1. A horizontal plane is a plane parallel to the horizon. 2. A vertical plane is a plane perpendicular to a horizontal plane. 3. A horizontal line is a line parallel to a horizontal plane. 4. A vertical line is a line perpendicular to a horizontal plane. 5. A horizontal angle is an angle whose plane is horizontal. 6. A vertical angle is an angle whose plane is vertical. C 7. An angle of elevation is a verticle angle, one of whose sides is horizontal, and the inclined side above the horizontal side. Thus, BAC. D A 8. An angle of depression is a vertical angle, one of whose sides is horizontal, and the inclined side below the horizontal side. Thus, DCA. 80. Problems. 1. Wishing to know the height of a tree standing on a horizontal plane, I measured from the tree the horizontal line BA, 150 ft., and found the angle of elevation, BAC, to the top of the tree to be 35° 20′. Required the height of the tree. Ans. 106.335 ft. B 2. In surveying a tract of land, I found it impracticable to measure the side AB on account of thick brushwood lying between A and B. I therefore measured AE, 7.50 ch., and EB, 8.70 ch., and found the angle AEB 38° 46'. Required AB. Ans. 5.494 ch. E A B B D 3. One side of a triangular field is double another, their included angle is 60°, and the third side is 15 ch. Required the longest side. Ans. 17.32 ch. 4. Wishing to know the width of a river, from the point A on one bank to the point C on the other bank, I measure the distance AB, 75 yd., and find the angle BAC= = 87° 28′ 30′′, and the angle ABC= 47° 38' 25". Required AC, the width of the river. Ans. 78.53 yd. 5. I find the angle of elevation, BAC, from the foot of a hill to the top to be 46° 25′ 30′′. Measuring back from the hill, AD-500 ft., I find the angle of elevation D E B A 6. From the foot of a tower standing at the top of a declivity, I measured AB 45 ft., and the angle ABD 50° 15'. I also measured, in a straight line with AB, BC 68 ft., and the angle BCD = 30° 45'. Required AD, the .height of the tower. Ans. 82.94 ft. 7. Wishing to know the height of a tower standing on a hill, I find the angle of elevation, BAC, to the top of the hill to be 44° 35', and the angle of elevation to the top of the tower to be 59° 48'. B D C B B D Measuring the horizontal line AE, 275 ft., I find the angle of eleva tion to the top of the tower to be 46° 25'. Required the height of the tower. Ans. 317.143 ft. 33° 45', BDC 22° 30′. Re Ans. DA=710.15 yd., DC=1042.5 yd., DB 934.28 yd. 1. a, NT CM = cos a, covers a, OR cot a, CR sec a, CS NO' D Remark. Describing the circumference through A, B, D, and drawing AE and BE, EAB BDC, EBA ADC. Let a = the angle OCT 600 yd., BC=400 yd., = RELATIONS OF CIRCULAR FUNCTIONS. 81. Fundamental Formulas. MO tan =cosec a. By articles 39-46, sin (90° — a) = cos a, cos (90° — a) = sin a, etc. From the diagram we have D = vers a, = a, E the arc OT, and CO CT CN= sin R с S B T M B MT2 + CM2 = CT2. Substituting the values of MT, CM, and CT, we have (1) sin2 a + cos2 a = 1. Hence, The square of the sine of any arc plus the square of its co-sine is equal to 1. From (1) we have, by transposition, (2) sin2 a (3) Hence, 1. The square of the sine of any arc is equal to 1 minus the square of its co-sine. cos2 a= 1— cos2 a, 1-sin2 a. .'. 2. The square of the co-sine of any arc is equal to 1 minus the square of its sine. From the diagram we have MOCO- CM. Substituting the values of MO, CO, and CM, we have (4) vers a 1 vers (90° — a) (5) Covers α = Hence, The versed-sine of any arc is equal to 1 minus its co-sine. COS a. 1- sin a. Hence, The co-versed-sine of any arc is equal to 1 minus its sine. From the diagram we have CM : CO:: MT : OR, or cos a: 1 :: sin a tan a. (6) tan a= cos a Hence, The tangent of any arc is equal to its sine divided by its co-sine. S. N. 7. |