So a bearing of N80°W = 90+80 =170°

and a bearing of N85°W = 90+85 = 175°

Let a = vector indicating airplane's speed and direction.

Let w = vector indicating wind's speed and direction.

Let g = vector indicating plane's speed and direction from the ground

a + w = g

w = g - a

w = 650 (cos(175), sin(175)) - 680 (cos(170), sin(170))

w = (650cos(175)-680cos(170), 650sin(175)-680sin(170))

w = (22.142718, -61.429528)

|w| = √(22.142718^2 + 61.429528^2) = 65.298445

w = 65.298445 (cos(θ), sin(θ))

sin(θ) = -61.429528/65.298445 = -0.94075

cos(θ) = 22.142718/65.298445 = 0.3391

tan(θ) = -61.429528/22.142718

Since cos(θ) > 0 and sin(θ) < 0, then terminal point of θ is in quadrant IV

arctan has range between -90 and 90 (i.e. values in quadrant IV and quadrant I)

So we can just take arctan to find θ

(Otherwise, when θ is in QII or QIII, then we have to add 180 to arctan)

θ = arctan (-61.429528/22.142718) = -70.18

Changing this back into a bearing, we get S19.82°E

So windspeed = 65.3 km/h at S19.82°E

This this correct? Is there a simpler way to solve this problem.