Answer:
Magnitude of the vector is [tex]23.75\ \text{units}[/tex] and the direction is [tex]35.23^{\circ}[/tex]
Explanation:
Magnitude of first vector = [tex]|A| = 13\ \text{units}[/tex]
Angle = [tex]\theta_1=27^{\circ}[/tex]
Magnitude of second vector = [tex]|B| = 11\ \text{units}[/tex]
Angle = [tex]\theta_2=45^{\circ}[/tex]
x component of first vector
[tex]A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}[/tex]
y component of first vector
[tex]A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}[/tex]
x component of second vector
[tex]B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}[/tex]
y component of first vector
[tex]B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}[/tex]
Adding the magnitudes
[tex]C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}[/tex]
[tex]C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}[/tex]
Magnitude of the sum of the vectors would be
[tex]|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}[/tex]
The direction would be
[tex]\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}[/tex]
The magnitude of the vector is [tex]23.75\ \text{units}[/tex] and the direction is [tex]35.23^{\circ}[/tex]
