3. The sides of a right-angled triangle are 6250 and 1250 links; required the hypotenuse, angles, and area. Ans. Hyp. 6374 links, angles 78° 41′ and 11° 19′, area 39a. Or. 10p. OBLIQUE-ANGLED PLANE TRIANGLES. CASE I. and II. Given two sides and an angle opposite one of them, or given two angles and a side opposite one of them, to determine the triangle. RULE.—As a side sine of its opposite angle: : any other side : sine of its opposite angle. And conversely, As sine of an angle opposite side sine of any other angle: to its opposite side. 1. In an obtuse-angled scalene triangle are given the angle A = 32° 15', the angle B 114° 24', and the side AB 98 chains, to find the sides AC and BC, and the other angle. B = By construction. = = = Make AB 98 chains, draw BC, making the angle B 114° 24', also draw AC making the angle A = 32° 15', and meeting BC in C; then AC and BC, being measured, will be found respectively 162.34 and 95.12, and the angle c = 33° 21'. By calculation. The angle c 180° (114° 24′ + 32° 15') = 33° 21' C = Since the angle B is greater than 90°, its sine cannot be found in the tables, but by taking it from 180°, its supplement 65° 36' is obtained, the sine of which is also equal to the sine of the angle B. 2. In an acute-angled scalene triangle are given AB = 60 chains, = = BC 95.12 chains, and the angle C 33° 21'; required the greater side AC, and the angles A and B. By construction. Make BC= 95.12 chains, draw AC, making the angle c 33° 21'; with the side AB = 60, as radius, and centre B,'describe an arc cutting AC in A; then ABC is the triangle required. AC 108-87 chains, angle A = 60° 38', angle B = 86° 1' = B A To find AC. 1.77815 As sin. c 60 1.77815 1.97827 :: sin. B = 86° 1' 9.99895 9.94029 :AC = The sum of the angles C and A subtracted from 180°, leaves the angle B 86° 1'. 3. In the obtuse-angled scalene triangle ABC are given the two greater sides BC 9512 chains, BA 60 chains, and the angle c opposite the lesser of them = 33; 21'; required the side AC and the angles A and B. 95.12 chains; draw CA indefinite, making the angle c = 33° 21′; then with BA 60 chains as radius, and centre B, de B scribe an arc cutting CA in A; draw BA; then ABC is the triangle required. AC 50-03 chains; A = 119° 22′; and B = 27° 17'. 4. In a plane triangle AB 98, and Bo= 95·12 chains, and angle = c = 33° 21'; required the other parts. C Z 5. In a plane triangle are given two angles 79 23′ and 54° 22′, and a side 1250 links opposite the first angle, to find the other parts. Ans. Sides 1033.6 and 918.7 links. CASE III. Given two sides and their included angle to find the rest. RULE.-As the sum of the given sides: their difference: co-tangent of half the included angle: tangent of half the difference of the required angles. This angle, added to half the complement of the included angle, gives the greater required angle, and subtracted gives the lesser. The other side is then found by Case I. 1. Given the side AB = 9800 links, the side BC = 9512, and their included angle B = 114° 24', to find the other side and angles. By construction. C Make AB = 9800 links, and the angle B = 114° 24'; draw BC, which make = 9512 links then ABC is the triangle required. The angles A and c measure 32° 15′ and 33° 21', and AC 16234 links. 2. Given the two sides 103 and 126 chains, and their contained angle 56° 30′, to determine the triangle. Ans. The angles 72° 20′ and 51° 10', the side 110-3 chains. 3. Two sides of a triangle are 34500 links and 17407, and their included angle 37° 20'; required the other angles and side. Ans. The angles 27° 4′ and 115° 36,' the side 23200 links. CASE. IV. Given the three sides to find the angles. RULE I. From half the sum of the three sides subtract the side opposite the angle sought, add the logarithms of the half sum and remainder, and increase the index of the sum by 20; from the sum thus increased, subtract the sum of the logarithms of the sides containing the angle sought; the remainder, divided by 2, is the log. cosine of half the angle sought. RULE II. From half the sum of the three sides subtract each of the sides containing the angle sought, add the logarithms of the two remainders, and increase the index of the sum by 20; also from half the sum of the three sides subtract the side opposite the angle sought, and add the logarithms of the half sum and remainder; then half the difference between these logarithmic sums is the log. tangent of half the angle sought. The remaining angles may be found by Case I. NOTE.-Rule II. is to be preferred when the angle sought is very small or near 180°. 1. Given the three sides AC = 98, AB = 162.34, and BC= 95.12 chains, to find the angles of the triangle. The remaining angles may be found by Case I. = 4.16679 2)18.92207 9.46104 2. In a plane triangle are given the three sides 7000, 10400, and 14202 links to find the angles. Ans. 27° 59', 44° 12', and 107° 49'. 3. When the sides of the triangle are 2253, 2240, and 2400 links, what are the angles? Ans. 57° 27', 57° 59', and 64° 34'. SECTION IV. THE MEASUREMENT OF ANGLES. The measurement of horizontal and vertical angles is usually taken with a theodolite. |