# A vector A has a magnitude of 48.0 m and points in a direction 20° below the positive x axis. A second vector, B, has a magnitude of 75 m and points in a direction 60.0° above the positive x axis?

## Using the component method of vector addition, how do you find the magnitude and direction of the vector C=A +B?

Aug 5, 2016

The magnitude of $C$ is 90.808 m, nearly, and the direction of $C$

makes ${31.234}^{o}$, with the positive direction of the x-axis.

#### Explanation:

The vector

$A = \left(x , y\right) = r \left(\cos \theta , \sin \theta\right) = 48 \left(\cos \left(- {20}^{o}\right) , \sin \left(- {20}^{o}\right)\right)$ and,

likewise, the vector

$B = 75 \left(\cos {60}^{o} , \sin {60}^{o}\right)$.

So, $A = \left(45.105 , - 16.417\right) \mathmr{and} B = \left(37 , 5 , 64.952\right)$, nearly.

Now, the components of $C = A + B$ are

$\left(45.105 , - 16.417\right) + \left(37 , 5 , 64.952\right)$

$= \left(45.105 + 37.5 , - 16.417 + 64.952\right)$

$= \left(82.605 , 48.535\right) = r \left(\cos \theta , \sin \theta\right) = \left(x , y\right)$

$= 96.808 \left(\cos {31.234}^{o} , \sin {31.234}^{o}\right)$,

using $r = \sqrt{{x}^{2} + {y}^{2}} , \cos \theta = \frac{x}{r} \mathmr{and} \sin \theta = \frac{y}{r}$ .

The magnitude of $C$ is 90.808 m, nearly, and the direction of $C$

makes ${31.234}^{o}$, with the positive direction of the x-axis.

Aug 5, 2016

Magnitude of resultant vector is $95.81 m$ and direction is ${30.44}^{o}$ above $x$-axis

#### Explanation:

As vector $A$ has magnitude of $48.0 m$ and points in a direction ${20}^{o}$ below positive $x$-axis (i.e. $- {20}^{o}$) and vector $B$ has magnitude of $75.0 m$ and points in a direction ${60}^{o}$ above positive $x$-axis.

Hence the angle between two vectors is $60 - \left(- 20\right) = {80}^{o}$ and hence

Magnitude of resultant vector is sqrt(48^2+75^2+2xx48xx75xxcos80^o

= $\sqrt{2304 + 5625 + 7200 \times 0.17365}$

= $\sqrt{2304 + 5625 + 1250.28} = \sqrt{9179.28} = 95.81 m$

Direction wll be given by $\alpha = {\tan}^{- 1} \left(\frac{75 \sin {80}^{o}}{48 + 75 \cos {80}^{o}}\right)$

= ${\tan}^{- 1} \left(\frac{75 \times 0.9848}{48 + 75 \times 0.17365}\right)$

= ${\tan}^{- 1} \left(\frac{73.86}{48 + 13.02375}\right) = {\tan}^{- 1} \left(\frac{73.86}{61.02375}\right)$

= ${\tan}^{- 1} 1.21035 = {50.44}^{o}$

But this is w.r.t. $A$ and hence direction is ${50.44}^{o} - {20}^{o} = {30.44}^{o}$ above $x$-axis