Resolve the following forces into their vertical and horizontal components:

Kijenzi cha mlalo ($F_x$) = $100 \sin 60^\circ$

$$F_x = 100 \times 0.8660$$ $$F_x = 86.60 \, \text{N}$$

Kijenzi cha wima ($F_y$) = $100 \cos 60^\circ$

$$F_y = 100 \times 0.5000$$ $$F_y = 50.00 \, \text{N}$$

Vijenzi vya (a) ni: Mlalo = $\boxed{\text{86.60 N}}$, Wima = $\boxed{\text{50.00 N}}$

Kijenzi cha mlalo ($F_x$) = $100 \sin 85^\circ$

$$F_x = 100 \times 0.9962$$ $$F_x = 99.62 \, \text{N}$$

Kijenzi cha wima ($F_y$) = $100 \cos 85^\circ$

$$F_y = 100 \times 0.0872$$ $$F_y = 8.72 \, \text{N}$$

Vijenzi vya (b) ni: Mlalo = $\boxed{\text{99.62 N}}$, Wima = $\boxed{\text{8.72 N}}$

Kijenzi cha mlalo ($F_x$) = $120 \cos 40^\circ$

$$F_x = 120 \times 0.7660$$ $$F_x = 91.92 \, \text{N}$$

Kijenzi cha wima ($F_y$) = $120 \sin 40^\circ$

$$F_y = 120 \times 0.6428$$ $$F_y = 77.14 \, \text{N}$$

Vijenzi vya (c) ni: Mlalo = $\boxed{\text{91.92 N}}$, Wima = $\boxed{\text{77.14 N}}$

Kijenzi cha mlalo ($F_x$) = $120 \cos 10^\circ$

$$F_x = 120 \times 0.9848$$ $$F_x = 118.18 \, \text{N}$$

Kijenzi cha wima ($F_y$) = $120 \sin 10^\circ$

$$F_y = 120 \times 0.1736$$ $$F_y = 20.83 \, \text{N}$$

Vijenzi vya (d) ni: Mlalo = $\boxed{\text{118.18 N}}$, Wima = $\boxed{\text{20.83 N}}$