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BD 39.29 (4) sin 60.6=
BO- BC ?
log BC= 1.6542
BC=45.1 (5) a=BC=45.10, b=AD-CD=61.91-22.14=39.77.
6. Exercises.-(1) Two angles and the included side of a triangle are 35°, 25°, and 18° inches. What are the missing parts? (2) Find the area of a triangle whose sides are 25, 29, and 31.
339.8 Answer. (3) Find the missing parts of the triangle shown in figure 129.
(4) Find the length of the altitude to the 29 side in exercise (2).
Answer. (5) Find the angles of the triangles in exercises (4) and (5).
(6) An airplane has an air speed of 217.6 mph. and a heading of 110°. A 47.5 mph, wind is blowing from 20°. Find the ground speed and track.
122.3°, 222.7 mph. Answer. (7) An airplane flies over a straight road. Two observers on the road 15 miles apart, find its angles of elevation at the same time and record them as 14° and 10.3°. How high is the airplane?
Purpose and scope--
85 86 87 88 89 90 91 92 93 94
85. Purpose and scope.—a. Spherical trigonometry is presented for the purpose of providing the necessary mathematical background in the study of celestial navigation.
6. The scope includes the basic elements of the right and oblique spherical triangles together with the application of spherical triangles to the solution of the astronomical triangle as used in celestial navigation. The paragraphs below contain explanations with illustrations designed to give the student a working knowledge of necessary fundamentals of spherical trigonometry.
86. Definitions and geometric properties of a sphere.—a. In order to solve right spherical triangles it is necessary that new terms be introduced.
b. The definitions and properties which follow should be learned thoroughly as they are essential to an understanding of spherical trigonometry.
(1) Sphere.- A sphere is a closed surface in three dimensions all of whose points are equidistant from a fixed point called the center. Usually the earth is considered as a true sphere, although this is not strictly true.
(2) Circle. -A plane intersects the surface of a sphere in a circle.
FIGURE 130.--Spherical triangles.
(3) Great and small circles.-A plane passing through the center of a sphere intersects the surface of a sphere in a great circle. Other circles on the sphere are called small circles.
(4) Hemispheres.-A plane through the center of a sphere divides the sphere into two equal parts, called hemispheres.
(5) Spherical triangle.—A spherical triangle is that portion of the surface of a sphere which is bounded by three great circular arcs.
(6) Great circle distance. -Distance on a great circle of a sphere is usually measured in arc units, for example, degrees, minutes, and seconds.
(7) Spherical triangle rules.-Referring to triangle ABC, figure 130, the following rules, which were given in section IX, apply:
(a) (a+b+c) is less than 360°.
(c) If the length of two sides of a spherical triangle are equal, the opposite angles are equal, and conversely.
87. Special rules for solving spherical triangles.—This section deals only with right spherical triangles except for a special solution of an oblique triangle by right triangle formulas. As in the case of plane right triangles, the right angle is lettered C. A right spherical triangle can be solved if any two parts other than the right angle are given, even if the two given parts are angles.
a. The nature of the angles and arcs of a spherical triangle can be made clear by a study of figure 131.
(1) Radii drawn to A, B, and C, form three planes OBC, OBA, and OAC.
(2) The plane DEF is drawn perpendicular to the line OA and thus angles OED and OEF are right angles.
(3) The dihedral angle between planes OAB and OAC is angle DEF and is also angle A of the spherical triangle.
(4) This is true because the measure of a dihedral angle between two planes is the angle between two lines in the two planes which meet at the vertex and are perpendicular to the vertex line, as EF and DE.
(5) The distance OD in the figure was made unity or 1 for convenience. - 6. In the right triangle OED DE
Since DFE was drawn perpendicular to 0A, it is perpendicular to both plane OAC and plane OAB. Plane OBC is perpendicular to plane OAC since the spherical triangle is a right spherical triangle with the right angle at C. Therefore DF is perpendicular to both EF and OF. In triangle OBF
sin A.sin:c=sin a g. In triangle DEF
EF cos A=
cos c. tan 6 DE
=cot c . tan b
h. In a similar manner, by drawing a plane perpendicular to OB so that a triangle with angle EDF=B can be drawn, a similar development would give five more formulas, making a total of 10 formulas for the solution of the right spherical triangle. These formulas are ás follows: (1) sin a=sin c.sin A
(6) sin b=sin c.sin B (2) tan b=tan cócos A
(7) tan a=tan cócos B (3) tan a=sin b-tan A
(8) tan b=sin a.tan B (4) cos A=sin B.cos a
(9) cos B=sin A.cos b (5) cos c=cos a•cos b
(10) cos c=cot A.cot B 88. Napier's rules.-a. General.--The ten formulas just given are required in the solution of right spherical triangles. They may be obtained from two simple rules which were discovered by John Napier. These rules which follow serve as an aid in remembering the formulas.
b. Rules.-Consider the right spherical triangle ABC, figure 132, lettered as usual. All parts of the triangle except C=90° are arranged in circular order, figure 133, as they occur in the triangle, using
90°-A, 90°-C, 90°-B instead of merely the letters A, C, B as in figure 132. With the circle labeled as in figure 132, looking at the