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Also if log sin a is positive, the triangle is impossible.

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173. In questions concerning geodesy or navigation, the earth may be regarded as a sphere.

The shortest path between any two points is the arc of a great circle which joins them, and the angles between this arc and the meridians of the points determine the bearings of the points from each other.

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Thus, if Q and Q' are the points, and PQ and PQ' their meridians, the angle PQQ' determines the bearing of Q' from Q, and the angle PQ'Q determines the bearing of Q from Q'.

If the latitudes and longitudes of Q and Q' are known, the arc QQ' and the angles PQQ' and PQ'Q may be determined by the solution of a spherical triangle.

For if EE' is the equator, and PG the meridian of Greenwich, we have

angle QPQ' angle Q'PG— angle QPG

=

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Thus, in the triangle PQQ', two sides and their included angle are known, and the remaining elements may be computed.

Note. When QQ' has been found in angular measure, its length in miles may be calculated by the method of Art. 132. In the following problems the diameter of the earth is taken as 7912 miles.

1. Boston lies in lat. 42° 21' N., longitude 71° 4′ W.; and the latitude of Greenwich is 51° 29' N. Find the shortest distance in miles between the places, and the bearing of each place from the other.

2. Calcutta lies in lat. 22° 33′ N., lon. 88° 19′ E.; and Valparaiso lies in lat. 33° 2' S., lon. 71° 42' W. Find the shortest distance in miles between the places, and the bearing of each place from the other.

3. Sandy Hook lies in lat. 40° 28' N., lon. 74° 1′ W.; and Queenstown lies in lat. 51° 50' N., lon. 8° 19' W. In what latitude does a great circle course from Sandy Hook to Queenstown cross the meridian of 50° W.?

If the latitude of a place is known, and the altitude and declination of the sun, the solution of a spherical triangle serves to determine the hour of the day at the time and place of observation.

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Thus let O be the position of the observer; P the celestial north pole; EE' the celestial equator; HH' the horizon; Z the zenith; S the sun's position; PSM a meridian passing through the sun's position; and ZSN a great circle passing through Z and S.

Then SM is the sun's declination, SN its altitude, and EZ the latitude of the place of observation.

Then in the spherical triangle SPZ, we have

SP=PM-SM 90°- the sun's declination,

SZ

=

ZN-SN = 90°- the sun's altitude, and PZ EPEZ 90° - the latitude of the place.

That is, the three sides of the triangle SPZ are known, and the angle SPZ may be computed.

If 24 hours is multiplied by the ratio of this angle to 360°, we have the time required for the sun to move from S to the meridian EP.

Hence, if this time is subtracted from 12 o'clock, if the observation is made in the morning, or added, if made in the afternoon, we obtain the hour of the day at the time and place of observation.

If the Greenwich time of the observation is noted on a chronometer, the difference between this and the local time as calculated above serves to determine the longitude of the place of observation.

In reducing time to longitude, it should be borne in mind that 24 hours of time correspond to 360° of longitude; that is, one hour of time corresponds to 15° of longitude, one minute to 15', and one second to 15".

4. A mariner observes the altitude of the sun to be 14° 18', its declination being 18° 36' N. If the latitude of the vessel is 50° 13' N., and the observation is made in the morning, find the hour of the day. If the observation is taken at 9 A.M., Greenwich time, what is the longitude of the vessel?

5. What will be the altitude of the sun at 4 P.M. in San Francisco, lat. 37° 48' N., its declination being 12° S.?

6. In Melbourne, lat. 37° 49' S., the altitude of the sun is observed to be 25° 46'. If the sun's declination is 3° S., and the observation is made in the morning, find the hour of the day.

7. At what hour will the sun rise in Boston, lat. 42° 21' N., when its declination is 15° N.?

Note. At sunrise the sun's altitude is 0, so that the arc SZ becomes 90°.

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tan (90°+4)=-cot A,

sec (90°+4)=-esc A, esc (90°+4)= A)

cot (90°+4)=-tan A, A) sec A.

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