Three forces act on a particle at a point, as shown in diagram. Two of the forces have magnitude 12 N each, and the third force has magnitude 14 N. The angles between the forces are as indicated (75°). Find the magnitude and direction of the resultant force.

• Force $F_1 = 12 \, \text{N}$: The diagram shows an angle of $75^\circ$ with the positive y-axis in the first quadrant. The angle with the positive x-axis is $\theta_1 = 90^\circ - 75^\circ = 15^\circ$. $F_{1x} = 12 \cos(15^\circ) \approx 12 \times 0.9659 = 11.5908 \, \text{N}$ $F_{1y} = 12 \sin(15^\circ) \approx 12 \times 0.2588 = 3.1056 \, \text{N}$

• Force $F_2 = 12 \, \text{N}$: The diagram shows an angle of $75^\circ$ with the positive y-axis in the second quadrant. The angle with the positive x-axis is $\theta_2 = 90^\circ + 75^\circ = 165^\circ$. $F_{2x} = 12 \cos(165^\circ) = 12 (-\cos(15^\circ)) \approx 12 \times (-0.9659) = -11.5908 \, \text{N}$ $F_{2y} = 12 \sin(165^\circ) = 12 (\sin(15^\circ)) \approx 12 \times 0.2588 = 3.1056 \, \text{N}$

• Force $F_3 = 14 \, \text{N}$: The diagram shows an angle of $75^\circ$ with the negative y-axis in the third quadrant. The angle with the negative x-axis is $90^\circ - 75^\circ = 15^\circ$. The angle with the positive x-axis is $\theta_3 = 180^\circ + 15^\circ = 195^\circ$. $F_{3x} = 14 \cos(195^\circ) = 14 (-\cos(15^\circ)) \approx 14 \times (-0.9659) = -13.5226 \, \text{N}$ $F_{3y} = 14 \sin(195^\circ) = 14 (-\sin(15^\circ)) \approx 14 \times (-0.2588) = -3.6232 \, \text{N}$

Rounding to three significant figures, the magnitude is $R \approx \mathbf{13.8 \, \text{N}}$.

The angle $\theta$ of the resultant force with the positive x-axis is given by $\tan \theta = \frac{R_y}{R_x}$.

Since $R_x$ is negative and $R_y$ is positive, the resultant force is in the second quadrant.
The reference angle $\alpha_{ref} = \arctan(|-0.19138|) \approx 10.83^\circ$.
The angle from the positive x-axis (measured counter-clockwise) is $\theta = 180^\circ - \alpha_{ref}$.

Rounding to three significant figures, the direction is $\theta \approx \mathbf{169^\circ}$ counter-clockwise from the positive x-axis.

The magnitude of the resultant force is $\boxed{\text{13.8 N}}$ and its direction is $\boxed{\text{169}^\circ}$ counter-clockwise from the positive x-axis.