BY CONSTRUCTION Draw the line A B, which make equal to 157 ; take the length of A C 88 in the compasses, and with one foot on A describe the arc C; then, with the length of C B 110 in the compasses, and one foot in B, describe an arc cutting the former in C, to which draw the lines A C and B C; then A B C is the triangle required; the angle A will measure 42° 44', the angle B 32° 53', and the angle C 104° 23'. D 157 By Rule I. a + b + c 2 S S By Rule II. a = IIO b= 88 C= 157 2)355 s = 177.5 6 89.5 log. 1.951823 S-a= 67:5 20.5 log. I'311754 S-6= 89.5 SCE 2) 22.843683 20:5 For 2 B For LC Constants 11.421841 11:421841 II.421841 s — a 67.5 log. I•829304 S 6 89.5 log. 1.951823 s - C 20°5 log. 1.311754 A 21° 22' 18" L. tan. 9:592537 } B16°26'35" tan. 9•470018 °C 52°11'7" tan. 10:110087 А 42 44 36 B 32 53 10 C 104 22 14 The three angles of a triangle determine it in species only, and not in magnitude : they may define it as an acute-angled triangle, a right-angled triangle, or an obtuse-angled triangle. Moreover, since the sum of the three angles of every triangle is equal to 180°, it is evident that an indefinite number of similar triangles may be constructed, the angles of which shall be equal respectively to three given angles the sum of which is 180°. Examples for Practice 1. Given one side 129, an adjacent angle 56° 30', and the opposite angle 81° 36'; required the third angle and the remaining sides. Ans. The third angle is 41° 54', and the remaining sides are 108.7 and 87.08. 2. Given one side 96.5, another side 59.7, and the angle opposite the latter side 31° 30'; required the remaining angles and the third side. Ans. This question is ambiguous, the given side opposite the given angle being less than the other given side (see Example II., p. 53); hence, if the angle opposite the side 96.5 be acute, it will be 57° 38', the remaining angle 90° 52', and the third side 114:2; but if the angle opposite the side 96.5 be obtuse, it will be 122° 22', the remaining angle 26° 8', and the third side 50-32. 3. Given one side 110, another side 102, and the contained angle 113° 36'; required the remaining angles and the third side. Ans. The remaining angles are 34° 37' and 31° 47'; and the third side is 177.5 4. Given the three sides respectively, 120.6, 125.5, and 14:67; required the angles. Ans. The angles are 51° 53', 54° 58', and 73° 9'. 5. Given one side 684:5, another side 496-7, and the angle opposite the latter side 40° 58'; required the remaining angles and the third side. Ans. If the angle opposite the former side be acute, the remaining angles will be 64° 37' and 74° 25', and the third side 729.8; but if obtuse, the angles will be 115° 23' and 23° 39', and the third side 303.9. 6. Given one side 117.8, another side 96-55, and the contained angle 67° 30'; required the remaining angles and the third side. Ans. The remaining angles are 64° 41' and 47° 49', and the third side 120.4. 7. Given the three sides 87.6, 66-2, and 41:3; required the angles. Ans. The angles are 26° 49', 45° 20', and 106° 51'. APPLICATION OF TRIGONOMETRY The following methods of ascertaining the height or distance of an object being frequently useful, they may properly be introduced here before the present subject is dismissed. To find the Height of an Accessible Object Measure the horizontal distance, between the eye and the object, of the point immediately under it, and observe the angle of elevation with a quadrant or sextant : thus will be obtained the base and angles of a rightangled triangle, the perpendicular of which being found will be the height of the object above the horizontal plane, to which add the height of the eye. ! Or, by removing either towards or from the object, until the angle of elevation be 45°, the horizontal distance, added to the height of the eye, will give the height of the object. If the height of the object be known, and the angle of elevation observed, the horizontal distance of the eye may be found; for in this case there will be given the perpendicular and angles of a right-angled triangle to