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