# What is the bearing that the control tower should use to locate the​ aircraft?

## A​ DC-9 aircraft leaves an airport from a runway whose bearing is $N 42 \text{ degrees } E$. After flying for three fourths of a mile, the pilot requests permission to turn 90° and head toward the southeast. The permission is granted. After the airplane goes $1$ mile

May 13, 2018

The bearing of the plane is 095° or S 85°E

#### Explanation:

The distance of $\frac{3}{4}$ mile after take-off, the $1$ mile flown by the plane and the direct line from the control tower to the DC-9 form a right-angled triangle. It is offset from North by an angle of 42°

Use trigonometry to find the angle at the control tower.

Opposite side = $1$ mile and adjacent side is $\frac{3}{4}$mile

$T a n \theta = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

theta = 53.1°

Taking the bearing of the runway into consideration, the bearing of the plane from North is 42°+53.1° = 095.1°

Bearings are usually given with $3$ figures, so 095°

The other format would be in relation to the N-S line.

The place is on a bearing of S 85° E

May 13, 2018

$\text{bearing to locate aircraft} = {95}^{\circ} 7 ' 48 ' '$

#### Explanation:

$\text{Let the control tower be C}$

$\text{The aircraft turning point A}$

$\text{The 1 mile point B}$

$\text{Pythagoras:-}$

$\therefore {a}^{2} = {b}^{2} + {c}^{2}$

$\therefore {a}^{2} = {\left(\frac{3}{4}\right)}^{2} + {\left(1\right)}^{2}$

$\therefore {a}^{2} = \frac{9}{16} + 1$

$\therefore \sqrt{{a}^{2}} = \sqrt{1.5625}$

$\therefore a \text{=1.25 miles =BC}$

$\text{sine rule:-}$

$\therefore \sin \frac{B}{A C} = \sin \frac{A}{B C}$

$\therefore \sin \frac{B}{0.75} = \sin \frac{A}{1.25}$

$\therefore \sin B = \frac{0.75 \times \sin A}{1.25}$

$\therefore \sin B = \frac{0.75 \times 1}{1.25}$

$\therefore \sin B = 0.6$

$\therefore \angle B = {36}^{\circ} 52 ' 12 ' '$

$\therefore \text{Bearing C-A} = {42}^{\circ}$

$\text{Bearing A-C} = {42}^{\circ} + {180}^{\circ} = {222}^{\circ}$

$\therefore \angle a t A = {90}^{\circ}$

$\therefore \text{Bearing A-B} = {222}^{\circ} - {90}^{\circ} = {132}^{\circ}$

$\therefore \text{Bearing B-A} = {132}^{\circ} + {180}^{\circ} = {312}^{\circ}$

$\therefore \text{Bearing B-C"=312^@-36^@52'12} = {275}^{\circ} 7 ' 48 ' '$

$\therefore \text{Bearing C-B} = {275}^{\circ} 7 ' 48 ' ' - {180}^{\circ} = {95}^{\circ} 7 ' 48 ' '$

$\therefore \text{Bearing to locate aircraft} = {95}^{\circ} 7 ' 48 ' '$