The thrust of an airplane's engines produces a speed of 361 mph in still air. The wind velocity is given by (42, -7). In what direction should the airplane head to fly due south? Give your answer as an angle from due south rounded to three significant digits.
I tried drawing a rectangle with sides of 361 and 7√37 (I got this by doing √(42² + (-7)²) )
solving for the hypotenuse, I did √(361² + (7√37)²) = 363.5024
and then i did cos(theta) = adjacent /hypotenuse = 361/363.5024
with cos = 6.73
but this answer was counted as wrong :/
What am I doing wrong?
Thanks :]
I tried drawing a rectangle with sides of 361 and 7√37 (I got this by doing √(42² + (-7)²) )
solving for the hypotenuse, I did √(361² + (7√37)²) = 363.5024
and then i did cos(theta) = adjacent /hypotenuse = 361/363.5024
with cos = 6.73
but this answer was counted as wrong :/
What am I doing wrong?
Thanks :]
-
OK so you want a picture something like this:
.....A
......
......
......
......
......
B...C
.....D
A, C and D are all meant to be on a straight vertical line. AB is the plane engine velocity vector (length 361) and BD is the wind vector (42, -7), which I assume means +42 in the x (east) direction and -7 in the y (south) direction, as I've drawn it there. AD is the final velocity vector, which needs to be due south. C is the point due east of B on the AD line, so there are right angles at C.
If CAB is the angle at A, then sin(CAB) = BC / AB = 42 / 361
=> CAB = 6.68 degrees
Interestingly the wind north/south component (the -7) doesn't affect the answer at all. All we need is the x (east-west) component of the wind to cancel out the x (east-west) component of the plane engine velocity.
I think your method was wrong because you can't assume that the wind speed is perpendicular to the plane speed vector, so you can't add them using pythagoras' theorem.
.....A
......
......
......
......
......
B...C
.....D
A, C and D are all meant to be on a straight vertical line. AB is the plane engine velocity vector (length 361) and BD is the wind vector (42, -7), which I assume means +42 in the x (east) direction and -7 in the y (south) direction, as I've drawn it there. AD is the final velocity vector, which needs to be due south. C is the point due east of B on the AD line, so there are right angles at C.
If CAB is the angle at A, then sin(CAB) = BC / AB = 42 / 361
=> CAB = 6.68 degrees
Interestingly the wind north/south component (the -7) doesn't affect the answer at all. All we need is the x (east-west) component of the wind to cancel out the x (east-west) component of the plane engine velocity.
I think your method was wrong because you can't assume that the wind speed is perpendicular to the plane speed vector, so you can't add them using pythagoras' theorem.