Two boys pull a toy with a force 10N and 15N inclined at 60° to each other. Did their resultant in magnitude and direction

We use the formula for the magnitude of the resultant of two forces $F_1$ and $F_2$ acting at an angle $\theta$ to each other:

$$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta}$$

Given $F_1 = 10 \text{ N}$, $F_2 = 15 \text{ N}$, and $\theta = 60^\circ$.

$$R = \sqrt{(10 \text{ N})^2 + (15 \text{ N})^2 + 2(10 \text{ N})(15 \text{ N}) \cos(60^\circ)}$$ $$R = \sqrt{100 \text{ N}^2 + 225 \text{ N}^2 + 2(150 \text{ N}^2)(0.5)}$$ $$R = \sqrt{100 \text{ N}^2 + 225 \text{ N}^2 + 150 \text{ N}^2}$$ $$R = \sqrt{475 \text{ N}^2}$$ $$R \approx 21.79 \text{ N}$$

Let $\alpha$ be the angle the resultant force makes with the $10 \text{ N}$ force. We can use the Law of Sines:

$$\frac{F_2}{\sin\alpha} = \frac{R}{\sin\theta}$$

Rearrange to solve for $\sin\alpha$:

$$\sin\alpha = \frac{F_2 \sin\theta}{R}$$

Substitute the values:

$$\sin\alpha = \frac{15 \text{ N} \times \sin(60^\circ)}{21.79 \text{ N}}$$ $$\sin\alpha = \frac{15 \times 0.8660}{21.79}$$ $$\sin\alpha = \frac{12.99}{21.79}$$ $$\sin\alpha \approx 0.5961$$

Now, find $\alpha$:

$$\alpha = \arcsin(0.5961)$$ $$\alpha \approx 36.59^\circ$$

The resultant force has a magnitude of approximately $21.79 \text{ N}$ and is directed at an angle of approximately $36.59^\circ$ with respect to the $10 \text{ N}$ force.

The final answer is $\boxed{\text{Magnitude: } 21.79 \text{ N, Direction: } 36.59^\circ \text{ with the } 10 \text{ N force}}$.