When forces are parallel and act in the same direction, $\theta = 0^\circ$. Step 2: Substitute the values into the formula. $$\begin{align} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(0^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (1)} \\ &= \sqrt{289 \text{ N}^2} \\ &= 17 \text{ N}\end{align}$$ Alternatively, for forces in the same direction, $R = F_1 + F_2$. $$R = 7 \text{ N} + 10 \text{ N} = 17 \text{ N}$$ The resultant force is $\boxed{\text{17 N}}$. b) The forces are parallel and act in opposite directions. Step 1: Determine the angle between the forces. When forces are parallel and act in opposite directions, $\theta = 180^\circ$. Step 2: Substitute the values into the formula. $$\begin{align} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(180^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (-1)} \\ &= \sqrt{149 \text{ N}^2 - 140 \text{ N}^2} \\ &= \sqrt{9 \text{ N}^2} \\ &= 3 \text{ N}\end{align}$$ Alternatively, for forces in opposite directions, $R = |F_2 - F_1|$. $$R = |10 \text{ N} - 7 \text{ N}| = 3 \text{ N}$$ The resultant force is $\boxed{\text{3 N}}$. c) The two forces are inclined at an angle of $60^\circ$ to each other. Step 1: Determine the angle between the forces. $\theta = 60^\circ$. Step 2: Substitute the values into the formula. $$\begin{align} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(60^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (0.5)} \\ &= \sqrt{149 \text{ N}^2 + 70 \text{ N}^2} \\ &= \sqrt{219 \text{ N}^2} \\ &\approx 14.80 \text{ N}\end{align}$$ The resultant force is $\boxed{\text{14.80 N}}$. d) The two forces are inclined at an angle of $160^\circ$. Step 1: Determine the angle between the forces. $\theta = 160^\circ$. Step 2: Substitute the values into the formula. $$\begin{align} R &= \sqrt{(7 \text{ N})^2 + (10 \text{ N})^2 + 2(7 \text{ N})(10 \text{ N}) \cos(160^\circ)} \\ &= \sqrt{49 \text{ N}^2 + 100 \text{ N}^2 + 140 \text{ N}^2 (-0.9397)} \\ &= \sqrt{149 \text{ N}^2 - 131.558 \text{ N}^2} \\ &= \sqrt{17.442 \text{ N}^2} \\ &\approx 4.18 \text{ N}\end{align}$$ The resultant force is $\boxed{\text{4.18 N}}$. Send me the next one 📸