The problem in the image is asking in which direction the man in the figure will be pulled by his dogs, assuming both dogs are pulling with equal strength. It also asks to represent the resulting force using a vector.

Steps to Solve:

1. Determine the forces: The two dogs are pulling with equal force. Let's denote this force as $F$ for each dog.

2. Analyze the angles: The angle between the two force vectors (from the dogs) is given as $40^\circ$.

   • The first dog pulls along a horizontal axis.
   • The second dog pulls at an angle of $40^\circ$ above this horizontal axis.

3. Decompose the forces:

   • The force from the dog pulling horizontally is $F_1 = F \hat{i}$, where $\hat{i}$ is the unit vector in the horizontal direction.
   • The force from the second dog has two components: a horizontal component $F_2 \cos(40^\circ) \hat{i}$ and a vertical component $F_2 \sin(40^\circ) \hat{j}$.

4. Resultant force:

   • Add the horizontal components: $F_{\text{horizontal}} = F_1 + F_2 \cos(40^\circ) = F + F \cos(40^\circ)$.
   • The vertical component is just from the second dog: $F_{\text{vertical}} = F_2 \sin(40^\circ) = F \sin(40^\circ)$.

5. Magnitude of the resultant force: The magnitude of the resultant vector $\vec{F}_{\text{resultant}}$ is found using Pythagoras’ theorem:

   = \sqrt{\left( F + F \cos(40^\circ) \right)^2 + \left( F \sin(40^\circ) \right)^2}.$$

6. Direction of the resultant force: The direction of the resultant force can be found by calculating the angle $\theta$ it makes with the horizontal axis: [ \theta = \tan^{-1} \left( \frac{F_{\text{vertical}}}{F_{\text{horizontal}}} \right) = \tan^{-1} \left( \frac{F \sin(40^\circ)}{F +