find the base or distance required. A B = B C cot. A. A To find the Height of an Inaccessible Object Measure the angle of elevation at a convenient distance from the given object; then remove in a direct line from the object, and again observe the angle of elevation, the distance between the two stations being carefully measured ; hence will be given one side and the angles of an oblique-angled triangle, with which, find either of the two other sides. Now that side will be the hypotenuse of a rightangled triangle, the perpendicular of which being found, and the height of the eye added to Ć 120 Fathoms it, their sum will be the height of the object. Example.--Wanting to know the height of a lighthouse above the level of the sea, and not being able to measure its horizontal distance, I took the 21020' angle of elevation, and found it to be 31° 45', and after removing from it 120 fathoms, I observed the angle of elevation to be 21° 20'; required the height of the lighthouse. In the triangle ABC are given the angle A CB 21° 20'; the angle CAB, which is the difference between the angles A B D and ACD; and the side C B i2o fathoms, to find the side A B. AB sin. A CB CB sin. A CB .. AB sin. CAB L ABD ...... 31° 45° 2 АС В 21 20 Z СА В .... IO 25 Side C B 120 .... log. 2:079181 9.560855 11•640036 9.257211 2:382825 L. CAB 10° 20' L. sin. Side A B 241.45 In the right-angled triangle A B D are given the angle A B D 31° 45', and the hypotenuse A B 241.45, to find the perpendicular A D. AD sin. A BD... AD= A B x sin. ABD log. A D = log. A B + L. sin. ABD -- 10 Hypotenuse A B 241.45 log. 2.382825 L. sin. 9•721162 2.103987 Hence the height of the lighthouse is 127.05 fathoms, or 762-3 feet above the level of the sea. To find the Distance of Two Points on a Horizontal Plane, which are Inaccessible Let A and B be the objects : from C, one extremity of a measured distance, observe the angles B C A and B CD; from D, the other extremity of the measured distance, observe the angles A D C and ADB; then by The Rule of Sines, p. 50, compute the sides D A and D B; and from the sides D A and D B, and included angle A D B, compute by Rule on p. 54 the required distance A B. Example.—To find the distance between a windmill (B) and house (A), I took two stations, C and D, on the opposite side of a river, distant 1,267 yards from each other, and such, that from each of them the other station and the windmill (B) and house (A) were seen. At C the angles BCA 53° 38', and B C D 34° 50' were observed ; at D the angles A D C 43° 44' and A D B 58° 38' were observed. Required the distance between the house and mill. Ans. Side A D= 1709-7 yards ; side D B = 1065.1 yards ; LABD = 83° 9' 26". Dist. between house and mill 1470-3 yards. Examples for Practice I. Find the height of a cliff in feet, the angle of elevation of its top at a distance of one nautical mile being 5o. Ans. 531.9 feet. 2. The angle of elevation of a tower was 26° 30', and 150 yards nearer to it the angle of elevation was 51° 30'; find the height of the tower, and the distance of its base from the last station. Ans. 123.94 yards; distance 98-59 yards. 3. Two ships are 950 yards apart; an observer on each finds the angle between the other and a buoy to be 57° 30'; find the distance of the buoy from each ship. Ans. 884:05 yards. 4. From a ship a point of land bore N. 20° W., and after sailing N. 60° W. 3 miles it bore N. 37° E. ; find the distance of the ship from the point of land when the second bearing was taken. Ans. 2.299 miles. 5. Two landmarks are distant from a ship 1,550 yards and 975 yards, they bear respectively N. 73° E. and S. 30° E.; find the distance between the landmarks. Ans. 1,635 yards. 6. From the top of a ship's mast 75 feet above the water, the angle of depression of a floating mark was 12° 20'; required the distance of the top of the mast from the floating mark. Ans. 351•1 feet. 7. Three ships not in one line are distant from each other 750 yards, 930 yards, and 1,006 yards; find the greatest difference of bearing between them. Ans. 72° 403'. |