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# Global Navigation Help (page 2)

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By — McGraw-Hill Professional
Updated on Aug 30, 2011

#### 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 great-circle route from point R to point S in the scenario of Problem 11-7. 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 11-7:

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 great-circle route from point S to point R in the scenario of Problem 11-7. 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 11-7:

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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