Solution 2
We multiply the angular distance, 1.95 rad, by 6371 kilometers. This gives us 12,400 kilometers, accurate to three significant figures.
Practice 3
Suppose an aircraft pilot wants to fly a greatcircle route from point R to point S in the scenario of Problem 117. What should the initial azimuth heading be as the aircraft approaches cruising altitude after taking off from point R ? Express the answer to the nearest degree.
Solution 3
In order to determine this, we must figure out the measure of angle ψ _{r} in degrees. The initial azimuth heading is 360° – ψ _{r} . This is because side s of Δ _{sph} QRS runs directly northward, or toward azimuth 0°, from point R . The spherical law of sines tells us this about Δ _{sph} QRS :
(sin q )/(sin ψ _{q} ) = (sin r )/(sin ψ _{r} )
We can solve this equation for _{r} by manipulating the above expression and then finding the arcsine:
sin ψ _{r} = [(sin r )(sin ψ _{r} )]/(sin q )
ψ _{r} = arcsin {[(sin r )(sin ψ _{r} ])/(sin q )}
We already have the following information, having solved Problem 117:
q = 1.95 rad
ψ q = 109.5°
r = 102.00°
Plugging in the numbers gives us this:
ψ _{r} = arcsin {[sin 102.00°)(sin 109.5°)]/(sin 1095 rad)}
= arcsin [(0.97815 × 0.94264)/0.92896]
= arcsin 0.99255
= 83.00°
This means that the pilot’s initial heading, enroute on a great circle from point R to point S , should be 360° – 83.00°, or 277° to the nearest degree. This is 7° north of west.
Practice 4
Suppose an aircraft pilot wants to fly a greatcircle route from point S to point R in the scenario of Problem 117. What should the initial azimuth heading be as the aircraft approaches cruising altitude after taking off from point S ? Express the answer to the nearest degree.
Solution 4
This is similar to the previous problem. We must figure out the measure of angle ψ _{s} in degrees. The initial azimuth heading is equal to ψ _{s} because side r of Δ _{sph} QRS runs directly northward, or toward azimuth 0°, from point S. According to the spherical law of sines:
(sin q )/(sin ψ _{q} ) = (sin s )/(sin ψ _{s} )
We can solve this equation for ψ _{s} by manipulating the above expression and then finding the arcsine:
sin ψ _{s} = [(sin s )(sin ψ _{q} )]/(sin q )
ψ _{s} = arcsin {[(sin s )(sin ψ _{q} ])/(sin q )}
We already have the following information, having solved Problem 117:
q = 1.95 rad
ψ _{q} = 109.5°
s = 40.00°
Plugging in the numbers gives us this:
ψ _{s} = arcsin {[sin 40.00°)(sin 109.5°)]/(sin 1.95 rad)}
= arcsin [(0.64279 × 0.94264)/0.92896]
= arcsin 0.65226
= 40.71°
This means that the pilot’s initial heading, enroute on a great circle from point S to point R, should be 41° to the nearest degree. This is 41° east of north.
Practice problems for these concepts can be found at: Global Trigonometry Practice Test
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