Find the sum of the given vectors displaystyle{50}{N}angle{320}^{circ}+{170}{N}angle{150}^{circ} Rx = ? Ry = ? R = ? theta = ?

Step 1
Assume the vectors as vector a and vector b. Represent the vectors on the graph.
${\displaystyle \overrightarrow{a}=50N\mathrm{\angle }{320}^{\circ }}$
${\displaystyle \overrightarrow{b}=170N\mathrm{\angle }{150}^{\circ }}$

Step 2
Now, resolve the vectors a and b to get $R_x\text{and}{R}_y.$

${\displaystyle \left|{\overrightarrow{a}}_x\right|=-50\cos {50}^{\circ }}$
${\displaystyle \left|{\overrightarrow{a}}_y\right|=50\sin {50}^{\circ }}$
${\displaystyle \left|{\overrightarrow{b}}_x\right|=170\cos {60}^{\circ }}$
${\displaystyle \left|{\overrightarrow{b}}_y\right|=-170\sin {60}^{\circ }}$
${\displaystyle R_x=\left|{\overrightarrow{a}}_x\right|+\left|{\overrightarrow{b}}_x\right|}$
${\displaystyle R_x=-50\cos {50}^{\circ }+170\cos {60}^{\circ }}$
${\displaystyle R_x=52.86}$
${\displaystyle R_y=\left|{\overrightarrow{a}}_y\right|+\left|{\overrightarrow{b}}_y\right|}$
${\displaystyle R_y=50\sin {50}^{\circ }-170\sin {60}^{\circ }}$
${\displaystyle R_y=-108.92}$
Step 3
Finally, determine the resultant vector R and the direction theta using the resultant vectors $R_x\text{and}{R}_y.$
${\displaystyle R=\sqrt{R_x^2+R_y^2}}$
${\displaystyle =\sqrt{{\left(52.86\right)}^2+{(-108.92)}^2}}$
${\displaystyle =\sqrt{14657.746}}$
$=121.07$
Hence, $R=121.07$
${\displaystyle \theta ={\tan }^{-1}\left(\frac{R_y}{R_x}\right)}$
${\displaystyle ={\tan }^{-1}\left(\frac{-180.95}{52.86}\right)}$
${\displaystyle =-{64.11}^{\circ }}$
${\displaystyle ={295.89}^{\circ }}$
${\displaystyle \approx {296}^{\circ }}$
Hence, theta ${\displaystyle ={296}^{\circ }}$ from horizontal axis.