# How do you find the magnitude and direction for U: magnitude 140, bearing 160° V: magnitude 200, bearing 290°?

Jul 3, 2016

$U = \left(- 140 \cos {20}^{o} , 140 \sin {20}^{o}\right) = \left(- 131.56 , 47.883\right)$ and
$V = \left(- 200 \cos {10}^{o} , - 200 \sin {10}^{o}\right) = \left(- 196.96 , - 34.730\right)$.

#### Explanation:

If m is the magnitude and b is the bearing (with respect to x-axis),

the vector is $\left(m \cos b , m \sin b\right)$.

Here, $U = \left(140 \cos {160}^{o} , 140 \sin {160}^{o}\right)$

$= \left(- 140 \cos {20}^{o} , 140 \sin {20}^{o}\right)$

$= \left(- 131.56 , 47.883\right)$ and

$V = \left(299 \cos {190}^{o} , 200 \sin {190}^{o}\right)$

$= \left(- 200 \cos {10}^{o} , - 200 \sin {10}^{o}\right)$

$= \left(- 196.96 , - 34.730\right)$.