Chapter 5.2, Problem 31E

(a)

To determine

To find: The angle of inclination for Brownsville on the first day of summer (day 172) and on the first day of winter (day 355), if the latitude of Brownsville, Texas is ${26}^{\circ }$ .

Answer to Problem 31E

  ${85.3}^{\circ }\text{ on the first day of summer, 70}{\text{.7}}^{\circ }\text{ on the first day of winter}$

Explanation of Solution

Given information: The sun’s angle of inclination at noon equals ${90}^{\circ }-L-23.5\times \cos \left[\frac{\left(N+10\right)360}{365}\right]$ . In this expression, L is the latitude of the building site and N is the number of days elapsed in the year.

Calculation:

Angle of inclination = ${90}^{\circ }-L-23.5\times \cos \left[\frac{\left(N+10\right)360}{365}\right]$

Where L is the latitude and N is the number of days into the year (out of 365)

For Brownsville, Texas $l={26}^{\circ }$

First day of summer, N = 172

  $\begin{array}{l}={90}^{\circ }-{26}^{\circ }-{23.5}^{\circ }\times \cos \frac{\left(172+10\right)\times 360}{365}\\ ={64}^{\circ }-{23.5}^{\circ }\times \cos 179.5\\ ={64}^{\circ }-\left({23.5}^{\circ }\times -0.906\right)\\ ={64}^{\circ }-\left(-{21.3}^{\circ }\right)={85.3}^{\circ }\end{array}$

For the first day of winter N = 355

  $\begin{array}{l}={90}^{\circ }-{26}^{\circ }-{23.5}^{\circ }\times \cos \frac{\left(355+10\right)\times 360}{365}\\ ={64}^{\circ }-{23.5}^{\circ }\times \cos 360\\ ={64}^{\circ }-{23.5}^{\circ }\times \left(-0.2837\right)\\ ={64}^{\circ }-\left(-{6.66}^{\circ }\right)\approx {70.7}^{\circ }\end{array}$

  ${85.3}^{\circ }\text{ on the first day of summer, 70}{\text{.7}}^{\circ }\text{ on the first day of winter}$

(b)

To determine

To find: The angle of inclination for Nome on the first day of summer and on the first day of winter, if the latitude of Nome, Alaska, is ${64}^{\circ }$ .

Answer to Problem 31E

  ${41.2}^{\circ }\text{ on the first day of summer, 2}{\text{.5}}^{\circ }\text{ on the first day of winter}$

Explanation of Solution

Given information: The sun’s angle of inclination at noon equals ${90}^{\circ }-L-23.5\times \cos \left[\frac{\left(N+10\right)360}{365}\right]$ . In this expression, L is the latitude of the building site and N is the number of days elapsed in the year.

Calculation:

Angle of inclination =

  ${90}^{\circ }-L-23.5\times \cos \left[\frac{\left(N+10\right)360}{365}\right]$

Where L is the latitude and N is the number of days into the year (out of 365)

For Nome, Alaska $l={64}^{\circ }$

First day of summer, N = 122

  $\begin{array}{l}={90}^{\circ }-{64}^{\circ }-{23.5}^{\circ }\times \cos \frac{\left(122+10\right)\times 360}{365}\\ ={26}^{\circ }-{23.5}^{\circ }\times \cos 130.2\\ ={26}^{\circ }-23.5\times -0.6455\\ ={26}^{\circ }+15.17={41.2}^{\circ }\end{array}$

For the first day of winter N = 355

  $\begin{array}{l}={90}^{\circ }-{64}^{\circ }-{23.5}^{\circ }\times \cos \frac{\left(355+10\right)\times 360}{365}\\ ={26}^{\circ }-{23.5}^{\circ }\times \cos 360\\ ={26}^{\circ }-{23.5}^{\circ }\times 1\\ ={26}^{\circ }-{23.5}^{\circ }={2.5}^{\circ }\end{array}$

  ${41.2}^{\circ }\text{ on the first day of summer, 2}{\text{.5}}^{\circ }\text{ on the first day of winter}$

(c)

To determine

To find: city that has the greater change in the angle of inclination.

Answer to Problem 31E

  $\text{They both have the same change of }{38.7}^{\circ }$

Explanation of Solution

Given information: The sun’s angle of inclination at noon equals ${90}^{\circ }-L-23.5\times \cos \left[\frac{\left(N+10\right)360}{365}\right]$ . In this expression, L is the latitude of the building site and N is the number of days elapsed in the year.

Calculation:

Texas change:

  $79.2-40.5=38.7$

Alaska change:

  $41.2-2.5=38.7$

Thus,

  $\text{They both have the same change of }{38.7}^{\circ }$