# There is a tunnel which has a maximum depth below ground of 196.9 ft. The downward incline is at an angle of 4.923 degrees. The upward slop is also at an angle of 4.923 degrees to the horizontal. How far apart are the entrances?

Oct 19, 2016

The entrances are approximately 4564.947 ft horizontally apart.which is approx. $4564 \text{ ft "11 1/3" inch}$
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The tunnel length is approximately 4588.841 feet long.

#### Explanation:

$\textcolor{red}{\text{Assumption: the tunnel is symmetrical}}$

It is often helpful to draw a quick sketch.

$\textcolor{b l u e}{\text{Determine distance between entrances}}$

Trigonometry is all about ratios. From the diagram the ratio of:

$\frac{h}{L} = \tan \left(\theta\right)$

We are given that $\theta = {4.923}^{o}$

So $2 L = 2 \times \frac{h}{\tan} \left({4.923}^{o}\right) \approx \textcolor{b l u e}{4564.947 \text{ to 3 decimal places}}$
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$\textcolor{b l u e}{\text{Determine length of the tunnel}}$

We can use Pythagoras or Trig.

$\cos \left(\theta\right) = \frac{L}{S} \text{ " =>" } 2 S = 2 \times \frac{L}{\cos} \left(\theta\right)$ ...............(1)

To increase precision I am going to integrate the trig. You will see what I mean as we go along.

From the previous calculation we know that $L = \frac{h}{\tan} \left(\theta\right)$...(2)

Substitute for $L$ in equation(1) using equation(2)

$2 S = 2 \times \frac{h}{\tan \left(\theta\right) \times \cos \left(\theta\right)}$

$\implies \textcolor{b l u e}{2 S \approx 4588.841 \text{ ft to 3 decimal places}}$