2021 May TZ1 P2 Q8

Two straight fences meet at point \(\mathrm{A}\) and a field lies between them.
A horse is tied to a post, \(\mathrm{P}\), by a rope of length \(r\) metres. Point \(\mathrm{D}\) is on one fence and point \(\mathrm{E}\) is on the other, such that \(\mathrm{PD} = \mathrm{PE} = \mathrm{PA} = r\) and \(\mathrm{D\hat{P}E} =\theta\)  radians. This is shown in the following diagram.

The length of the arc \(\mathrm{DE}\) shown in the diagram is \(28\) m

1. Write down an expression for \(r\) in terms of \(\theta\).

   [1]

2. Show that the area of the field that the horse can reach is \(\frac{392}{\theta^2}(\theta + \sin{\theta})\).

   [4]

3. The area of field that the horse can reach is \(460\) m\(^2\) . Find the value of \(\theta\).

   [2]

4. Hence, find the size of \(\mathrm{D\hat{A}E}\).

   [2]

A new fence is to be constructed between points \(\mathrm{B}\) and \(\mathrm{C}\) which will enclose the field, as shown in the following diagram.

Point \(\mathrm{C}\) is due west of \(\mathrm{B}\) and \(\mathrm{AC} = 800\) m. The bearing of \(\mathrm{B}\) from \(\mathrm{A}\) is \(195^{\circ}\).

5. 1. Find the size of \(\mathrm{A\hat{B}C}\).

   2. Find the length of new fence required.

      [5]