Chapter 9.1, Problem 35PS

Solution

1. To determine the circumference of the Tropic of Cancer.
2. To determine the distance between two points directly across on the Tropic of Cancer.

Given:

The radius of the Earth is $3960$ miles, the Tropic of Cancer lies ${23.5}^{\circ }$ North of the Equator.

Calculation:

Consider the given diagram of the Tropic of Cancer and Equator as shown:

  

Here, $R=3960$ is the Radius of the Earth, $r$ is the radius of the plane of the Tropic of Cancer. The Tropic of Cancer lies ${23.5}^{\circ }$ North of the Equator. Hence, $\angle ABC={66.5}^{\circ }$ .

In $△ABC$ :

  $\sin B=\frac{AC}{BC}$

  $\Rightarrow \sin B=\frac{r}{R}$

  $\Rightarrow r=R\sin B$

  $\Rightarrow r=3960*\sin ({66.5}^{\circ })$

  $\Rightarrow r=3631.56$

1. Hence, the circumference of the Tropic of Cancer, C

  $C=2*\pi *3631.56$

  $=22817.76$ miles.

2. The distance between two points directly across on the Tropic of Cancer is the diameter of the plane of the Tropic of Cancer, D

  $D=2*3631.56$

  $D=7263.12$ miles.