Chapter 7.1, Problem 79AYU

Problems 89 and 90 require the following discussion:

The shortest distance between two points on Earth’s surface can be determined from the latitude and longitude of the two locations. For example, if location 1 has $\left(\text{lat,}\text{lon}\right)=\left({\alpha }_1,{\beta }_1\right)$ and location 2 has $\left(lat,lon\right)=\left({\alpha }_2,{\beta }_2\right)$, the shortest distance between the two locations is approximately $d=r{\cos }^{-1}\left[\left(\cos {\alpha }_1\cos {\beta }_1\cos {\alpha }_2\cos {\beta }_2\right)+\left(\cos {\alpha }_1\sin {\beta }_1\cos {\alpha }_2\sin {\beta }_2\right)+\left(\sin {\alpha }_1\sin {\alpha }_2\right)\right]$, where $r=radiusofEarth\approx 3960$ miles and the inverse cosine function is expressed in radians. Also, $N$ latitude and $E$ longitude are positive angles, and $S$ latitude and $W$ longitude are negative

Source: www.infoplease.com

Shortest Distance from Chicago to Honolulu Find the shortest distance from Chicago, latitude ${41}^{\circ }50\text{'}N$, longitude ${87}^{\circ }37\text{'}W$, to Honolulu, latitude ${21}^{\circ }18\text{'}N$, longitude ${157}^{\circ }50\text{'}W$. Round your answer to the nearest mile.

To determine

To calculate: The Shortest Distance from Chicago to Honolulu.

Answer to Problem 79AYU

The Shortest Distance from Chicago to Honolulu $\approx 4252\text{ miles}$.

Explanation of Solution

Given:

Chicago, latitude $41°50\prime N$, longitude $87°37\prime W$ to Honolulu, latitude $21°18\prime N$, longitude $57°50\prime W$.

${\alpha }_1=41°50\prime N, {\beta }_1=87°37\prime W$

${\alpha }_2 =21°18\prime N, {\beta }_2=157°50\prime W$

$r=$ radius of the earth $\approx 3960\text{ miles}$

$\text{N}$ latitude and $\text{E}$ longitude are positive angles, and $\text{S}$ latitude and $\text{W}$ longitude are negative Angles.

Formula used:

Shortest distance $d= r{\text{cos}}^{-1} \left[\text{cos}{\alpha }_1\text{cos}{\beta }_1\text{cos}{\alpha }_2\text{cos}{\beta }_2+\text{ cos}{\alpha }_1\text{sin}{\beta }_1\text{cos}{\alpha }_1\text{sin}{\beta }_1+\text{ sin}{\alpha }_1\text{sin}{\alpha }_2\right]$.

Calculation:

Using the formula,

$d=3960{\text{cos}}^{-1}[\text{cos}(41°50\prime N)\text{cos}(-87°37\prime W)\text{cos}\left(21°18\prime N\right)\text{cos}(-157°50\prime W)+ \text{cos}(41°50\prime N)\text{sin}(-87°37\prime W)\text{cos}\left(21°18\prime N\right)\text{sin}(-157°50\prime W)+\text{ sin}(41°50\prime N)\text{sin}\left(21°18\prime N\right)]$

$=3960{\text{cos}}^{-1}[\text{cos}\left(41.83°\right)\text{cos}(-87.61°)\text{cos}\left(21.3°\right)\text{cos}(-157.83°)+\text{ cos}\left(41.83°\right)$

$\text{sin}(-87.61°)\text{cos}(21.3°)\text{sin}(-157.83°)+\text{ sin}\left(41.83°\right)\text{sin}\left(-21.3°\right)]$

$=3960{\text{cos}}^{-1}[0.7451\text{ x }0.0417\text{ x }0.9316\text{ x }\left(-0.9260\right)+0.7451\text{ x }0.9991\text{ x }0.9316\text{ x }0.3773+ 0.6669\text{ x }0.363]$

$=3960{\text{cos}}^{-1}\left[-0.0268+0.2616+0.242\right]$

$=3960{\text{cos}}^{-1}\left[0.4768\right]$

$=3960\text{ x }61.52°$

$=3960\text{ x }{1.073}^c$

$\approx 4252\text{ miles}$