1. ## Vectors

A plane flies on a bearing of 120 degrees at a constant speed of 550 km/h. If the velocity of the wind is 50 km/h on a bearing 220 degrees, what is the velocity of the plane with respect to the ground?

Textbook Answer: 543.5 km/h at a bearing of 125.2 degrees

My answer: 543.6 km/h at a bearing of 175 degrees

My work:

Let vector w represent 50 km/h
Let vector v represent 550 km/h
Let vector r represent the resultant

I'm using component vectors. . .

w = (50cos220 , 50sin220)
w = (-38.3 , -32.1)

v = (550cos120 , 550sin120)
v = (-275 , 476.3)

r = (-38.3 , -32.1) + (-275 , 476.3)
r = (-313.3 , 444.2)

To find the resultant:
| r | = (-313.3)^2 + (444.2)^2
| r | = (sqrt)295 470.53
| r | = 543.6 km/h

To find the angle:
tanx = 444.2/-313.3
x = 55 degrees

120 + 55 = 175 (To find the bearing)

----------------What did I do wrong? How can I get the 125 degree instead of 175 degrees?

2. Originally Posted by Macleef
...
To find the angle:
tanx = 444.2/-313.3
x = 55 degrees

120 + 55 = 175 (To find the bearing)

----------------What did I do wrong? How can I get the 125 degree instead of 175 degrees?
With $\tan(x)=\frac{44.2}{-313.3}$ you calculate the bearing directly:

$\tan(x)=\frac{44.2}{-313.3}~\implies~ x=-54.8^\circ$

Since bearings are given as positive numbers add 180° and you'll get the result given by your book.