The sketch shows a plot of land . (a) Using a scale of 1 cm to 10m, draw an accurate diagram of the plot ; (b) Construct : (i) The locus \(l_{1}\) of points...

The supplied reference value \(|TC|=60\text{ m}\) is not consistent with the stated plot dimensions. The \(6\text{ cm}\) measurement is the radius of the circle centred at \(A\), so it represents \(|AT|=60\text{ m}\), not \(|TC|\). Using \(\angle A=40^\circ\), \(\angle B=95^\circ\), and \(|AB|=85\text{ m}\), \(|TC|\) is approximately \(72\text{ m}\).

(a) Scale drawing

With scale \(1\text{ cm}:10\text{ m}\), draw \(|AB|=8.5\text{ cm}\). Construct a \(40^\circ\) angle at \(A\) and a \(95^\circ\) angle at \(B\); their rays meet at \(C\). The remaining angle is

\[ \angle C=180^\circ-40^\circ-95^\circ=45^\circ. \]

(b), (c) and (d): completed construction

Locus \(l_1\). Points equidistant from the two lines \(AC\) and \(BC\) lie on an angle bisector of \(\angle ACB\). Since only points inside the plot are relevant, draw the internal angle bisector of the \(45^\circ\) angle at \(C\). It makes angles of \(22.5^\circ\) with both \(AC\) and \(BC\).

Locus \(l_2\). Points \(60\text{ m}\) from \(A\) lie on a circle with centre \(A\) and radius \(60\text{ m}\). At the given scale, this radius is \(6\text{ cm}\). Only the arc inside the triangular plot is needed.

Locating \(T\) and finding \(|TC|\). The tree is the intersection of the internal angle bisector and the \(60\text{ m}\) arc. To check the measurement numerically, first find \(AC\):

\[ AC=\frac{85\sin95^\circ}{\sin45^\circ}\approx119.8\text{ m}. \]

In triangle \(ACT\), \(|AT|=60\text{ m}\) and \(\angle ACT=22.5^\circ\). Applying the cosine rule:

\[ 60^2=119.8^2+TC^2-2(119.8)(TC)\cos22.5^\circ. \]

The intersection that is inside the plot gives

\[ |TC|\approx71.8\text{ m}\approx\boxed{72\text{ m}}. \]

Flagpole region. “Nearer \(AC\) than \(BC\)” means the region on the \(AC\) side of \(l_1\). “More than \(60\text{ m}\) from \(A\)” means outside \(l_2\). Therefore, shade the part of the triangle that is on the \(AC\) side of the angle bisector and outside the \(60\text{ m}\) arc, as shown in green.

Examination reminder: A circle of radius \(60\text{ m}\) tells you that \(AT=60\text{ m}\). Do not automatically assign that radius to another line such as \(TC\); \(TC\) must be measured or calculated from the intersection.