Chapter 9, Problem 18TP

Solution

a.

To determine the depth of water.

The depth of water is $17.1\text{ feet}\text{.}$

Given:

It is given that the rope length is $50\text{ feet}$ and the rope makes $20°$ angle with the bottom of lake.

Calculation:

The rope position with respect to the bottom of the lake is shown as:

  

Now, calculating the depth of the water $h,$ using trigonometric function as:

  $\begin{array}{l}\text{ s}in\theta =\frac{\text{Height}}{\text{Hypotenuse}}\\ \sin \theta =\frac{AB}{AC}\\ \sin {20}^{\circ }=\frac{h}{50}\\ h=50\times 0.342\\ h=17.1\text{ feet}\end{array}$

Hence, the depth of the water is $17.1\text{ feet}\text{.}$

b.

To determine the what angle $\theta$ does the rope make with the lake bottom.

The rope makes with the lake bottom the angle $\theta$ is $42.3°.$

Given:

It is given that rope length is $50\text{ feet}$ and the horizontal distance from the boat to the anchor is $37\text{ feet}\text{.}$

Calculation:

The new rope position with respect to the bottom of the lake is shown as:

  

Now, calculating the angle $\theta$ using trigonometric function as:

  $\begin{array}{l}\cos \theta =\frac{\text{Base}}{\text{Hypotenuse}}\\ \cos \theta =\frac{DE}{DF}\\ \cos \theta =\frac{37}{50}\\ \theta ={\cos }^{-1}\left(\frac{37}{50}\right)\\ \theta ={\cos }^{-1}\left(0.74\right)\\ \theta =42.3°\end{array}$

Hence, the rope makes with the lake bottom the angle $\theta$ is $42.3°.$

c.

To describe how $\theta$ changes as the boat travels to deeper water.

  $\theta$ increases as the boat travels to deeper water.

Given:

It is given that the length of anchor rope is constant and assume that rope is always taut.

Interpretation:

When the boat travels to deeper water with a constant length of anchor rope the value of $\theta$ increases. The same length of rope needs to reach the bottom of the lake resulting in the less horizontal distance and hence the angle with respect to the bottom of the lake will increase.