A plane flies to city A, located 175km away in a direction 30 degrees north of east. Next it flies for?

Question

150km 20 degrees west of north, to city B. Finally the plane floes 190km due west to city C. Find the location of city C relative to the location of the starting point. The starting point is at the origin. The answer in the back of the book is 245 km at 21.4 degrees W of N. Will someone please help me how they arrived at this answer?

Answer 1

change them all to vectors using the angles and magnitudes (distance traveled), then add them and get the magnitude and angle (with respect to the nearest axis) of the resulting vector.

or you could spend the rest of your life trying to figure it out using geometry.

Answer 2

Draw the diagram. Then draw a line from b perpedicular to the y-axis at D. Call the origin O. In the triangle ODB we have BD/150 =sin 20, so BD = 150sin 20 = 51.303.
So DC =BD + 190 = 241.303
OD = 150 cos 20 = 140.954
Since ODC is also a right triangle we can us the pythgorean theorem to get OC = sqrt(241.303^2+140.958^2) = 279.456
The angle DOC is arctan (140.954/241.303)= 30.29 degrees west of North.

Unless you have given me the wrong numbers, your answer book is wrong. If you draw the picture to scale, you can see that the angle cannot possibly be 21.4 degrees.

Answer 3

Resolve the east/west and north/south travel.
To A: 175cos30 = 151 east and 175sin30 = 88 north
To B: 150cos20 = 141 west and 150sin20 = 51 north
To C: 190 west

Total north = 88 + 51 =139
Total west = 190 + 141 - 151 = 180

180 km west and 139 km north.

Alternatively, C is invtan(180/139) = 52 deg west of north and
180sin52 = 142 km away


That hurt, nemahknat