1 To find the two other angles. Assum of the given sides 290 =arith.comp.of log. 7.537602 :diff. of the sides 16=log. -1 1.204120 :: tan.of 1 sum op. angs.= 3(180° —- 40°33' 12") 69° 43' 24" = log-1 ' = 10.432446 : tan. of diff. of op. angs. log. of which is sum rejec. 10 = 9.174168=log.tan. 8° 29' 37" Add and subt. with } sum of angs. 69° 43' 24" sum=greater angle 78° 13' 1" diff. = lesser angle 61° 13' 47" = * This notation" = log. - 1” signifies whose log. is. : opp. side 153† :: sin 40° 33′ 12′′ = log.-19.813017 side opp. 40° 30′ 12′′ log. 1 sum rejec. 102.006958 The length of the lake is 101.616 yards. log of 101.616 This case of the solution of a triangle combined with that exhibited at Art. 65, serves to determine the horizontal distance between two inaccessible objects. Let the distance between two towns which are in sight be required. Measure a line upon the ground (which is called the base line) of 2 miles. Take the angles at each extremity formed by this base line and a line to each of the towns. Two triangles will be formed in each of which a side, viz. the base line, and two adjacent angles will be given. Let the angles in the triangle of which the upper town is the vertex be 159° and 14°; and those in that of which the lower town is the vertex be 25° and 1.59% 2 Miles 149. 149°. Calculate the distance from one extremity of the base * A more direct mode of finding this side when it is the only part required, is given at Art. 123. + 153 is known to be the side opposite 78° 13' 1" because the greater angle of a triangle is always opposite the graster sida (Goom. B. 1, Prop. 13.) line, say the upper extremity, to each of the towns as in Art 65, Then you will know two sides of a triangle the third side of which is the distance between the towns required. The included angle between these two sides is 159c 25° 134°. Having then two sides and the included angle, the remainder of the solution is the same as in the last case, We leave it as an exercise for the learner. PART II. SPHERICAL TRIGONOMETRY. 77. A spherical triangle is formed by three arcs of great circles. The planes of these arcs produced, form a trisolid angle, the vertex of which is at the centre of the sphere, (Geom. B. 9, Prop, 1.) The angle which two planes make with each other, is called a diedral angle. This angle is equal to the angle formed by two lines, drawn one in each plane and perpendicular to the common intersection of the two planes at the same point. (Geom. B. 6, Def. 4.) The angles of a spherical triangle are the angles formed by the planes of the arcs, (Geom. B. 9, Def. 1.) These are the diedral angles of the trisolid angle, mentioned above. The sides of the spherical triangle are the arcs of the great circles, by which it is bounded. These arcs subtend the plane angles of the trisolid angle, and consequently measure them. The arcs are given in degrees, and since they contain the same number as the plane angles which they subtend, these plaue angles may be employed in a demonstration, instead of 1 the arcs, or sides of the spherical triangle; and for a like reason, the diedral angles of the trisolid angle may be employed instead of the angles of the triangle. If we suppose the trisolid angle, which has its vertex at the centre of a small sphere, to be produced so as to cut out a triangle upon a larger concentric* sphere, the sides of the triangle upon the larger sphere, containing respectively the same number of degrees as the plane angles of the trisolid angle, will contain the same number of degrees respectively as the sides of the spherical triangle cut out by the trisolid angle on the smaller sphere. So that as the number of degrees in the angles and sides which are given or required, and not their absolute length, is taken into consideration in the solution of spherical triangles, the size of the sphere need not be regarded. 78. Let abc be a spherical с triangle right angled at A. Let o be the centre of the sphere, and let the planes A. of the arcs which are the sides of the triangle be produced so as to form the trisolid angle whose vertex is P a 6 a B M at o. D a The plane angle coB will contain the same number of degrees as the side a of 5 the spherical triangle; the plane angle coa, the same number as the side b; and AOB the same number as c; so that these plane angles may be marked a, b, and c as in the figure. It has already been mentioned that the diedral angles of the trisolid angle correspond in the same manner to the angles of the spherical triangle; and that these diedral an . Having the same cantre. gles are measured by the angle of two lines, drawn one in each plain, perpendicular to the common intersection of the two planes at the same point. In order to draw these lines so as to be used most conveniently in the following demonstration, take o м the radius of the tables; draw м P perpendicular to o A, it will be perpendicular to the plane A O B; (Geom. B. 6, Prop. 17.) since the two planes A o B and A O C are perpendicular to each other, ▲ being by hypothesis a right angle; from P draw P D perpendicular to o e, a line of the plane A O B; join DM; M D will be perpendicular to o B (Geom. B. 6, Prop. 6.); м D and D P being both perpendicular to o B at the same point D, the angle M D P is the diedral angle of the planes A O B and B oc; or M DP — the angle B of the spherical triangle; o м being equal to radius, m d is the sine of the plane angle a, and M P is the sine of the plane angle b; in the triangle M D P, right angled at P, we have the proportion (Art. 38.) R: Sin D:: MD: MP substituting for D its equal B, for м D, its value sin a, and for M P, its value sin b, we have That is, radius is to the sine of either of the oblique angles of a right angled spherical triangle as the sine of the hypothenuse is to the sine of the side opposite that angle. 79. The solution of astronomical problems forms one of the most useful and agreeable applications of the theory of spherical trigonometry. To such inquiries the theory itself, no doubt, owes its origin, as well as many of the successive improvements which it has gradually received, so that a specimen of its use in the solution of astronomical problems may reasonably be looked for in a book on trigonometry. To illustrate therefore the above and subsequent formulæ of spherical trigonometry we shall accordingly introduce a few great circles of the celestial sphere. They are so well known that to define them is perhaps superfluous. The equator is that great circle the plane of which is perpendicular to the axis of the earth. The axis being the line about which the earth performs its diurnal rotation. This |
