a)

The component of force in the $x$-direction can be found using $F \times \cos\theta$. Remember to set your calculator to use degrees and not radians.

\begin{align} \text{component in the }x \text{-direction } & = F \cos \theta \\
                    & = \var{force} \times \cos \var{angle} \\
                     & = \var{precround(force*cos(radians(angle)),3)}\end{align}

b)

Now we need to make sure we find the angle between the force and the direction we are resolving in. Therefore $\theta = 90 - \var{angle} = \var{yangle}$.

\begin{align} \text{component in the y-direction } & = F \cos \theta \\
                    & = \var{force} \times \cos \var{yangle} \\
                     & = \var{precround(force*cos(radians(yangle)),3)}\end{align}

Since $\sin \theta = \cos(90-\theta)$, we could also use $\sin \var{angle}$ in our calculations instead of $\cos(90 - \var{angle})$.

a)

We need to find the angle $\theta_x$ of $F$ relative to the $x$-axis and then use $F \times cos\theta_x$.

We consider the angle between the positive $x$-axis and $F$, i.e. $\theta_x = 90 + \var{theta}$.

\begin{align}
\text{component in the } x \text{-direction} & = F \cos\theta_x \\
& = \var{force} \times \cos \var{angle} \\
& = \var{precround(force*cos(radians(angle)),3)}
\end{align}

b)

The positive $y$-direction is vertically upwards and we need the angle relative to the positive $y$-direction therefore $\theta_y = 180 - \var{theta}$.

\begin{align}
\text{component in the } y \text{-direction} & = F \cos\theta_y \\
& = \var{force} \times \cos \var{yangle} \\
& = \var{precround(force*cos(radians(yangle)),3)}
\end{align}

Notice that both these answers are negative as the force acts in the opposite direction to the positive. You could also answer these by resolving in the negative $x$ or $y$ direction and changing the sign of your solution. For example in part b) you could use $\theta_y = \var{theta}$ which gives $F \cos \theta_y = \var{precround(-force*cos(radians(yangle)),3)}$ and then change the sign.

a)

We resolve in the positive $x$-direction so the answer will be negative. We take $\theta = \var{theta}^{\circ}$ as the angle is already between the force and the $x$-axis.

\begin{align} \text{component in } x \text{-direction} & = F \cos \theta \\
& = \var{force} \times \cos \var{theta} \\
& = \var{precround(force*cos(radians(theta)),3)}
\end{align}

b)

We need the angle between the force and the direction we are resolving so take $\theta = 90 - \var{theta}= \var{90-theta}^{\circ}$.

\begin{align}
\text{component in } y \text{-direction} & = F \cos \theta \\
& = \var{force} \times \cos \var{yangle} \\
& = \var{precround(force*cos(radians(yangle)),3)}
\end{align}

a) - c)

Resolve each force from the positive $x$-direction (pointing to the right). For the force $P$ this is at $90^{ \circ}$ to the $x$-axis therefore has a contribution of $P \times \cos 90^{\circ} = 0$ to the sum of components.

The force $F$ is at $(180-\var{theta1})^{\circ}=\var{90 + 90 - theta1}^{\circ}$ to the positive $x$-direction therefore has a contribution of $F \times \cos \var{180-theta1}^{\circ} = \var{force1} \times \cos \var{180-theta1}^{\circ} = \var{precround(force1*cos(radians(180-theta1)),3)}$ to the sum of components. This will be negative as you can imagine if you were moving in the positive $x$-direction this force is acting in the opposite direction and pulling you back!

The force $Q$ is at $\var{theta2}^{\circ}$ to the positive $x$-direction therefore has a contribution of $Q \times cos \var{theta2}^{\circ} = \var{force3} \times \cos\var{theta2}^{\circ} = \var{precround(force3*cos(radians(theta2)),3)}$. This is positive as it is acting in the same direction as the positive.

Therefore the sum of components in the $x$-direction is $0 - \var{precround(-force1*cos(radians(180-theta1)),3)} +  \var{precround(force3*cos(radians(theta2)),3)} = \var{precround(force1*cos(radians(180-theta1)) + force3*cos(radians(theta2)),3)}$.

d) - g)

Resolve each force from the positive $y$-direction (upwards). For the force $P$ this is acting completely in the positive direction, at no angle. Therefore it's contribution is $\var{force2}$. Note that this is the same as $\var{force2} \times \cos 0^{\circ}$. 

