A, B, C and D are points on horizontal ground. AB = 35 m and BC = 68 m. The bearing of B from A is $137^\circ$ and the bearing of C from B is $192^\circ$. Angle ABD = $85^\circ$ and angle BCD = $70^\circ$.

1. Find the angle between $BA$ and the north line at $B$.

The bearing of $B$ from $A$ is $137^\circ$.

Since the north lines at $A$ and $B$ are parallel,
$\angle\text{(north at }B, BA) = 180^\circ - (180^\circ - 137^\circ)=43^\circ.$
Equivalently, the bearing of $A$ from $B$ is
$137^\circ + 180^\circ = 317^\circ,$
so the angle between $BA$ and the north line at $B$ is
$360^\circ - 317^\circ = 43^\circ.$

2. Find the angle between the north line at $B$ and $BC$, then add the two angles.

The bearing of $C$ from $B$ is $192^\circ$.

So the angle between south and $BC$ is
$192^\circ - 180^\circ = 12^\circ.$
Hence the angle between the north line at $B$ and $BC$ on the side containing $A$ is
$180^\circ - 12^\circ = 168^\circ.$
Therefore
$\angle ABC = 168^\circ - 43^\circ = 125^\circ.$

1. Identify the appropriate rule and substitute the two sides with the included angle.

In $\triangle ABC$, $AB=35$, $BC=68$ and $\angle ABC=125^\circ$.

Using the cosine rule,
$AC^2=AB^2+BC^2-2(AB)(BC)\cos\angle ABC$
$AC^2=35^2+68^2-2(35)(68)\cos125^\circ$

2. Calculate $AC$ and give the answer in metres.

$AC^2=35^2+68^2-2(35)(68)\cos125^\circ$
$AC^2=8579.22\ldots$
$AC=\sqrt{8579.22\ldots}=92.624\ldots$

So, to 3 significant figures,
$AC=92.6\text{ m}$

1. Find the angle at $B$ in triangle $BCD$.

From part (a), $\angle ABC = 125^\circ$.

Given $\angle ABD = 85^\circ$,
$\angle DBC = 125^\circ - 85^\circ = 40^\circ.$

2. Use the sine rule to calculate $DC$.

In $\triangle BCD$,
$\angle BCD = 70^\circ, \quad \angle DBC = 40^\circ.$

Using the sine rule,
$\frac{DC}{\sin 40^\circ}=\frac{BC}{\sin 70^\circ}.$

Since $BC=68$,
$DC=\frac{68\sin 40^\circ}{\sin 70^\circ}=46.5\text{ m} \quad \text{to 3 s.f.}$

1. Identify $BX$ as the perpendicular shortest distance from $B$ to the path $DC$.

Since $X$ is the position on $DC$ where the distance from $B$ is shortest,
$BX \perp DC.$
So triangle $BCX$ is right-angled at $X$. Also,
$\angle BCX = \angle BCD = 70^\circ.$

2. Use the sine ratio to calculate the perpendicular distance $BX$.

In right-angled triangle $BCX$,
$\sin 70^\circ = \frac{BX}{BC}$
$BX = 68\sin 70^\circ$
$BX = 63.899\ldots$
$BX = 63.9\text{ m} \quad \text{to 3 significant figures}.$

1. Find the height of the drone above B using the angle of elevation from C.

Let the height of the drone above $B$ be $h$ metres.

Since $BC=68$ m and the angle of elevation from $C$ is $40.5^\circ$,
$\tan 40.5^\circ=\frac{h}{68}$
$h=68\tan 40.5^\circ$

2. Identify the point on DC giving the greatest angle of elevation.

The angle of elevation is greatest when the horizontal distance from $B$ is smallest.

From part (d), the shortest distance from $B$ to $DC$ is
$BX=63.9\text{ m}.$

3. Calculate the greatest angle of elevation.

Let the greatest angle of elevation be $\theta$.

$\tan \theta=\frac{68\tan 40.5^\circ}{63.9}$
$\theta=\tan^{-1}\left(\frac{68\tan 40.5^\circ}{63.9}\right)$
$\theta=42.3^\circ\text{ correct to 1 decimal place}$