# Vector Problem ? Find windspeed

An airplane heads N80W with an airspeed of 680.0 km/h. Measurements made from the ground indicate that the plane’s groundspeed is 650.0 km/h at N85W. Find the windspeed and wind direction. [7 marks]

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.

#### Susanne217

An airplane heads N80W with an airspeed of 680.0 km/h. Measurements made from the ground indicate that the plane’s groundspeed is 650.0 km/h at N85W. Find the windspeed and wind direction. [7 marks]

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.
May I surgest you take a piece of paper and draw the situation like a simple geometric math problem! Then its often easier to solve.

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