The force $F$ is at $(90 - \var{theta1})^{\circ} = \var{90 - theta1}^{\circ}$ to the positive $y$-direction therefore has a contribution of $F \times \cos \var{90 - theta1}^{\circ} = \var{force1} \times \cos \var{90 - theta1}^{\circ} = \var{precround(force1*cos(radians(90-theta1)),3)}$ to the sum of components. This is positive as it is acting in the same direction to the positive.

The force $Q$ is at $(90+\var{theta2})^{\circ}=\var{90 + theta2}^{\circ}$ to the positive $y$-direction therefore has a contribution of $Q \times \cos \var{90 + theta2}^{\circ} = \var{force3} \times \cos \var{90 + theta2}^{\circ} = \var{precround(force3*cos(radians(90+theta2)),3)}$ to the sum of components. This is negative as it is acting downwards, in the opposite direction to the positive.

Therefore the sum of components in the $y$-direction is $\var{force2} + \var{precround(force1*cos(radians(90-theta1)),3)} - \var{-precround(force3*cos(radians(90+theta2)),3)} = \var{precround(force2 + force1*cos(radians(90-theta1)) + force3*cos(radians(90+theta2)),3)}$.

a)

Resolve each force from the positive $x$-direction (pointing to the right).

For the force $Q$ this is at $180^{ \circ}$ to the positive $x$-axis so makes a contribution to the sum of components of $Q \times \cos 180^{\circ} = -\var{q}$ 

The force $F$ is at $(90+\var{theta1})^{\circ}=\var{90 + theta1}^{\circ}$ to the positive $x$-direction therefore has a contribution of $F \times \cos \var{90+theta1}^{\circ} = \var{f} \times \cos \var{90+theta1}^{\circ} = \var{precround(f*cos(radians(90+theta1)),3)}$ to the sum of components. This will be negative as you can imagine if you were moving in the positive $x$-direction this force is acting in the opposite direction and pulling you back!

The force $P$ is at $\var{theta2}^{\circ}$ to the positive $x$-direction therefore has a contribution of $P \times \cos \var{theta2}^{\circ} = \var{p} \times \cos\var{theta2}^{\circ} = \var{precround(p*cos(radians(theta2)),3)}$. This is positive as it is acting in the same direction as the positive.

Therefore the sum of component in the $x$-direction is $-\var{q} - \var{precround(-f*cos(radians(90+theta1)),3)} +  \var{precround(p*cos(radians(theta2)),3)} = \var{precround(-q+f*cos(radians(90+theta1)) + p*cos(radians(theta2)),3)}$.

b)

Resolve each force from the positive $y$-direction (upwards). For the force $Q$ this is acting at a right angle to the positive direction. Therefore it's contribution is $\var{q} \times \cos 90^{\circ}= 0$. There is no upward or downward pull from force $Q$.

The force $F$ is at $(180 - \var{theta1})^{\circ} = \var{180 - theta1}^{\circ}$ to the positive $y$-direction therefore has a contribution of $F \times \cos \var{180 - theta1}^{\circ} = \var{f} \times \cos \var{180 - theta1}^{\circ} = \var{precround(f*cos(radians(180-theta1)),3)}$ to the sum of components. This is negative as it is acting in the opposite direction to the positive (downwards).

The force $P$ is at $(90-\var{theta2})^{\circ}=\var{90 - theta2}^{\circ}$ to the positive $y$-direction therefore has a contribution of $P \times \cos \var{90 - theta2}^{\circ} = \var{p} \times \cos \var{90 - theta2}^{\circ} = \var{precround(p*cos(radians(90-theta2)),3)}$ to the sum of components. This is positive as it is acting upwards.

Therefore the sum of components in the $y$-direction is $0 - \var{precround(-f*cos(radians(180-theta1)),3)} + \var{precround(p*cos(radians(90-theta2)),3)} = \var{precround(f*cos(radians(180-theta1)) + p*cos(radians(90-theta2)),3)}$.