Chapter 4.8, Problem 30E

a.

To determine

Find the height h of the slide.

Answer to Problem 30E

  $\boxed{25.98}$

Explanation of Solution

Given information:

The designers of a water park have sketched a preliminary drawing of a new slide

Find the height h of the slide.

Calculation:

Consider the following figure

   Since the sine of the angle is given by the ratio of the length of the perpendicular to the hypotenuse, therefore in right angled triangle $\Delta ACB$

  $\sin A=\frac{BC}{AB}$

Put the values

  $\sin 60°=\frac{h}{30}$

  $h=30\times \sin 60°$

  $=30\times 0.866$

  $=25.98$

Hence, the height $h$ of the slides is $\boxed{25.98}$ feet

b.

To determine

Find the angle of depression $\theta$ .

Answer to Problem 30E

  $\boxed{\theta ={\cot }^{-1}\left(\frac{d}{25.98}\right)}$

Explanation of Solution

Given information:

The designers of a water park have sketched a preliminary drawing of a new slide

Find the angle of depression $\theta$ from the top of the slide to the end of the slide at the ground in terms of the horizontal distance $d$ a rider travels.

Calculation:

Consider the following figure

  Since the tangent of the angle is given by the ratio of the length of the perpendicular to the base, therefore in right angled triangle $\Delta DEF$

  $\tan D=\frac{EF}{DE}$

Put the values

  $\tan D=\frac{d}{h}$

  $=\frac{d}{25.98}$

Since $D=90°-\theta$ , therefore

  $\tan \left(90°-\theta \right)=\frac{d}{25.98}$

  $\cot \theta =\frac{d}{25.98}$ , since $\tan \left(90°-\theta \right)=\cot \theta \mathrm{.....}\left(1\right)$

  $\theta ={\cot }^{-1}\left(\frac{d}{25.98}\right)$

Hence, the angle of depression from the top of the slide to the end of the slide at the ground is

  $\boxed{\theta ={\cot }^{-1}\left(\frac{d}{25.98}\right)}$

c.

To determine

Find an interval for how far a rider travels horizontally.

Answer to Problem 30E

  $\boxed{\text{45 feet to 55}.\text{71 feet}}$

Explanation of Solution

Given information:

The designers of a water park have sketched a preliminary drawing of a new slide

Safety restrictions require the angle of depression to be no less than $\text{25}°$ and no more than $30°$ . Find an interval for how far a rider travels horizontally.

Calculation:

Consider the following figure

  Assume that the angle of depression of the ride is bounded by safety restrictions to be less than $\text{25}°$ and not more than $30°$

When the minimum angle of depression is $\text{25}°$ , then from (1)

  $\cot \theta =\frac{d}{25.98}$

  $d=25.98\cot 25°$

  $=25.98\times 2.1445$

  $=55.71$

When the maximum angle of depression is $30°$ , then from (1)

  $\cot \theta =\frac{d}{25.98}$

  $d=25.98\cot 30°$

  $=44.99$

  $\approx 45$

Hence, the rider travels horizontally between $\boxed{\text{45 feet to 55}.\text{71 feet}}$