Question

A ship travels 200 miles due west, then adjusts its course 30° north of west. The ship continues on this course for 30 miles. Approximately how far is the ship from where it began?

Answer 1

Answer:

$174.66\ mi$  or $175\ mi$

Step-by-step explanation:

we know that

Applying the law of cosines

$c^{2}=a^{2}+b^{2}-2(a)(b)cos(C)$

In this problem we have

$a=200\ mi$

$b=30\ mi$

$C=30\°$

substitute the values

$c^{2}=200^{2}+30^{2}-2(200)(30)cos(30\°)$

$c^{2}=30,507.695$

$c=174.66\ mi$

Answer 2

Answer:

Approximately the ship is 174.67 m far from where it began

Step-by-step explanation:

Let positive x axis represent east and positive y axis represent north.

A ship travels 200 miles due west.

Displacement, s₁ = -200 i

Then adjusts its course 30° north of west. The ship continues on this course for 30 miles.

Displacement, s₂ = 30 cos 30 i + 30 sin 30 j = 25.98 i + 15 j

Total displacement = s₁ + s₂ = -200 i +25.98 i + 15 j = -174.02 i + 15 j

$\texttt{Magnitude =}\sqrt{(-174.02)^2+(15)^2}=174.67m$

Approximately the ship is 174.67 m far from where it began