SOLUTION OF RIGHT ANGLED PLANE TRIANGLES BY MEANS OF LOGARITHMS. THE solution of right-angled triangles may be embraced in four cases: 1°. When the hypothenuse and one of the acute angles aré given. Chap. II., Theor. I. By multiplying the first member by radius to make the equation homogeneous, R.BC AC.sin. A (1), .. (a.) As radius is to the sine of the acute angle, so is the hy pothenuse to the side opposite; and, (b.) As radius: the cosine of the acute angle:: the hypothenuse to the side adjacent. 2°. When the hypothenuse and one side are given. From the above equation (1) we have, (c.) As the hypothenuse: the given side::radius: the sine of the angle opposite the given side. (d.) As radius: the sine of the angle opposite the required side ::the hypothenuse: the side required. or 3°. When a side and one of the acute angles are given. (Chap. II., Theor. I.) BC AB.tan. A, R.BC AB. tan. A. (e.) As radius: the tangent of the given angle:: the side adjacent: the side opposite the given angle. (f.) As sine of the angle opposite the given side: radius :: given side: the hypothenuse. 4°. When the two sides are given. Using the same formula as the last, we have, (g.) As the base: the perpendicular:: radius: the tangent of the angle at the base. H (h.) As sine the angle at the base: radius :: the perpendicular :the hypothenuse. EXAMPLES. 1. In the right-angled triangle ABC are given the hypothenuse AC=250, and the angle A=73° 44′ 23′′, to solve the triangle. Ans. C-16° 15' 37". a=240. 2. Given the angle A=50° 30', the hypothenuse = 142.27, to solve the triangle. Ans. C=39° 30'. a=109.78. c= 90.5. 3. Given the hypothenuse = 47.467, and one side = 37.29, to solve the triangle. 4. Given the perpendicular=222, and the angle opposite the base-25° 15', to solve the triangle. Ans. 64° 15', 104.7, and 245.45. 5. Given the base=123, and perpendicular-765, to solve the triangle. Ans. 80° 51'57". 9° 8' 3". 774.82. SOLUTION OF OBLIQUE ANGLED PLANE TRIANGLES. THE solution of all oblique-angled plane triangles may be included under four cases: 1o. When one side and two angles are given. Since the sum of the three angles of every plane triangle is equal to 180°, therefore, if we subtract the sum of any two angles of the triangle from 180°, we shall have the third angle. Then, from Chap. II., Theor. II., we have the following RULE. As sine of the angle opposite the given side: sine of the angle opposite the required side:: the given side: required side. 20. When two sides and an angle opposite one of them are given. From the same theorem as above we derive the following RULE. As the side opposite the given angle: the other given side:: sine of the given angle: the sine of the angle opposite the other giver side. Having found a second angle, the third is obtained as above, and the side is found by Case 1°. 3°. When two sides and their included angle are given. First, subtract the included angle from 180° for the sum of the other two angles. Then, by Theor. III., we have the following RULE. As the sum of the two given sides: their difference :: tangent of half the sum of the opposite angles: tangent half their differ ence. Then to half the sum of the two angles add half their dif ference for the greater angle, and from half their sum take half their difference for the less angle. The third side is found by Case 1o. 4°. When the three sides are given. The student may use any of the formulas under Theorem V., from 50 to 60, for the solution of the problem; but perhaps from Theorem IV. we may have a more convenient As the base: the sum of the other two sides :: the difference of those sides: the difference of the segments made by a perpendicular let fall from the vertical angle. Then, to half the base add half the difference of the segments for the greater segment, and from half the base take half the difference of the segments for the less segment. Now, As the greater of the other two given sides of the triangle: the greater segment:: radius: the cosine of the less angle; and, As the less of the two given sides: the less segment :: radius the cosine of the greater angle at the base. EXAMPLES. 1. In the triangle ABC are given the side c=93.37, the angle A=30° 20′, and the angle C-99° 30′, to find the other parts. Ans. a 47.81. b=72.697. B-50° 10'. 2. In the triangle ABC are given the side c=364, the angle A=57° 15', and the angle B=35° 30', to find the other parts. Ans. a= -306.49. b=211.62. C-87° 15'. 3. In the triangle ABC are given the side a=51.234, the side b=42.356, and the angle A=55°, to find the other parts. Ans. c-61.992. B-42° 37' 33". 4. In the triangle ABC are given the side b=50.24, the side c=43.25, and their included angle A=40° 15', to find the other parts. Ans. B=81° 24' 25", C-58° 20′ 35′′, and a=32.829. |