How do I work out this bearings problem?A helicopter flies 15km on a bearing of 053, then 7km on a bearing of 112. How far north is the helicopter from its starting position?
This problem needs some thought since when you use directions North and bearing 0 degrees is at the top of the page But, in math, if you use degrees, 0 is usually to the right. To translate, so your calculator will give the correct answer, subtract the bearings from 90 degrees. So 90 degrees minus 53 degrees equals 37 degrees. The distance to the east is equal to cosine of 37 degrees times 15km = 12 km Sine of 37 degrees times 15km equals 9km So the first leg he ended up 12 km east and 9 km north of original position. Then to translate the 112 degrees, subtract it from 90 to get -22 degrees Cosine of -22 times 7km equals 6.5 km and Sin -22 times 7km equals -2.6km. The second leg took him 6.5km farther east and 2.6km south since the 2.6 was negative, it is in the direction south. Next, add the two together Cos + Cos for the travel east and you get 12 + 6.5 so he ended up 18.5km east of the starting point Add the Sines for the north/south directions. 9 -2.6=6.4 km north of starting position.
