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(6) 43°12' N and 25°28' N. 1,064 nautical miles. Answer. (c) 14°27' N and 16°14' S. (d) 25°11' N and 76° 59' S. 6,130 nautical miles. Answer.
c. Time differences.—(1) Since the earth turns on its axis approximately once each day, it is apparent that 360° of turning must correspond to 24 hours of time. For convenience, the earth's surface is divided into 24 time belts of roughly 15° longitude each: that is, 15° of longitude corresponds to 1 hour of time. It is also evident that 1° longitude corresponds to 4 minutes of time. The reference time belt again starts at Greenwich, England. Sometimes for further convenience, the time belts do not directly follow the meridians but take geographical and population factors into consideration. In the United States we have four time belts, making a difference in time between the East and West Coasts of 3 hours.
(2) True time is not determined by belts, but must be determined for each separate location from the exact longitudinal position.
(a) Example 1: Find the exact time difference between 15°13' E and 74°43' E. Solution: The difference in longitude is obtained by subtraction.
59°30' or 59.5° difference in longitude Since each degree corresponds to 4 minutes time, multiplying 59.5 by 4 gives 238 minutes time difference. Changing to hours and minutes, the time difference is 3 hours 58 minutes.
(6) Example 2: Find the exact time difference between 17°10' W and 21°35' E.
Solution: Since the longitudes are on opposite sides of the Prime Meridian, they must be added.
38°45' or 383°=difference in longitude Each degree corresponds to 4 minutes difference in time. Multiplying 3894° by 4 gives 155 minutes as the time difference. Changing to hours and minutes the time difference is 2 hours 35 minutes.
(c) Exercises.—Find the exact difference in time between the following:
1. 15°13' E and 120°13' E.
SECTION X :
Paragraph Purpose and scope.-
63 Development and basic formulas ---Functions of 30°, 45°, and 60° angles Complementary angles ---Use of tables.--Methods of solving right triangles--Interpolation.. Use of right-triangle methods in solving obtuse triangles.---Miscellaneous exercises-------
63. Purpose and scope.—Trigonometry is based on the properties of similar triangles. It is applied whenever angles enter into the solution of the problem and is important in navigation and bombing. This section is limited to a discussion of the properties of right triangles.
64. Development and basic formulas.—.. Take any acute angle DAE, figure 10, and from one side drop any number of perpendiculars to the other side; a series of similar triangles are formed. ZA is the same angle and the series of angles at C, C', C'"', etc. are all 90°, hence all angles of the triangles are equal.
C C C"
b. From geometry it is known that the corresponding sides are in
: BC B'C' B'C' proportion; that is = 8 AB-AB-AB'i' and
Pir, and so for the other sides.
:- AB AC - AB Similarly each of the ratios BA B and án has a series of other
ratios equal to it. These ratios of the sides remain unchanged as long as the angle remains unchanged, but they change as the angle
.. AᏴ AB BC BC AᏟ , , AᏟ ... changes. The six possible ratios
vie ratios AC BC AB AC AB' and BC are therefore the functions of angle A, because each is completely determined by angle A.
c. Names of functions.—(1) By definition, the following ratios, figure 111, in a right triangle have been given names.
FIGURE 111.-Standard lettering used for identification of angle functions.
side opposite a (2) The foregoing relations must be learned by the student. It should also be noted here that the lettering in figure 52 is standard and variations thereof are not used.
65. Functions of 30°, 45°, and 60° angles.-By using geometry, it is rather easy to find given functions for 30°, 45°, and 60° angles.
a. The functions of the 45° angle may be found as follows: Example: In the isosceles right triangle ABC, figure 112,
A=45°, B=45°, a=b
FIGURE 112.--Isosceles right triangle.
Solution: Since a2 +62=c?, it follows
2a2=c2 or cravi This gives the sides of the triangle in terms of one unknown and therefore the following is true:
sin A=% -002 z 1.4142=0.7071
b. In a like manner, the other five functions may be found. The functions of the 30° and 60° angle may be found as follows:
Example: In the equilateral triangle ABD, figure 113, BC is the perpendicular bisector of the base AD and also the bisector of ZABD.
From the foregoing values of a, b, and c the functions of a 60° angle may be found as follows:
c13 (13_V3 —.8660
sin 60o== 2 == =.8660
cos 60°= = 2 =.5000
(1) cos 45° (2) tan 45o
1.000 Answer. (3) cot 45° (4) sec 45°
1.141 Answer. (5) csc 45° (6) tan 60°
1.732 Answer. (7) cot 60° (8) sec 60°
2.000 Answer. (9) Find all six functions of 30°.
66. Complementary angles.—a. Two angles whose sum is 90° are said to be complementary and either is the complement of the other. By inspection of the previously defined functions, the following relationships are seen to exist.