the magnitude and direction of two vectors are shown in the diagram. what is the magnitude of their sum?

Explanation:

Step1: Write vectors in component - form

For the vector with magnitude $A = 2$ and direction $\theta_1=135^{\circ}$, its components are $A_x = 2\cos135^{\circ}=2\times(-\frac{\sqrt{2}}{2})=-\sqrt{2}$ and $A_y = 2\sin135^{\circ}=2\times\frac{\sqrt{2}}{2}=\sqrt{2}$. For the vector with magnitude $B = 4$ and direction $\theta_2 = 45^{\circ}$, its components are $B_x=4\cos45^{\circ}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$ and $B_y = 4\sin45^{\circ}=4\times\frac{\sqrt{2}}{2}=2\sqrt{2}$.

Step2: Find the components of the sum vector

The $x$ - component of the sum vector $\vec{R}$ is $R_x=A_x + B_x=-\sqrt{2}+2\sqrt{2}=\sqrt{2}$. The $y$ - component of the sum vector $\vec{R}$ is $R_y=A_y + B_y=\sqrt{2}+2\sqrt{2}=3\sqrt{2}$.

Step3: Calculate the magnitude of the sum vector

The magnitude of a vector $\vec{R}$ with components $(R_x,R_y)$ is given by $R=\sqrt{R_x^{2}+R_y^{2}}$. Substitute $R_x=\sqrt{2}$ and $R_y = 3\sqrt{2}$ into the formula: $R=\sqrt{(\sqrt{2})^{2}+(3\sqrt{2})^{2}}=\sqrt{2 + 18}=\sqrt{20}=2\sqrt{5}$.

Answer:

$2\sqrt{5}$