# A \vec A makes an angle of 20 ^o and \vec B makes an angle of 110^o with the x-axis. The...

## Question:

A {eq}\vec A {/eq} makes an angle of {eq}20 ^o {/eq} and {eq}\vec B {/eq} makes an angle of {eq}110^o {/eq} with the {eq}x {/eq}-axis. The magnitudes of these vectors are {eq}3 m {/eq} and {eq}4 m {/eq} respectively. Find the resultant.

Vectors have magnitude and direction. Adding two or more vectors is not quite as straightforward as adding scalar quantities. We need to decompose first the vectors into its components and add them, respectively.

We need to find first the components of the resultant displacement. These are given below.

{eq}\begin{align*} R_x = A_x + B_x &= A\cos\theta + B\cos\theta \\ &= 3~\mathrm{m}~\cos 20^\circ + 4~\mathrm{m}~\cos 110^\circ \\ &\approx 1.45~\mathrm{m} \\ \ \\ R_y = A_y + B_y &= A\sin\theta + B\sin\theta \\ &= 3~\mathrm{m}~\sin 20^\circ + 4~\mathrm{m}~\sin 110^\circ \\ &\approx 4.78~\mathrm{m} \end{align*}{/eq}

The magnitude of the resultant displacement can de be solved using the Pythagorean Theorem.

{eq}\begin{align*} R &= \sqrt{{R_x}^2 + {R_y}^2} \\ \ \\ &= \sqrt{(1.45~\mathrm{m})^2 + (4.78~\mathrm{m})^2} \\ \ \\ &\approx 5.0~\mathrm{m} \end{align*}{/eq}

Lastly, its direction is given by:

{eq}\theta = \arctan \left ( \dfrac{R_y}{R_x} \right ) =\arctan \left ( \dfrac{4.78~\mathrm{m}}{1.45~\mathrm{m}} \right ) \approx 73.1^\circ{/eq}