A yacht is moving at 10kmh in a south easterly direction and encounters a 3kmh current from the north. Find the actual speed and direction of the yacht.

Answer 1

Using trig:
let the speed be x km/h
after making your diagram,
x^2 = 10^2+3^2 - 2(10)(3)cos 135°
= 151.426...
x = 12.3 km/h

angle:
using the sine law:
sinØ/3 =sin135/12.3055
sinØ = .17238...
Ø = 9.9266..
direction angle would be S (45-9.93)° E
direction would be S 35.07° E

If you know vectors

r = (10cos315, 10sin315) + (3cos270, 3sin270)
= (10√2/2, -10sin√2/2) + (0, -3)
= (7.07107, -7.07107) + (0,-3)
= (7.07107, -10.07107)

magnitude = √(7.07107^2+(-10.07107)^2)
= 12.3 , (same as above)

tanØ = -10.07107/7.07107 = -1.42426..
Ø = -54.93°
which would put in quad IV as 305.07° or S 35.07° E