Chapter 9, Problem 26RP

To determine

Tocalculate: Thedistance the boat will move forward.

Answer to Problem 26RP

The distance the boat will move forward is $12$ feets.

Explanation of Solution

Given information:

  $\begin{array}{l}AB=25\text{'},\\ AC=15\text{'}.\end{array}$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Assume that the dock is perpendicular to the water. Two right angles are formed.

Distance y is that horizontal distance when $AB=25\text{'}$ .

Distance z is that horizontal distance when $AB=17\text{'}$ .

Distance y and z can be calculated by applying Pythagoras Theorem.

In right angled $\Delta ABC$ , we get

  $\begin{array}{l}{\left(15\right)}^2={\left(y\right)}^2+{\left(25\right)}^2\\ \left(225\right)+y^2=625\\ y^2=400\\ y=20\end{array}$

Also,

  $\begin{array}{l}{\left(15\right)}^2+{\left(z\right)}^2={\left(17\right)}^2\\ 225+z^2=289\\ z^2=64\\ z=\sqrt{64}\\ z=8\end{array}$

  $\begin{array}{l}y-z=x\\ 20-8=x\end{array}$

  $x=12$ feets