Two forces are acting on an object. Force F1 has a magnitude of 75 N and acts at 53 degrees west of north. Force F2 has a magnitude of 100 N and acts at 37 degrees east of north. Calculate the resulta

We are given the following values from question 40: $\sin(53^\circ) = \cos(37^\circ) = 0.8$ and $\cos(53^\circ) = \sin(37^\circ) = 0.6$.
Let North be the positive y-axis and East be the positive x-axis.

For force $F_1 = 75 \, \text{N}$ at $53^\circ$ west of north:
The angle with the positive y-axis towards the negative x-axis is $53^\circ$.
The x-component is negative, and the y-component is positive.

$$F_{1x} = -F_1 \sin(53^\circ) = -75 \, \text{N} \times 0.8 = -60 \, \text{N}$$ $$F_{1y} = F_1 \cos(53^\circ) = 75 \, \text{N} \times 0.6 = 45 \, \text{N}$$

For force $F_2 = 100 \, \text{N}$ at $37^\circ$ east of north:
The angle with the positive y-axis towards the positive x-axis is $37^\circ$.
The x-component is positive, and the y-component is positive.

$$F_{2x} = F_2 \sin(37^\circ) = 100 \, \text{N} \times 0.6 = 60 \, \text{N}$$ $$F_{2y} = F_2 \cos(37^\circ) = 100 \, \text{N} \times 0.8 = 80 \, \text{N}$$

The resultant force $\vec{R}$ has components $R_x$ and $R_y$:

$$R_x = F_{1x} + F_{2x} = -60 \, \text{N} + 60 \, \text{N} = 0 \, \text{N}$$ $$R_y = F_{1y} + F_{2y} = 45 \, \text{N} + 80 \, \text{N} = 125 \, \text{N}$$

The magnitude of the resultant force is:

$$|\vec{R}| = \sqrt{R_x^2 + R_y^2} = \sqrt{(0 \, \text{N})^2 + (125 \, \text{N})^2} = \sqrt{15625 \, \text{N}^2} = 125 \, \text{N}$$

Since $R_x = 0$ and $R_y$ is positive, the resultant force is directed purely along the positive y-axis, which corresponds to North.

The given graphical scale is $1 \, \text{cm}$ for $10 \, \text{N}$ force.
Length of the arrow line = $\frac{\text{Magnitude of resultant force}}{\text{Force per cm}}$

$$\text{Length} = \frac{125 \, \text{N}}{10 \, \text{N/cm}} = 12.5 \, \text{cm}$$

The possible length of the arrow line is $12.5 \, \text{cm}$ and its direction is North.

Comparing this with the given options:
A. $12.5 \, \text{cm}$ to the North
B. $17.3 \, \text{cm}$ to $45^\circ$ North of East
C. $1.25 \, \text{cm}$ to the North
D. $1.73 \, \text{cm}$ to $45^\circ$ North of East

The calculated result matches option A.

The final answer is $\boxed{\text{A. } 12.5 \, \text{cm} \text{ to the North}}$.