Chapter A.2, Problem 55AYU

To determine

Find the distance from deck and bridge.

Answer to Problem 55AYU

  $\boxed{12.24miles}$ , $\boxed{15miles}$

Explanation of Solution

Given information:

A deck of a destroyer is $100$ feet above the sea level. How far can a person see from the deck?

How far can a person see from the bridge which is $150$ feet above the sea level?

Consider radius of earth is $3960miles$ and $1mile=5280feet$ .

Calculation:

Consider the illustration,

  

Draw two radii from the center of the earth, one through the deck and other to the farthest point a person can see from the deck.

We know that $1mile=5280feet$

  $100feet=\frac{100}{5280}miles$

Applly pythagorean Theorem to the right triangle,

  $\begin{array}{l}{d}^2+{3960}^2={\left(3960+\frac{100}{5280}\right)}^2\\ d^2={\left(3960+\frac{100}{5280}\right)}^2-{3960}^2\\ d^2=149.99723\\ \sqrt{d^2}=\sqrt{149.99723}\\ d=12.24miles\end{array}$

Now find distance from the bridge,

  $\begin{array}{l}{d}^2+{3960}^2={\left(3960+\frac{150}{5280}\right)}^2\\ d^2={\left(3960+\frac{150}{5280}\right)}^2-{3960}^2\\ d^2=225.00080\\ \sqrt{d^2}=\sqrt{225.00080}\\ d=15miles\end{array}$

Hence, the distance from the deck and bridge is $\boxed{12.24miles}$ , $\boxed{15miles}$ .