A farmer owns two triangular plots of land $ABC$ and $ADC$ as shown in the diagram above. It is given that $AB = 7\ \text{m}$, $AC = 8\ \text{m}$ and $AD = 6\ \text{m}$, angle $BAD$ is a right angle and angle $BAC = \theta$ radians.

1. Express the area of triangle $ABC$ in terms of $\theta$.

Area of $\triangle ABC = \frac{1}{2}(AB)(AC)\sin(\angle BAC)$
$= \frac{1}{2}(7)(8)\sin\theta$
$= 28\sin\theta$

2. Determine the angle $CAD$ and express the area of triangle $ADC$.

Since $\angle BAD = \frac{\pi}{2}$ and $\angle BAC = \theta$, we have $\angle CAD = \frac{\pi}{2} - \theta$.
Area of $\triangle ADC = \frac{1}{2}(AD)(AC)\sin(\angle CAD)$
$= \frac{1}{2}(6)(8)\sin\left(\frac{\pi}{2} - \theta\right)$

3. Simplify the area of triangle $ADC$ and find the total area $A$.

Using the identity $\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta$,
Area of $\triangle ADC = 24\cos\theta$
Total area $A = \text{Area of } \triangle ABC + \text{Area of } \triangle ADC$
$A = 28\sin\theta + 24\cos\theta$ (Shown)

1. Identify the expression for $A$ from part (a)(i)

From (a)(i), $A = 28\sin\theta + 24\cos\theta$.

2. Find the value of $R$

$R = \sqrt{28^2 + 24^2} = \sqrt{784 + 576} = \sqrt{1360} \approx 36.878$

3. Find the value of $\alpha$

$\tan\alpha = \frac{24}{28} = \frac{6}{7}$
$\alpha = \tan^{-1}\left(\frac{6}{7}\right) \approx 0.7086$ radians

4. State the final expression

$A = 36.9\sin(\theta + 0.709)$ (to 3 s.f.)

1. State the greatest possible value of $A$ using the $R$ value from part (a)(ii)

From (a)(i) and (a)(ii), $A = 28\sin\theta + 24\cos\theta = \sqrt{1360}\sin(\theta + 0.7086)$.
The greatest possible value of $A$ is $R$.
$\text{Greatest } A = \sqrt{1360} \approx 36.9\ \text{m}^2$

2. Set up the equation to find the corresponding $\theta$

The maximum value occurs when $\sin(\theta + 0.7086) = 1$.
$\theta + 0.7086 = \frac{\pi}{2}$

3. Solve for $\theta$

$\theta = \frac{\pi}{2} - 0.70862 \theta \approx 0.862\ \text{radians}$

1. Express the areas of triangles $ABC$ and $ADC$ in terms of $\theta$

Area of $ABC = \frac{1}{2}(7)(8)\sin\theta = 28\sin\theta$

Angle $DAC = \frac{\pi}{2} - \theta$
Area of $ADC = \frac{1}{2}(6)(8)\sin\left(\frac{\pi}{2} - \theta\right) = 24\cos\theta$

2. Set up the equation based on the given condition

Area of $ABC = 2 \times \text{Area of } ADC$
$28\sin\theta = 2(24\cos\theta)$
$28\sin\theta = 48\cos\theta$

3. Solve the trigonometric equation for $\theta$

$\frac{\sin\theta}{\cos\theta} = \frac{48}{28}$
$\tan\theta = \frac{12}{7}$
$\theta = \tan^{-1}\left(\frac{12}{7}\right) \approx 1.0432 \approx 1.04 \text{ radians}$