The diagram below shows a fenced field $PQRS$. It is given that $PS = 40\text{ m}$, $SR = 100\text{ m}$, both angles $PQR$ and $PSR$ are right angles and angle $QRS = \theta$, where $\theta$ is an acute angle in degrees.

1. Draw perpendiculars to form right-angled triangles and find the lengths of the segments related to $SR$.

Draw a vertical line from $S$ to meet $QR$ at $A$. In $\triangle SAR$, $\angle SAR = 90^\circ$.
$SA = SR \sin \theta = 100 \sin \theta$
$AR = SR \cos \theta = 100 \cos \theta$

2. Draw a horizontal line from $S$ to $PQ$ and find the lengths of the segments related to $PS$.

Draw a horizontal line from $S$ to meet $PQ$ (or its extension) at $B$. In $\triangle PBS$, $\angle PBS = 90^\circ$.
Since $\angle PSR = 90^\circ$ and the angle of $SR$ with the horizontal is $\theta$, the angle of $PS$ with the vertical is $\theta$. Thus, $\angle PSB = 90^\circ - \theta$ and $\angle SPB = \theta$.
$SB = PS \sin \theta = 40 \sin \theta$
$PB = PS \cos \theta = 40 \cos \theta$

3. Express $QR$ and $PQ$ in terms of $\theta$ and calculate the total perimeter.

From the rectangle $SBAQ$ (where $QA = SB$ and $QB = SA$):
$QR = QA + AR = SB + AR = 40 \sin \theta + 100 \cos \theta$
$PQ = QB - PB = SA - PB = 100 \sin \theta - 40 \cos \theta$

Perimeter $P = PQ + QR + RS + SP$
$P = (100 \sin \theta - 40 \cos \theta) + (40 \sin \theta + 100 \cos \theta) + 100 + 40$
$P = 140 + 140 \sin \theta + 60 \cos \theta$ (shown)

1. Set up the R-formula equation for the trigonometric part of $P$

We want to express $60 \cos \theta + 140 \sin \theta$ in the form $R \cos(\theta - \alpha)$.
Expanding $R \cos(\theta - \alpha)$ gives $R \cos \theta \cos \alpha + R \sin \theta \sin \alpha$.
Comparing coefficients:
$R \cos \alpha = 60$
$R \sin \alpha = 140$

2. Calculate the value of $R$

$R = \sqrt{60^2 + 140^2}$
$R = \sqrt{3600 + 19600} = \sqrt{23200}$
$R = 20\sqrt{58} \approx 152.315$

3. Calculate the value of $\alpha$ and write the final expression

$\tan \alpha = \frac{140}{60} = \frac{7}{3}$
$\alpha = \tan^{-1}\left(\frac{7}{3}\right) \approx 66.801^\circ$
Therefore, $P = 140 + 152.3 \cos(\theta - 66.8^\circ)$ (to 1 d.p.)

1. Set up the equation using the R-formula result from part (b)

From part (b), $P = 140 + \sqrt{23200} \cos(\theta - 66.801^\circ)$.
Given $P = 250$,
$140 + \sqrt{23200} \cos(\theta - 66.801^\circ) = 250$
$\sqrt{23200} \cos(\theta - 66.801^\circ) = 110$
$\cos(\theta - 66.801^\circ) = \frac{110}{\sqrt{23200}} \approx 0.72217$

2. Solve for $\theta$ and select the acute angle

$\theta - 66.801^\circ = \cos^{-1}(0.72217)$
$\theta - 66.801^\circ = \pm 43.765^\circ$
$\theta = 66.801^\circ + 43.765^\circ \text{ or } 66.801^\circ - 43.765^\circ$
$\theta = 110.566^\circ \text{ or } 23.036^\circ$
Since $0^\circ \le \theta \le 90^\circ$, we reject $110.566^\circ$.
$\theta \approx 23.0^\circ \text{ (to 1 d.p.)}$