To find the Perpendicular BC : As the angle A = 40:54:40" Log. co-tang. Ar. compt. = 9.937802 Given the Hypothenuse and One Side, to find the Angles and the Example. Other Side. Let the hypothenuse A C, of the annexed triangle ABC, be 330. 4, and the base AB 280.3; required the angles A and C, and the perpendicular BC? By making the hypothenuse A C radius; the perpendicular B C becomes the sine of the angle A, and the base A B the co-sine of the same angle.Hence, To find the Angle A: As the hypothenuse A C = 330. 4 Log. Ar. compt. = 7.480960 Is to radius 90: Log. sine = 10.000000 The base A B being made radius; the perpendicular BC becomes the tangent of the angle A, and the hypothenuse AC the secant of that angle. -Hence, Is to the base AB 280. 3 Log. = 10.000000 2.447623 9.795208 2.242831 Remark. The perpendicular BC may be found independently of the angles by the following rule (deduced from Euclid, Book I. Prop. 47, and Book II. Prop. 5), viz., To the log. of the sum of the hypothenuse and given side, add the log. of their difference; then, half the sum of these two logs, will be the log, of the required side :-as thus ; Given the Base and the Perpendicular, to find the Angles and the Example. Hypothenuse. Let the base AB, of the annexed triangle ABC, be 262. 5, and the perpendicular BC 210.4; required the angles, and the hypothenuse A C? By making the base A B radius; the perpendicular BC becomes the tangent of the angle A, and the hypothenuse AC the secant thereof. -Hence, Is to the base AB = 262.5 Log. = So is the angle A38:42:47" Log. secant =. To the hypothenuse A C = 336.4 = Log. = The perpendicular B C being made radius; the base A B will be the tangent of the angle C, or co-tangent of the angle A, and the hypothenuse A C will be the secant of C, or the co-secant of the angle A.-Hence, To find the Angle A :— As the perpendicular B C = 210. 4 Log. Ar. compt. = Is to radius = 90: Log. sine = = To the angle A = 38:42:47" Log. co-tangent = To find the Hypothenuse A C : As radius 90: Log. sine = Is to the perpendicular B C 210. 4 Log. = . 7.676954 10.000000 2.419129 So is the angle A = 38:42:47" Log. co-secant = To the hypothenuse AC 336.4 = Log. = The angle A subtracted from 90: leaves the angle C; thus 90: - 38: 42:47" = 51:17:13 the measure of the angle C. Remark. The hypothenuse A C may be found independently of the angles by the following rule, deduced principally from Euclid, Book I. Prop. 47; Book II. Prop. 5; and Book VI. Prop. 8, viz., From twice the log. of the base subtract the log. of the perpendicular, and add the corresponding natural number to the perpendicular; then, to the log. of this sum add the log. of the perpendicular, and half the sum of these two logs. will be the log. of the hypothenuse. As thus : Solution of Oblique-angled Plane Triangles by Logarithms. PROBLEM I. Given the Angles and One Side of an Oblique-angled Plane Triangle, to find the other Sides. RULE. As the Log. sine of any given angle, is to its opposite given side; so is the log. sine of any other given angle to its opposite side. Note. When a log. sine, or log. co-sine, is the first term in the proportion, the arithmetical complement thereof may be taken directly from the Table of secants by using a log. co-secant in the former case, and a log. secant in the latter. PROBLEM II. Given two Sides and an Angle opposite to one of them, to find the other Angles and the third Side. RULE. As any given side of a triangle is to the log. sine of its opposite given angle, so is any other given side to the log. sine of the angle opposite thereto. The angles being thus found, the third side is to be computed by the preceding Problem. Example. Let the side A B, of the triangle A B C, be 436.7, the side AC 684. 5, and the angle B 100:7:35"; required the angles A and C, and the side BC? |