Construct the circle as in the first example D of this case. Then in the triangle ADC, of which the angles and the side AC are known; A find AD. In the triangle ABD, of which B the sides AB, AD, and the included angle BAD are known, find the angle ABD=SBC, the supplement of which ABS=DBC. In the triangle ABS, of which the angles and side AB are known, And SA, SB. And in the triangle ASC, of which the angles and AC are known, find SC Example If three objects, A, B, C, as seen from S, are situate in one straight line, (as in the above figure) and AB=2550 yards, AC=7000 yards, the angles ASB=256 15, and BSC=35° 30'. Required the distances SĂ, SB, SC? Answer, AS=5936.16, CS=7609.58, BS=5669.92. N. B. When the station S is in one of the sides of the triangle, or in one of the sides produced, the solution presents no difficulty.. The above examples are not worked at length, as after the full explanation given it was not thought necessary to do so, MISCELLANEOUS EXAMPLES ON HEIGHTS AND DISTANCES. 1. Suppose the breadth of a well at the top to be 6 feet, and the angle formed by its side, and a visual diagonal line from the edge at the top, to the opposite side at the bottom to be 18° 30'; what is the depth of the well ? Answer, 17.932 feet. 2. From the top of a tower 180 feet high, I took the angle of depression* of two trees, A and B, which stood in a direct line, on the same horizontal plane with the bottom of the tower, and found that of the nearer to be 55°, and that of the farther 3]0; what was the distance between the trees ? Answer, 173.532 feet. 3. Let A and B be two stations 500 yards apart, and C a church separated from AB by a river; also let the angle A be 75° 19', and the angle B 59° 43'; how far is the church from each station ? Answer, 610.972 yards, and 684.412 yards. 4. Required the distance between two objects, A and B, separated from each other by a marsh, from the following data : A point C is taken, from which both A and B are accessible, and AC, is found=500 yards, BC=450, and the angle ACB=66° 30' ? Answer, 522.555 yards. The angle of depression of an object, is the angle formed by two lines one drawn from the eye parallel to the horizon, and the other drawn from the eyc to the object. 5. Wanting to know the distance between a house D, and . church C, both on the further side of a river, and separatecl from each other by a wood; I pitched upon two stations A and B, 300 yards apart, and found the angles to be as follows: ABC=53° 30', CBD=45° 15', CAD=37o, and DAB=58° 20'; what is the disa tance between the house and church? Answer, 479.304 yards. 6. To determine nearly the distance between two ships at sea, I carefully observed the interval of time between the flash and report of a gun from each, and measured also the angle which the two ships subtended. The intervals were 4 seconds and 6 seconds, and the angle 48, 12. Now supposing that sound moves uniformly at the rate of 1142 feet per second, and that the flash strikes the eye at the instant of its production ; required the distance of the ships? Answer, 5147.9 feet. 7. At the top of a castle which stood on a hill near the sea shore, the angle of depression of a ship at anchor was 4° 52', and at the bottom of the castle its depression was 4° 2'; required the height of the top of the building above the level of the sea, supposing the castle itself 54 feet high; required also the horizontal distance of the vessel ? Answer, height 314.2 feet; distance 3690.3 feet. a SPHERICAL TRIGONOMETRY 53. IMAGINE that the semi-circle PAp revolves P round its diameter Pp; and by its revolution de- d ld scribes the sphere APAP; it is evident that the А. A radius AC will describe a great circle, while the lines bd, ef, will describe smaller circles, di ľ minishing as they recede from the radius AC. 54. Any other great semi-circle equal to the first, the semicircle APA, for example, would have generated the same sphere by its revolution round the diameter AA. Hence, since there are an infinite number of equal diameters about which this revolution may take place, there are evidently an infinite number of great circles in any spheres, all of which are equal. There are alse an infinite number of smaller circles; but because of their inequality they are not used in Spherical Trigono metry. 55. The name of SPHERICAL TRIGONOMETRY is given to that science which instructs us to resolve triangles formed upon the surface of a sphere by three of its great circles. Take any spherical body, a billiard ball for example, trace upon it three circular arcs whose planes pass through its center ; these arcs, by their intersection, will form a spherical triangle, of which they will be the sides. Each of these arcs belong to a circle whose center is also the center of the ball, and through this center we may suppose that & diameter of the ball passes perpendicularly to the plane of the circle. 56. A diameter thus perpendicular to the plane of a great circle, is called the aris of this circle. The two extremities of the axis are called the poles of the circle. Hence the points P, p are the poles of the circle ACA. Each great circle has therefore its own two poles, since two great cireles cannot have the same axis. 57. And since the axis of a great circle must always be perpendicular to its plane, and pass through its centre, it evidently fol. lows 1st. That it forms as many right angles as there are radii in the plane of the circle. 