Answer:
129.5km
9.78° North of east
Explanation:
Let say HQ is at the origin (x = 0, y = 0)
the coordinates of the helicopter after the 1st flight can be expressed below as
X1= -120 cos45=84.85km
Y1= -120 sin45= -84.85km
its coordinate after the 2nd flight can be expressed as
X2= x1 - 125sin20
= -84.85-42.75
=-127.6km
Y2= y1- 125cos60
=-84.5+62.5
=-22km
The distance and it's direction for it to fly back to its HQ can be calculated as
S= √(22^2 + 127.6^2 )
=129.48km
Tan θ= y2/y1
=22/127.6= 0.1724
Tan-1(0.1724)
= 9.78° North of east
Hence, the distance and direction it should fly is 129.48km and 9.78° North of east respectively