Aship leaves the island of guam and sails a distance 255 km at an angle 49.0 o north of west. part a: in which direction must it now head so that its resultant displacement will be 125 km directly east of guam? (express your answer as an angle measured south of east) part b: how far must it sail so that its resultant displacement will be 125 km directly east of guam?

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

he made 11 calls in one day

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000