Answer:
130 km at 35.38 degrees north of east
Explanation:
Suppose the HQ is at the origin (x = 0, y = 0)
So the coordinates of the helicopter after the 1st flight is
[tex]x_1 = -120cos70^o = -41.04 km[/tex]
[tex]y_1 = -120sin70^o = -112.763 km[/tex]
After the 2nd flight its coordinate would be:
[tex]x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km[/tex]
[tex]y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km[/tex]
So in order to fly back to its HQ it must fly a distance and direction of
[tex]s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km[/tex]
[tex]tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71[/tex]
[tex]\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o[/tex] north of east