Chapter 9.3, Problem 40PS

Solution

To calculate: The latitude of $\theta$ is the circumference of the circle of latitude at $P$

half around the equator.

As a result, the angle at which the circumference equals half the radius is ${60}^o$

Given information: The formula for calculating a circle's circumference is $2\pi r$ , where $r$ is the radius.

Calculation:

Divide the formula for circumference by $2$ since the problem asks for half the distance around the equator.

Half distance around the equator

  $\begin{array}{c}=22\pi r\\ \text{}=\pi r\end{array}$

To get the latitude where the circle's circumference is half its equatorial distance, apply trigonometric characteristics.

  

Since the angle is formed by the radius of the circle of latitude's adjacent line and the hypotenuse, which is also the radius of the Earth, use $\cos \theta$ .

Equating the known values together before solving for the angle

  $\begin{array}{c}\cos \theta =\frac{\text{ adjacent }}{\text{ radius }}\\ \cos \theta =\frac{a}{r}\\ r\cos \theta =a\end{array}$

Replace with the known values. Given that the circumference of the latitude should be half that of the Earth, the following formula can be used to solve the problem

  $2\pi a=\pi r$

Substitute $a=r\cos \theta$

  $2\pi rcos\theta \text{}=\pi r\text{}$

Dive both side by $\pi$

  $2rcos\theta =r$

  $\begin{array}{c}\frac{2r\cos \theta }{2r}=\frac{r}{2r}\\ \cos \theta =\frac{1}{2}\end{array}$

Using a calculator to solve $\theta ={60}^o$

As a result, the angle at which the circumference equals half the radius is ${60}^o$