B = A — A = (A C ± A B) '— A B) ; and therefore B C = C ± A 13) • C — A B) Log. (A c A B) ± Log. (A c —A 13). Log. BC— 2 From which a c is easily determined.
Example the hypothenuse of a triangle be 274.5 yards, and its base 196.25; what are the two acute angles ? Example 3.—Suppose a ship sailed between south and east 510 miles, and thereby made her difference of latitude, or soothing, 315 miles ; upon what course did she sail ? _Example ship sailed from latitude 30' north, between the south and west, 135 leagues, till, by a good observation, she is found in latitude 15° 15'; required the course on which she sailed.
l'itom.pm the base and perpendicular, to find the angles and hypothenuse.
Example 1.-1n the right-angled triangle A B C, right angled at B, let the perpendicular, A B, be 650 feet, and the base, B c, 420 feet ; required the acute-angles, A and c, and the hypothenuse, a c.
Geometrically.
Make B C = 420, taken from a scale of equal parts ; from B raise the perpendicular A B = 650; from the same scale join A C to complete the triangle; then, with 60° taken from the line of chords, describe arches round the angles A and c; and their measures, applied to the same line of chords, will give the quantity of each angle.
By Calculation.
The base, B C, being radius, To find angle c.
As B c, 420 2.623249 is to A n, 650 2.812913 so is radius 10.000000 to tangent of c, 7' 52" 10.189664 The hvpothenuse may also be found, independently of the angles, for AC = from which AC is easily determined.
Example 2.—In the rectilinear triangle A n c, rectangular at B, suppose the side A B 495.45 yards, and the side u e 560.5 yards; what are the acute angles, A and C? Example 3.—Suppose three towns so situated, that A lies 3543 miles south from n, and C lies 501 miles west from i3 ; the bearing of A from C, and of C from A are required.
Example 4.—When the sun shines, if a steeple, 196 feet high, project a shadow 237 feet 9 inches, on the horizontal plane, what is the sun's altitude at that time ? With respect to taking angles by the theodolite, when we are not obliged to cross zero, subtract the less number of degrees from the greater, which will give the angle. But if we are under that necessity, subtract the greater number from 360 and add the less to the remainder for the angle.
Thus, suppose the index to stand at in looking at one point or object, and at in looking at another ; or, sup pose first at 253°, and then at 145°, then, in either case, 253° — 145° = 108° for the angle contained between the two objects.
Again : suppose the index stands at and we are obliged to cross zero in order to come to the other object, and then the index to stand at 15° ; or suppose the index stands at 15° in looking at the first object, and at 254° in looking at the second, then, in either of these cases, 360°— 15° 106° + 15° = 121°.
Figure 7.—To find the height of an inaccessible object.
Let c be the apex of a steeple, standing on the summit of a hill, and let A B be a straight line parallel to the horizon, so that the point e may be seen by a spectator from the points A and B, at a convenient distance from the eye above ground : let A B be 367 feet, the angle D A c = 30', and the angle c B D = 51° 20' ; it is now required to find the height of the summit c, of the steeple above the horizon.
By subtracting 51° 20' from 180°, a remainder of 40' is left for the angle A B C; and since the three angles of a plane triangle are equal to two right angles, the third angle A c a may be found by subtracting the sum of the two angles C A B and c n A from 180°, viz. 44° 30' + 128° 10' from which will leave a remainder of 50'; so that in the right-angled triangle all the angles and one side are given, by which to find the side B C. Now, as the sides of a plane triangle are as the sines of the opposite angles, Thus, or thus.
As sine of n c A 50'....log. = 10'924520 .. 9'075480 Figure 8.—To find the distance between two inaccessible objects.
Two inaccessible objects, D and c, are both visible from each of the two places A and a, whose distance is known, and each of which is visible from the other. Required the distance c D.
Observe the < CAD = a, and < DAn = 13 from A, and < DBC = a, and < C B A = g from B. Let A n = a, Then, if we can find D A and A c, or D B and n C, the problem is reduced to this ; find the third side when the other two sides and the angle included by them are given.
C A sin. C n A sin. 13 Now, in A c n A, = A D Sin. A 0 13 SM. A C B.
BUtACB = ISO(ABC CAB) = ISO— (13 + a + Sill. A C B = Sill. (a (3 i3) a sin. 13 .*. c A = sin. (a + + 13) D A sin. A BD sin. (a + 13) and illADDA — A B Sln. A D B sin. A D D