2nd. That the arcs which measure these angles are all 90°, and therefore equal, since they belong to equal circles. Hence their chords are equal. These chords measure the distance from the pole of the circle to the different points of its circumference. Therefore the pole of any circle is equally distant from every point of its circumference. 58. Thus we may say that the pole of a circle is a point in the surface of the sphere 900 distant from every point of the circumference of this circle; or which amounts to the same thing, that the arc contained between the pole of a circle and any point of its circumference, is always 90°. Hence it is very easy to trace on the surface of a sphere, a circle whose poles are known; and reciprocally to find the poles of a given circle. The spherical compasses solve this double problem with the greatest facility. 59. Since all great circles of the sphere have a common center, the line of their intersection is necessarily a diameter of the sphere, which is at the same time the diameter of all these circles. But every diameter bisects its circle, therefore two or more great circles divide one another into equal parts. And consequently, 1st. If two great circles have already intersected each other, they will again intersect each other at 180° from the first point of inter section, and not sooner. 2nd. Hence it is not possible to include between only two arcs, even the smallest portion of a spherical surface, unless they are each 1809. Whatever be the number of degrees which the two arcs contain, they form on the surface of the sphere an angle, of which it is important to ascertain the measure; and this we shall proceed to in. vestigate. 60. Let there be two arcs AB and AD, each of which I suppose to contain 90°, and which by their concurrence at A, form a spherical angle BAD. D is clear 1st. That the radii BC, DC, are both perpendicular to AC. 2nd. That these two radii have the same inclination towards each other, as the circular planes to which they belong. Srd. That the measure of this inclination is the arc BD, since the angle BCD has its summit at the center. 61. Hence the measure of any spherical angle is the arc of a great circle comprised between its sides at 90° distance from the summit of that angle. In general, if from the summit of a spherical angle taken as a pole, we describe an arc of a great circle, the portion of this arc comprised between the sides of the angle, will always be its measure. Not but that it would be possible to measure the angle by the arcs of small circles; for it is easily perceived that the angle bod, for example, formed by the sines of the arcs Ab, Ad, is equal to the angle BCD; and consequently that the arc bd, which mea A It E sures it, is of the same number of degrees as the arc BD. But for the sake of uniformity, it is usual to measure the value of spherical angles upon great circles described from their summits as poles. 62. Hence spherical angles are always equal to those which at 90° distance from their summits are formed by the sines of the arcs composing their sides. But any angle formed by tuo sines must be less than 180°; therefore every spherical angle is less than 180o. 63. Since spherical angles are equal to the angles formed by the sines of their sides, it evidently follows 1st. That when one arc of a great circle falls upon another, the angles which result are equal to two right angles. Ind. That if we prolong these two arcs beyond their point of intersection, the vertical angles are equal, as well as those formed by their sines, if prolonged in a similar manner. 3rd. That the sum of the angles formed about the point of intersection, is 360°. 64. Suppose now that the three circles, of which the sides of a spherical triangle are a part, be entirely described; it is clear that they will form on the other hemisphere a triangle perfectly equal to the first. It is also evident that, drawing the respective chords of all the arcs which form these two spherical triangles, there will result two rectilinear triangles perfectly equal, the one above, and the other below the center of the sphere. But we know that in any right-lined triangle the sum of any two sides is greater than the third side. Therefore in any spherical triangle A ABC, the sum of any two arcs is also greater than the third arc or side. Again since a right line is the shortest line that can be drawn from one point to another, it follows that the arc of a great circle passing through two points of the surface of a sphere, is the shortest line that can be drawn between those points. Hence it measures the distance between them. 65. We have already said (59) that it was not possible to enclose on all sides between only two arcs, any portion of a spherical surface, unless each arc was 180°. Therefore every spherical triangle CAB, results from the intersection of two arcs, cut by a third, before these arcs re-unite. But they meet again only at 180° distance from the point where they first intersected one another; therefore A any a spherical triangle is less than 180°. 66. If we prolong the two arcs CA and CB of the triangle CAB, till they again meet in D, it is evident that AB will be less than the sum of the prolongations AD, DB. But this sum added to the two prolonged arcs, is only 360°. a side of B |