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

Approximately the ship is 174.67 m far from where it began

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 =}√((-174.02)^2+(15)^2)=174.67m`

Approximately the ship is 174.67 m far from where it began

Answer 2

`174.66\ mi` or
`175\ mi`

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`