#### Answer

(a) The lift force $L=5.78\times10^4N$
(b) Air resistance $R=2.08\times10^4N$

#### Work Step by Step

Let's consider the lift force $\vec{L}$
- Vertical component: $L_y=L\cos21=0.93L$
- Horizontal component: $L_x=L\sin21=0.36L$
(a) To find $L$, we can rely on the fact that we already know $W=mg=53800N$
The vertical forces do not affect the movement of the helicopter here. In other words, $\sum F_y=0$
And since there are only 2 vertical forces $m\vec{g}$ and $\vec{L}_y$, they cancel out each other. $$L_y=0.93L=53800N$$ $$L=57849.5N=5.78\times10^4N$$
(b) As the helicopter moves at a constant velocity, the horizontal forces also balance each other.
Here, there are 2 horizontal forces: $\vec{R}$ and $\vec{L}_x$
$$R=L_x=0.36L=20825.8N=2.08\times10^4N$$