heading 121 - 90 = 31 degrees south of east
South speed = 192 sin 31 + 15.9 = 114.8
East speed = 192 cos 31 = 164.5
tan of angle S of E = 114.8/164.5 = .6979
angle S of E = 34.9
so course made good is 90+34.9 =124.9 degrees
My Question is :
A plane with an air pf 192 miles per hour is headed on a bearing of 121 degrees. A north wind is blowing ( from north to south ) at 15.9 miles per hour. Find the ground speed and the actual bearing of the plane.
2 answers
V = 192[121o] + 15.9[180o]. mi/h.
X = 192*sin121 + 15.9*sin180 = 164.6. mi/h.
Y = 192*Cos121 + 15.9*Cos180 = -114.8 mi/h.
V = 164.6 - 114.8i = 200.7mi/h[-55.1o] = 200.7mi/h[55.1o] S. of E. = 200.7mi/h[145.1o] CW.
Tan A = X/Y
X = 192*sin121 + 15.9*sin180 = 164.6. mi/h.
Y = 192*Cos121 + 15.9*Cos180 = -114.8 mi/h.
V = 164.6 - 114.8i = 200.7mi/h[-55.1o] = 200.7mi/h[55.1o] S. of E. = 200.7mi/h[145.1o] CW.
Tan A = X/Y