The line segment $[SE]$, shown below, represents an airport runway - Leaving Cert Mathematics - Question 9 - 2021

To plot point $L$, we measure 1,250 m from point $S$ along the runway. Since 1 unit on the diagram corresponds to 250 m:

$ext{Distance in units} = \frac{1250}{250} = 5 ext{ units}$

Thus, point $L$ is located 5 units to the right of point $S$. The angle of 14° from point $L$ towards $E$ can be represented on the diagram accordingly.

Following the flight path:

1. Distance from $S$ to $L$ is 1,250 m.
2. For the angle of 14° from $L$, we apply the cosine rule to find the height above $E$:

Using: $h = \frac{1250}{\cos(14°)}$. Calculating:

$h \approx \frac{1250}{0.970} \approx 1288 ext{ m}$

Then, we add the distance to reach $E$:

Total distance = 1288 m + 1250 m = 2538 m.

To find the distance from point $B$ to point $C$, we can use the law of sines:

$\frac{x}{\sin(47°)} = \frac{200}{\sin(36°)}$

Solving for $x$:

$x = \frac{200 \cdot \sin(47°)}{\sin(36°)} \approx 324 ext{ km}$.

To calculate the total distance travelled:

1. First, the distance from $C$ to the point where the plane turned is 10 km.
2. The circular arc is calculated as:

$C = 2\pi(10) \cdot \frac{70}{360} \approx 12.21 ext{ km}$.

Thus, the total distance is:

$10 + 10 + 12.21 = 32.22 ext{ km}$.