Chapter 7.1, Problem 80AYU

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 $\left(lat,\text{lon}\right)=\left({\alpha }_1,{\beta }_1\right)$ 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 Honolulu to Melbourne, Australia Find the shortest distance from Honolulu to Melbourne, Australia, latitude $37°47\text{'}S$, longitude $144°58\text{'}E$. Round your answer to the nearest mile.

To determine

To calculate: The Shortest Distance from Honolulu to Melbourne.

Answer to Problem 80AYU

Solution:

The Shortest Distance from Honolulu to Melbourne $\approx 4169\text{ miles}$.

Explanation of Solution

Given:

Honolulu, latitude $21°18\prime N$, longitude $157°50\prime W$ to Melbourne, Australia, latitude $37°47\prime S$, longitude $144°58\prime E$.

${\alpha }_1=21°18\prime N, {\beta }_1=-157°50\prime W$

${\alpha }_2 =-37°47\prime S, {\beta }_2=144°58\prime E$

$r=$ 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,

$d=3960{\text{cos}}^{-1}[\text{cos}(21°18\prime N)\text{cos}(-157°50\prime W)\text{cos}\left(-37°47\prime S\right)\text{cos}(144°58\prime E)+ \text{cos}(21°18\prime N)\text{sin}(-157°50\prime W)\text{cos}\left(-37°47\prime S\right)\text{sin}(144°58\prime E)+\text{ sin}(21°18\prime N)\text{sin}\left(-37°47\prime S\right)]$

$=3960{\text{cos}}^{-1}[\text{cos}\left(21.3°\right)\text{cos}(-157.83°)\text{cos}\left(-37.78°\right)\text{cos}(144.96°)+\text{ cos}\left(21.3°\right)$

$\text{sin}(-157.83°)\text{cos}(-37.78°)\text{sin}(144.96°)+\text{ sin}\left(21.3°\right)\text{sin}\left(-37.78°\right)]$

$=3960{\text{cos}}^{-1}[0.9316\text{x}\left(-0.926\right)\text{x}0.7903\text{ x }\left(-0.8187\right)+0.9316\text{ x }0.3773\text{ x }0.7903\text{ x }0.5741+0.3632\text{ x }\left(-0.6126\right)$

$=3960{\text{cos}}^{-1}\left[0.5581+0.1594-0.2224\right]$

$=3960{\text{cos}}^{-1}\left[0.4951\right]$

$=3960\text{ x }60.32°$

$=3960\text{ x }{1.0523}^c$

$\approx 4169\text{ miles}$