Reasoning: In two-dimensional (2-D) motion, a force can be resolved into two mutually perpendicular components: one along the $\mathbf{x}$-axis (horizontal) and one along the $\mathbf{y}$-axis (vertical). Therefore, the number of perpendicular components is $\mathbf{2}$.

Reasoning: We know: $\widehat{\mathbf{i}}\mathbf{\times}\widehat{\mathbf{j}}\mathbf{=}\widehat{\mathbf{k}}$
But cross product is anti-commutative, so changing the order changes the sign: $\widehat{\mathbf{j}}\mathbf{\times}\widehat{\mathbf{i}}\mathbf{= -}\mathbf{\ }\widehat{\mathbf{k}}$

Reasoning: The $\mathbf{x}$-component is $\mathbf{F}_{\mathbf{x}}\mathbf{=}\mathbf{F}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{\theta}$.
$\mathbf{F}_{\mathbf{x}}\mathbf{= 10}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{30}^{\mathbf{o}}$ $\mathbf{\Longrightarrow}$ $\mathbf{F}_{\mathbf{x}}\mathbf{= 10 \times 0.866}$ $\mathbf{\Longrightarrow}$ $\mathbf{F}_{\mathbf{x}}\mathbf{= 8.66 \approx 8.7}\mathbf{\ }\mathbf{N}$

Reasoning: For perpendicular forces: $\mathbf{R =}\sqrt{\mathbf{A}^{\mathbf{2}}\mathbf{+}\mathbf{B}^{\mathbf{2}}}$
$\mathbf{R =}\sqrt{\mathbf{3}^{\mathbf{2}}\mathbf{+}\mathbf{4}^{\mathbf{2}}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R =}\sqrt{\mathbf{9 + 16}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R =}\sqrt{\mathbf{25}}$ $\mathbf{\Longrightarrow}$ $\mathbf{R = 5}\mathbf{\ }\mathbf{N}$

Reasoning: Dot product formula is: $\overrightarrow{\mathbf{A}}\mathbf{\ }\mathbf{\cdot}\mathbf{\ }\overrightarrow{\mathbf{B}}\mathbf{= AB}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{\theta}$
From this formula, we can find the angle: $\mathbf{\cos}\mathbf{\ }\mathbf{\theta}\mathbf{=}$ $\frac{\overrightarrow{\mathbf{A}}\mathbf{\ \ }\mathbf{\cdot}\mathbf{\ \ }\overrightarrow{\mathbf{B}}}{\mathbf{AB}}$
So, angle between two vectors is easily found by dot product.
Note: While the cross product formula involves $\mathbf{\sin}\mathbf{\ }\mathbf{\theta}$, it yields a vector quantity $\mathbf{(A \times B =}\left| \mathbf{A} \right|\mathbf{\ .}\mathbf{\ }\left| \mathbf{B} \right|\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{\theta}\mathbf{\ }\widehat{\mathbf{n}}\mathbf{)}$ solve for the angle, you must first calculate the magnitude of the resulting vector, adding unnecessary extra mathematical steps

Reasoning: Magnitude of dot product: $\left| \overrightarrow{\mathbf{A}}\mathbf{\ \ .\ \ }\overrightarrow{\mathbf{B}} \right|\mathbf{= AB}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{\theta}$
Magnitude of cross product: $\mathbf{|}\overrightarrow{\mathbf{A}}\mathbf{\times}\overrightarrow{\mathbf{B}}\mathbf{| = AB}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{\theta}$ Given: $\mathbf{AB}\mathbf{\ }\mathbf{\cos}\mathbf{\ }\mathbf{\theta}\mathbf{= AB}\mathbf{\ }\mathbf{\sin}\mathbf{\ }\mathbf{\theta}$
$\mathbf{\cos}\mathbf{\ }\mathbf{\theta}\mathbf{= sin}\mathbf{\ }\mathbf{\theta}$ ; $\left\lbrack \mathbf{\because}\mathbf{\ \ \ }\frac{\mathbf{\cos}\mathbf{\ }\mathbf{\theta}}{\mathbf{\sin}\mathbf{\ }\mathbf{\theta}}\mathbf{= tan}\mathbf{\ }\mathbf{\theta} \right\rbrack$
This is true when $\mathbf{\theta}\mathbf{=}\mathbf{45}^{\mathbf{o}}$.