Answer:
(a) Magnitude =3.18 km
(a) Angle =28.2°
Explanation:
(a) To find magnitude of this vector recognize
[tex]R_{x}=-2.8 km\\R_{y}=-1.5km[/tex]
Use Pythagorean theorem
[tex]R=\sqrt{(R_{x})^{2}+(R_{y})^{2} }\\ R=\sqrt{(-2.8)^{2}+(-1.5)^{2} }\\ R=3.18km[/tex]
(b)To find the angle use the trigonometric property
[tex]tan\alpha =\frac{opp}{adj}\\\ tan\alpha =\frac{R_{y} }{R_{x}}\\\alpha =tan^{-1}(\frac{(-1.5)}{(-2.8)})\\\alpha =28.2^{o}[/tex]
