An aircraft waiting to land is flying around a triangular circuit ABC.
A, B and C are vertically above three beacons, X, Y and Z respectively.
T is the control tower at the airport, and T, X, Y and Z lie in a horizontal plane.
BC = 18 km, CA = 22 km and AB = 24 km.

1. Find the distance travelled in one complete circuit

One round of circuit ABC is the perimeter of triangle ABC.

$18+22+24=64\text{ km}$

2. Use speed to calculate the time and convert to minutes and seconds

$\text{Time}=\frac{64}{200}\text{ h}=0.32\text{ h}$

$0.32\times 60=19.2\text{ min}$

$0.2\text{ min}=0.2\times 60=12\text{ s}$

So the time taken is $19$ min $12$ s.

1. Identify which angle is the largest

The largest side is $AB=24$ km.

Therefore the largest angle is opposite $AB$, which is $\angle ACB$.

2. Apply the cosine rule to find the largest angle

Using the cosine rule for $\angle ACB$,
$AB^2=AC^2+BC^2-2(AC)(BC)\cos \angle ACB$
$24^2=22^2+18^2-2(22)(18)\cos \angle ACB$
$576=484+324-792\cos \angle ACB$
$792\cos \angle ACB=232$
$\cos \angle ACB=\frac{232}{792}$

3. Evaluate the angle and round suitably

$\angle ACB=\cos^{-1}\left(\frac{232}{792}\right)$
$\angle ACB=82.962\ldots^\circ$

So the largest angle is $83.0^\circ$ to 1 decimal place.

1. Find angle XZT using the bearings from Z

Since Z is due west of T, T is due east of Z.

So the bearing of T from Z is $090^\circ$.

The bearing of X from Z is $042^\circ$.

Hence
$\angle XZT = 090^\circ - 042^\circ = 48^\circ.$

2. Find angle ZTX using the bearings from T

Since Z is due west of T, the bearing of Z from T is $270^\circ$.

The bearing of X from T is $338^\circ$.

Hence
$\angle ZTX = 338^\circ - 270^\circ = 68^\circ.$

3. Use the angle sum of triangle ZXT

In triangle ZXT,
$\angle ZXT = 180^\circ - 48^\circ - 68^\circ = 64^\circ.$

1. Identify the known side and relevant angles in triangle $ZXT$

Since $A$ is vertically above $X$ and $C$ is vertically above $Z$, the horizontal distance
$XZ=CA=22\text{ km}.$
From part (b)(i),
$\angle ZXT=64^\circ.$
Also, using the bearings,
$\angle ZTX=68^\circ,$
so the side $XZ$ is opposite $68^\circ$, and $TX$ is opposite $48^\circ$.

2. Apply the sine rule to calculate $TX$

By the sine rule,
$\frac{TX}{\sin 48^\circ}=\frac{22}{\sin 68^\circ}.$
Hence
$TX=\frac{22\sin48^\circ}{\sin68^\circ}=17.633\ldots$
$TX\approx 17.6\text{ km}.$