An aircraft travels 250 km on a heading of 330 degrees. It then turns due east and travels 420 km. How far and on what heading did it end up from it's starting point?
Please help!!!
If you know Trig (law of cosines, law of sines), very easy. make a sketch. You know two sides, and the included angle. Use law of cosines to find the third side of the triangle, then use law of sines to find the third angle (and use geometry of the vectors to figure true heading).
Now, if you dont know trig.
break both vectors into N, E components.
250:
250cos330 N + 250sin330 E
adding that to the second leg, we get the resultant to be:
250cos330 N + (250sin330+420)E
then, the angle is... Theta= arc tan Ecomponent/N component
and the magnitude is 250cos330/cosTheta
Displacement = 250km[330o] + 420km[0o].
Disp. = 250*Cos330+250*sin330 + 420,
Disp. = 216.5 - 234.9i + 420.
Disp. = 636.5 - 234.9ii = 678.5km[-20.3],
Disp. = 678.5km[20.3o] S. of E.
To determine the distance and heading of the aircraft from its starting point, we can break down the problem into two components: the north-south (or vertical) displacement and the east-west (or horizontal) displacement.
First, let's find the north-south displacement. We know that the aircraft travels 250 km on a heading of 330 degrees. The angle between the heading and the north direction is 60 degrees (330 degrees - 270 degrees). To calculate the north-south displacement, we can use trigonometry.
north-south displacement = distance * sin(angle)
north-south displacement = 250 km * sin(60 degrees)
north-south displacement ≈ 250 km * 0.866 (approximately)
north-south displacement ≈ 216.5 km
Now, let's find the east-west displacement. After turning due east, the aircraft travels 420 km. Since the aircraft is now traveling along the east direction, the east-west displacement is equal to the distance traveled, which is 420 km.
Next, let's combine the north-south and east-west displacements to find the total displacement. We can use the Pythagorean theorem to calculate the magnitude of the displacement:
displacement = sqrt((north-south displacement)^2 + (east-west displacement)^2)
displacement = sqrt((216.5 km)^2 + (420 km)^2)
displacement ≈ sqrt(46922.25 km^2 + 176400 km^2)
displacement ≈ sqrt(223322.25 km^2)
displacement ≈ 472.7 km
Finally, let's find the heading of the aircraft from its starting point. We can use inverse trigonometry to calculate the angle between the displacement and the north direction:
heading = arctan(north-south displacement / east-west displacement)
heading = arctan(216.5 km / 420 km)
heading ≈ arctan(0.5167)
heading ≈ 27.5 degrees
Therefore, the aircraft ended up approximately 472.7 km away from its starting point, with a heading of 27.5 degrees.