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circle, figure 133, any part may be called the middle part; the two parts next to it, adjacent parts, and the remaining two, opposite parts. Vapier's rules state:

(1) The sine of a middle part equals the product of the tangents of the adjacent parts.

Thus:

sin b=tan a·tan (90°- A) but

tan (90°— A)=cot A (by definition) therefore

sin b=tan a.cot A (2) The sine of the middle part equals the product of the cosines of the opposite parts. Thus:

sin (90°– A)=cos (90°-B).cos a but

cos (90° —B)=sin B (by definition) therefore sin (90°- A)=sin B.cos a but

sin (90°- A)=cos A (by definition) therefore

cos A=sin B.cos a C. An aid in memorizing.These rules may be easily remembered if it is observed that the letter “” is common in the words “sine" and “middle”; the letter “a” is common in the words “tangent” and “adjacent”; and the letter “o” is common to the words “cosine" and “opposite."

89. Formula exercises.-a. Examples.--In these exercises the object is to obtain the formulas to be used in the solutions, solved for the unknown part.

Example 1: Given a and c, find b.

Solution: Of the three parts, a, b, and 90°-C, the part 90o--c is the middle part in the circular diagram while a and b are opposite parts; Hence

sin (90°—c)=cos a.cos 6 or

cos c=cos a•cos b Since 6 is the unknown:

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b. Exercises. Obtain the formulas solved for the unknown part

necessary for the solution of the following right spherical triangles, where the parts given are indicated. (1) a and b

(6) b and B (2) a and A

(7) c and A (3) a and B

(8) c and B (4) b and c

(9) A and B (5) b and A

(10) Given a and c, find B 90. General laws.-a. Obtuse angles. In the solution of right spherical triangles, obtuse angles will be encountered. Obtuse angles are greater than 90° and less than 180°. Since trigonometric tables give only the functions of acute angles, it is necessary to learn certain rules which follow:

b. Rules.-(1) Acute angles will be called first quadrant angles or angles in the first quadrant.

(2) Obtuse angles will be called second quadrant angles or angles in the second quadrant.'

(3) If a trigonometric function of an obtuse angle is involved in the solution, use the same function of its supplementary angle. For example, instead of cos 110°, use cos (180°—110°)=cos 70°.

C. Laws of quadrants. The following laws of quadrants must be mastered by the student:

(1) An angle and its opposite side are in the same quadrant.

(2) If two of the three sides, a, b, c, are in the same quadrant, the third is in the first quadrant;. if two are in different quadrants, the third is in the second quadrant.

91. Solution of right spherical triangles.-a. Suggestions. The following suggestions will be found useful in the solution of any right spherical triangle.

(1) Take any two given parts and solve for an unknown part by use of Napier's rules.

(2) Use the law of quadrants to determine in what quadrant the unknown parts lie.

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b. Eramples: (1) Solve the right spherical triangle ABC if B= 113°40' and a -23°50'.

Solution:
(a) Using Napier's rules with the parts, a, b, 90°– B,

sin a=tan b.tan (90°— B)=tan b.cot B

tan b=sin a

Ocot B
Since B=113°40', use cot (180°— 113°40')=cot (66°20')

log sin 23°50'=19.6065—20 Subtract
log cot 66°20'=9.6418–10
log tan b=9.9647—10

b=180—°42°40'=137°20' The supplementary angle to 42°40' was used since b and B are in the same quadrant. (6) To find c, take the parts 90°-C, a, 90°—B giving sin (90°--B)=tan a.tan (90°—c)

cos B

R-tan a.cot c and cot (= tan a

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log cos 66°20'=19.6036—20 Subtract
log tan 23°50'=9.6452-10
log cot c=9.9584-10

c=180°— 47°44'=132°16' a is in the first quadrant; b is in the second quadrant. Hence c is in the second quadrant. (c) To find A, take the parts A, a, 90°-B, giving

sin (90°- A)=cos a.cos (90°-B)

cos A=cos a:sin B.
log cos 23°50'=9.9613—10 Add
log sin 66°20'=9.9618–10]
log cos A=19.9231-20

A=33°6'
A is in the first quadrant since a is.

(2) Solve the right spherical triangle if a=97o and b=-1020.

Solution: Since b is in the second quadrant, B is also. Similarly, A is in the second quadrant while c is in the first quadrant. Complete the solution.

92. Exercises. a. a=87°13'

b. a=131°19' b=95°42'

b=102°43'

C. a=84°27'

g. C=67°44' b=108°30'

A=112°39' d. b=112°11'

a=89°14' c=64°46'

B=75°13' e. b=64°27'

i. c=68°21' A=103°2'

B=122°23 f. A=97°35'

j. a=114° B=47°14'

c=60°49' 93. The astronomical triangle.—a. Definition.—The spherical triangle which is encountered in solving problems in celestial navigation is called the astronomical triangle. (See fig. 135.) It is also often called the PZS triangle since these letters are used for the vertices. P is for north pole or south pole; Z is the zenith or point directly above the observer, and S is the location of the star (which was sighted in making the observation).

- 90° Lat

Latitude

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20 - AHitide

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FIGURE 135.-The astronomical triangle. 6. Line of position.—The PZS triangle is shown in figure 136. A sight is taken on a star and its altitude above the horizon is measured and compared to the computed altitude based on an assumed position. This provides data to plot a "line of position” (LOP) on a navigation chart. Two such lines intersecting provide a "fix" or position of the plane on the chart.

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c. Determination of azimuth.The PZS triangle is oblique but a perpendicular can be dropped from S to PZ or to PZ extended, as in figure 136. Triangle SDZ is a right triangle and so is triangle SDP and either may be solved by the 10 right triangle formulas or by. oboz

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Napier's rules. If the latitude and longitude of the position are assumed (the usual procedure), then PZ is known, angle ZPS, called the hour angle of the star, is known and side PS is known. Then the altitude of the star may be computed and so may its azimuth, which is angle PZS. From the right triangle PDS:

CODZS

COP

COPSD

Co ZS

COZSD

COPS
FIGURE 137.–Reference spherical triangle for determination of azimuth.

sin R=cos (90°—PS).cos (90°—P) (See fig. 137)

sin R=sin PS:sin P From triangle PDS,

cos PS=cos R.cos PD
cos PD=cos Ps which gives PD

or

COS

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