Question

ann is flying a cessna and is descending at a 6° angle towards a runway. if she can see a gorge behind her at a 60° angle that is 200 yards away from the runway, how much further does she have to fly until she lands? round to the nearest tenth.

Explanation:

Step1: Find the height of the triangle

Let the height of the triangle (distance from the plane's current altitude to the runway level) be \(h\).
We know that in the right - triangle formed by the gorge, the runway and the vertical line from the plane's altitude to the runway level, \(\sin60^{\circ}=\frac{h}{200}\).
So, \(h = 200\times\sin60^{\circ}=200\times\frac{\sqrt{3}}{2}=100\sqrt{3}\) yards.

Step2: Find the distance the plane has to fly

Let the distance the plane has to fly be \(d\).
We know that \(\sin6^{\circ}=\frac{h}{d}\).
Since \(h = 100\sqrt{3}\), then \(d=\frac{100\sqrt{3}}{\sin6^{\circ}}\).
We know that \(\sin6^{\circ}\approx0.1045\) and \(\sqrt{3}\approx1.732\).
So \(d=\frac{100\times1.732}{0.1045}=\frac{173.2}{0.1045}\approx1657.4\) yards.

Answer:

\(1657.4\) yards