SECTION IV. SPHERICAL TRIGONOMETRY APPLIED. SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES. A GOOD general conception of the sphere is essential to a practical knowledge of spherical trigonometry, and this conception is best obtained by the examination of an artificial globe. By tracing out upon its surface the various forms of right-angled and oblique-angled triangles, and viewing them from different points, we may soon acquire the power of making a natural representation of them on paper, which will be found of much assistance in the solution and interpretation of problems. For instance, suppose one side of a right-angled spherical triangle to be 56°, and the angle between this side and the hypotenuse to be 24°. What is the hypotenuse, and what the other side and angle? A person might solve this problem by the application of the proper equations or proportions, without really comprehending it; that is, without being able to form a distinct notion of the shape of the triangle, and of its relation to the surface of the sphere on which it is situated. If we refer this triangle to the common geographical globe, the side 56° may be laid off on the equator, or on a meridian. In the first case, the hypotenuse will be the arc of a great circle drawn through one extremity of the side 56°, above or below the equator, and making wit it an angle of 24°; the other side will be an are of a meridian. In the second case, the side 56° falling on a meridian, the hypotenuse will be the arc of a great circle drawn through one extremity of this side, on the right or left of the meridian, and making with it an angle of 24°; the other side will be the arc of a great circle, at right angles to the meridian in which the given side lies. Generally speaking, the apparent form of a spherical triangle, and consequently the manner of representing it on paper, will differ with the position assumed for the eye in viewing it. From whatever point we look at a sphere, its outline is a perfect circle in the axis of which the eye is situated; and when the eye is, as will be hereafter supposed, at an infinite distance, this circle will be a great circle of the sphere. All great circles of the sphere whose planes pass through the eye, will seem to be diameters of the circle which represents the outline of the sphere. 56° We will now suppose the eye to be in the plane of the equator, and proceed to construct our triangle on paper. Let the great circle, PASA', represent the outline of the sphere, the diameter AA' the equator, and the diameter PS the central meridian, or the meridian in whose plane the eye is situated. Let AB = 56°, represent the given side, andA C,making with AB the angle BAC= A 56° B S V R IG 24°, the hypotenuse, then will BC, the arc of a meridian, be the other side at right angles to AB, and the triangle, ABC, corresponds in all respects to the given triangle. Again measure off 56° from P to Q, draw the arc DQ, make the arc A'G equal to 24°, and draw the quadrant PRG. The triangle PQR will also represent the given triangle in every particular. We know from the construction, that DV, = 24°, is greater than BC, and that AC is greater than AB, that is, greater than 56°. In like manner, we know that A', =24°, is greater than QR, and that PR is greater than PQ, because PR is more nearly equal to PG, =90°, than PQ is to PA, =90°. For illustration and explanation, we also give the following example: In a right-angled spherical triangle, there are given, the hypotenuse equal to 150° 33′ 20′′, the angle at the base, 23° 27′ 29′′, to find the base and the perpendicular. Let A'BC in the last figure, represent the triangle in which A'C 150° 33′ 20′′, the LBA'C= 23° 27' 29", and the sides A'B and BC are required. = This problem presents a right-angled spherical triangle, whose base and hypotenuse are each greater than 90°; and in cases of this kind, let the pupil observe, that the base is greater than the hypotenuse, and the oblique angle opposite the base, is greater than a right angle. In all cases, a spherical triangle and its supplemental triangle make a lune. It is 180° from one pole to its opposite, whatever great circle be traversed. It is 180° along the equator ABA', and also 180° along the ecliptic ACA'. The lune always gives two triangles; and when the sides of one of them are greater than 90°, we take the triangle having supplemental sides; hence in this case we operate on the triangle ABC. AC is greater than AB, therefore A'B is greater than the hypotenuse A'C. The ACB is less than 90°; therefore, the adjacent angle A'CB is greater than 90°, the two together being equal to two right angles. These facts are technically expressed, by saying, t the sides and opposite angles are of the same affect * Same affection: that is, both greater or both less than 90°. ferent affection: the one greater, the other less than 90°. Now, if the two sides of a right-angled spherical triangle are of the same affection, the hypotenuse will be less than 90°; and if of different affection, the hypotenuse will be greater than 90°. If, in every instance, we make a natural construction of the figure, and use common judgment, it will be impossible to doubt whether an arc must be taken greater or less than 90°. = We will now solve the triangle ACB. AC- 180° — 150° 33′ 20′′ 29° 26' 40". = To find BC, we use Eq. (3) or (13), Prop. 3, Sec. II., thus: To find AB, we use equation (1) or (11), thus: = 1. In the right-angled spherical triangle ABC, given AB 118° 21' 4", and the angle A = 23° 40′ 12′′, to find the other parts. B Ans. {40, 1160 171 55′′; the angle C, 100° 59′ 26′′; 2. In the right-angled spherical triangle ABC, given AB 53° 14′ 20′′, and the angle A 91° 25′ 53′′, to find the other parts. Ans. {AC, 91° 4' 9"; the angle C, 58° 15′ 8′′; and BC, 91° 47′ 10′′. 3. In the right-angled spherical triangle ABC, given AB 102° 50′ 25′′, and the angle A 113° 14′ 37′′, to find the other parts. Ans.{AC, 84° 51′ 36′′; the angle C, 101° 46′ 56′′; and BC, 113° 46′ 27′′. 4. In the right-angled spherical triangle ABC, given AB 48° 24′ 16′′, and BC 59° 38′ 27′′, to find the other parts. Ans. {AC, 70° 23′ 42′′; the angle 4, 66° 20′ 40′′; and the angle C, 52° 32′ 56′′. 5. In the right-angled spherical triangle ABC, given AB 151° 23′ 9′′, and BC 16° 35′ 14′′, to find the other parts. Ans. { AC, 147° 16' 51'; the angle C, 117° 37′ 25′′; and the angle A, 31° 52′ 49′′. 6. In the right-angled spherical triangle ABC, given AB 73° 4′ 31′′, and AC 86° 12′ 15," to find the other parts. Ans. {BC, 76° 51′ 20′′; the angle A, 77° 24′ 23′′; and the angle C, 73° 29′ 40′′. 7. In the right-angled spherical triangle ABC, given AC 118° 32′ 12′′, and AB 47° 26′ 35′′, to find the other parts. Ans. {BC, 1840 561 20'; the 30 le, A, 126° 19′ 2′′; and the angle C, 56° 58′ 44′′. 8. In the right-angled spherical triangle ABC, given AB 40° 18' 23", and AC 100° 3' 7", to find the other parts. Ans. { The angle A, 98° 38′ 53′′; the angle C, 41° 4' 6"; and BC, 103° 13′ 52". 9. In the right-angled spherical triangle ABC, given AC 61° 3′ 22′′, and the angle A 49° 28′ 12′′, to find the other parts. Ans. AB, 49° 36′ 6′′; the angle C, 60° 29′ 20′′; and BC, 41° 41' 32". 10. In the right-angled spherical triangle ABC, g |