A force of 40N acts at 35o to the horizontal together with a force of 25N acting at 110o to the horizontal. By considering force components OR by the triangle of forces determine the magnitude and direction of the resultant force.
1. If you use the parallelogram of forces you must calculate the length of a diagonal of the parallelogram. That is a side of a triangle of which you know 2 sides and the included angle = 180° - (110° - 35°) = 105°
Then use cosine rule
2. Consider the forces as vectors:
$\displaystyle \overrightarrow{F_1} = 40\ N(\cos(35^\circ)~, ~\sin(35^\circ))$
$\displaystyle \overrightarrow{F_2} = 25\ N(\cos(110^\circ)~, ~\sin(110^\circ))$
Now use vector addition to calculate the resulting force.
Use the components of the result vector to calculate the angle to the horizontal.
OK. As you may know the resulting vector is the sum of the two single forces:
$\displaystyle \overrightarrow{F_R} = \overrightarrow{F_1} +\overrightarrow{F_2} $
Plug in the terms of $\displaystyle \overrightarrow{F_1}$ and $\displaystyle \overrightarrow{F_2}$ of my previous post to calculate the components of $\displaystyle \overrightarrow{F_R}$
$\displaystyle \overrightarrow{F_R} = (40\ N \cdot \cos(35^\circ) + 25\ N \cdot \cos(110^\circ)~,~ 40\ N \cdot \sin(35^\circ) + 25\ N \cdot \sin(110^\circ)$
Let h and v denote the horizontal or vertical component of $\displaystyle \overrightarrow{F_R}$ then the magnitude of the resulting force is:
$\displaystyle |\overrightarrow{F_R}| = \sqrt{h^2 + v^2} \approx 52.37\ N$
The angle between the horizontal and $\displaystyle \overrightarrow{F_R}$ can be calculated by:
$\displaystyle \alpha = \arctan\left(\frac vh \right) \approx 62.458^\circ$
Abz, each force F applied at an angle has two components:
1. The horizontal component (denoted $\displaystyle F_{x}$)
2. The vertical component (denoted $\displaystyle F_{y}$)
Now, what we are trying to do is to find the sum of the horizontal components ($\displaystyle F_{x} = F_{x1} + F_{x2}$) and vertical components ($\displaystyle F_{y} = F_{y1} + F_{y2}$) of both of the given forces. Once we have these, we apply pythagorean theorem ($\displaystyle (F_{x})^2 +(F_{y})^2 = F^2$) to find the resultant force. This is what earboth has been doing. Do you have a problem finding $\displaystyle F_{x}$ or $\displaystyle F_{y}$ ? Do you understand what we're trying to do?