Answer
Horizontal component $=-25$
Vertical component $=25\sqrt{3}$
$\vec{v}=-20\vec{i}+25\sqrt{3}\vec{j}$
Work Step by Step
$|\vec{v}|=50$ $,$ $\theta=120^{\circ}$
Multiply the magnitude of the vector $\vec{v}$ by the cosine of $\theta$ to obtain its horizontal component:
Horizontal component $=|\vec{v}|\cos\theta=50\cos120^{\circ}=50\Big(-\dfrac{1}{2}\Big)=-25$
Multiply the magnitude of the vector $\vec{v}$ by the sine of $\theta$ to obtain its vertical component:
Vertical component: $=|\vec{v}|\sin\theta=50\sin120^{\circ}=50\Big(\dfrac{\sqrt{3}}{2}\Big)=25\sqrt{3}$
Write the vector in terms of $\vec{i}$ and $\vec{j}$. Do so by adding the product between its horizontal component and $\vec{i}$ and the product between its vertical component and $\vec{j}$:
$\vec{v}=-20\vec{i}+25\sqrt{3}\vec{j}$
