Hello, Mr_Green!

I must assume you are familiar with vectors.

A boat heads in the direction N 40° W at a speed of 24 knots.

The current runs at 6 knots in the direction N 25° E.

Find the resultant speed and direction of the boat. Code:

E 6·sin25° C
+ - - - - - *
: /
6·cos25° : / 6
: /
:25°/
: / 24·sin24°
B * - - - - - - + D
\ :
\ :
\ : 24·cos24°
24 \ :
\40°:
\ :
*
A

The boat is moving from point A to point B.

. . Its horizontal component is: DB = -24·sin24°

. . Its vertical component is: AD = 24·cos24°

The current is moving from point B to point C.

. . Its horizontal component is: EC = 6·sin25°

. . Its vertical component is: BE = 6·cos25°

The resultant vector is AC.

We must determine its magnitude and direction.

The horizontal component of AC is: .-24·sin24° + 6·sin25° .≈ .-7.226

The vertical component of AC is: .24·cos24° + 6·cos25° .≈ .27.363

. . . . . . . . . . . . . . . . . . . . . ________________

The magnitude is: .|AC| .= .√(-7.226)² + 27.363² .= .28.30103965

. . Therefore, the speed of the boat is about 28.3 knots.

The direction angle is given by: .tanθ .= .7.226/27.363 . → . θ .= .14.79292136°

. . Therefore, the direction is about: .N 14.8° W