The diagram shows a field ABCD on horizontal ground. Angle ABC = 115°, Angle CAD = 58°, AB = 9 m, BC = 6 m and CD = 11 m. Calculate

1. Apply the cosine rule to triangle $ABC$ to form an expression for $AC^2$.

In $\triangle ABC$,
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos \angle ABC$
$AC^2 = 9^2 + 6^2 - 2(9)(6)\cos 115^\circ$

2. Calculate and square-root to find $AC$.

$AC^2 = 81 + 36 - 108\cos 115^\circ$
$AC = \sqrt{81 + 36 - 108\cos 115^\circ}$
$AC = 12.75\ldots$
So, to 3 significant figures,
$AC = 13.0\text{ m}$

1. Apply the sine rule to triangle $ACD$ to find $\sin \angle ADC$.

In $\triangle ACD$, $AC=13.0\text{ m}$, $CD=11\text{ m}$ and $\angle CAD=58^\circ$.

By the sine rule,
$\frac{AC}{\sin \angle ADC}=\frac{CD}{\sin \angle CAD}$
$\frac{13.0}{\sin \angle ADC}=\frac{11}{\sin 58^\circ}$
$\sin \angle ADC=\frac{13.0\sin 58^\circ}{11}$

2. Calculate the acute possible value of the angle from the sine ratio.

$\sin \angle ADC=\frac{13.0\sin 58^\circ}{11}=1.002\ldots$
Using the unrounded value of $AC$ from part (a),
$\sin \angle ADC=0.920\ldots$
$\angle ADC=\sin^{-1}(0.920\ldots)=67.0^\circ\text{ approximately}$

3. Choose the obtuse angle required by the question.

Since the question asks for the obtuse angle,
$\angle ADC=180^\circ-67.0^\circ=113.0^\circ$
So,
$\angle ADC\approx 113^\circ.$

1. Find the included angle between $AC$ and $CD$ in triangle $ACD$.

In $\triangle ACD$,
$\angle ACD=180^\circ-58^\circ-113^\circ=9^\circ.$

2. Apply the triangle area formula using two sides and the included angle.

$\text{Area of }\triangle ACD=\frac12(13.0)(11)\sin 9^\circ$
$=12.8\text{ m}^2\text{, to 3 s.f.}$

1. Relate the shortest distance from $D$ to $AC$ to the perpendicular height of triangle $ACD$.

The shortest distance from $D$ to $AC$ is the perpendicular height, $h$, of $\triangle ACD$ when $AC$ is used as the base.

Using the area from part (c),
$12.8=\frac12\times 13.0\times h$

2. Solve for the perpendicular height.

$h=\frac{2\times 12.8}{13.0}$
$h=1.969\ldots$
$h\approx 1.97\text{ m}$