Question

virginia is flying a cessna and is descending at a 6° angle towards a runway. if she can see a dam behind her at a 45° angle that is 250 meters away from the runway, how much further does she have to fly until she lands? round to the nearest tenth.

Explanation:

Step1: Set up a right - triangle model

Let the distance Virginia has to fly until she lands be $x$ meters. Let the horizontal distance from her position (projected vertically) to the runway be $y$ meters. We know that the dam is 250 meters from the runway.

Step2: Use tangent functions

We have two right - triangle relationships. For the angle of descent of $6^{\circ}$, $\tan(6^{\circ})=\frac{h}{y}$ (where $h$ is the vertical height she needs to descend, but we don't need to find $h$ explicitly). For the $45^{\circ}$ angle to the dam, $\tan(45^{\circ}) = 1=\frac{h}{250 - y}$. So, $h = 250 - y$.
Since $\tan(6^{\circ})=\frac{h}{y}$, and $h = 250 - y$, we substitute $h$: $\tan(6^{\circ})=\frac{250 - y}{y}$.
We know that $\tan(6^{\circ})\approx0.1051$. So, $0.1051y=250 - y$.

Step3: Solve for $y$

Combining like terms: $0.1051y + y=250$, $1.1051y = 250$, $y=\frac{250}{1.1051}\approx226.22$ meters.

Step4: Find the distance $x$

We know that $\cos(6^{\circ})=\frac{y}{x}$. So, $x=\frac{y}{\cos(6^{\circ})}$. Since $\cos(6^{\circ})\approx0.9945$ and $y\approx226.22$ meters, $x=\frac{226.22}{0.9945}\approx227.5$ meters.

Answer:

$227.5$ meters