2. Given the base 186, and the perpendicular 152, to find the hypothenuse. 3. The hypothenuse being given equal 403, and one leg 321; required the other leg. Ans. 243.7. 4. What is the hypothenuse of a right-angled triangle, the base of which is 31.04, and perpendicular 27.2. Ans. 41.27. The following examples, in which trigonometry is applied to the mensuration of inaccessible distances and heights, will serve to render the student expert in solving the different cases, and also to elucidate its use. The Application of Plane Trigonometry to the mensuration of Distances and Heights. EXAMPLE 1. Fig. 54. Being on one side of a river and wanting to know the yards along the side of the river in a right line AB, and found the two angles* between this line and the object to be CAB = 74° 14', and CBA = 49° 23'. Required the distance between each station and the object. Calculation. The sum of the angles CAB and CBA is 123° 37', which subtracted from 180° leaves the angle ACB = 56° 23'. Then by Case 1; S. ACB : :S. CBA :: AB : AC and S. ACB : S. CAB :: AB : BC EXAMPLE 2. Fig. 55. Suppose I want to know the distance between two places A and B, accessible at both ends of the line AB, and that I measured AC = 735 yards, and BC = 840; also the angle ACB = 55° 40. What is the distance between A and B? Calculation. The angle ACB = 55° 40, being subtracted from ' 180°, leaves 124° 20'; the half of which is 62° 10. Then by Case 3. * The angles may be taken with a common surveyor's compass; or more accurately with an instrument called a theodolite. a CAB + CBA САВ— СВА BC + AC: BC - AC: :tang. -: tang. 2 2 1575 105 62° 10' 7° 12' CAB + CBA САВ— СВА To and from -62° 10', add and subtract2 2 7° 12' and we shall have CAB = 690 22', and CBA = 54° 58'. Then, S. ABC : S. ACB : : AC : AB EXAMPLE S. Fig. 56. Wanting to know the distance between two inaccessible objects A and B, I measured a base line CD = 300 yards: at C the angle BCD was 58° 20' and ACD 95° 20'; at D the angle CDA was 53° 30' and CDB 98° 45'. Required the distance AB. Calculation. 1. In the triangle ACD, are given the angles ACD= 95° 20', ADC = 53° 30, and the side CD= 300, to find AC = 465.98. 2. In the triangle BCD, are given the angle BCD = 58° 20', BDC = 98° 45', and side CD = 300, to find BC=761.47. 3. In the triangle ACB we have now given the angle ACB = ACD — BCD = 37°, the side AC = 465.98 and BC = 761.47, to find AB = 479.8 yards, the distance required. EXAMPLE 4. Fig. 57. Being on one side of a river and observing three ob apart I knew to be, AB= 3 miles, AC = 2, and BC = 1.8, I took a station D, in a straight line with the objects A and C, being nearer the former, and found the angle ADB = 17° 47'. Required my distance from each of the objects. Construction. With the three given distances, describe the triangle ABC; from B, draw BE parallel to CA, and draw BD making the angle EBD = 17° 47' (the given angle ADB) and meeting CA produced, in D: then AD, CD and BD will be the distances required. * Calculation. 1. In the triangle ABC we have all the sides given, to find the angle C= 104° 8'. 2. Subtract the sum of the angles D and C from 180°, the remainder 58° 5' will be the angle DBC; then in the triangle BCD we know all the angles and the side BC to find DC = 5.003 and BD = 5.715; therefore DA =DC - AC = 3.003. EXAMPLE 5. Fig. 58. From a station at D, I perceived three objects, A, B and C, whose distances from each other I knew to be as follow : AB = 12 miles, BC = 7.2 miles and AC = 3 miles; at D, I took the angle CDB= 25° and ADC = 19o. Hence it is required to find my distance from each of the objects. * DEMONSTRATION. By construction, the distances AB, BC and AC' are equal to the given distanees; also the angle (29.1) BDC = the angle DBE = the given angle. Construction. With the given distances describe the triangle ABC; at B, make the angle EBA = 199 = the given angle ADC, and at A, make the angle EAB = 25° = the givenangle BDC; draw AE, and BE meeting in E, and (by prob. 10.) describe a circle that shall pass through the points A, E and B: join CE and produce it to meet the circle in D, and join AD, BD, then will AD, CD and BD be the distances required. * Calculation. 1. In the triangle ABC, all the sides are given, to find the angle BAC = 35° 35'. 2. In the triangle AEB are given all the angles, viz. EAB = 25°, EBA= 19° and AEB = 136°, and the side AB = 12, to find AE = 5.624..' * DEMONSTRATION. The angle ADC standing on the same arc with the angle ABE is equal to it (21.3.) For the same reason the angle BDC is equal to the angle. BAE; but by construction the angles ABE and BAE are equal to the given angles; therefore the angles ADC and BDC are equal to the given angles. Note. When the given angles ADC, BDC are respectively equal to the angles ABC, BAC, the point E will fall on the point C, the cir. cle will pass through the points A, C, and B, and the point D may be any where in the arc ADB; consequently, in this case, the situation of the point D, or its distance from each of the objects A, B, C, cannot be determined. It may not be improper also to observe that even when the angle ADB, which is the sum of the given angles, is equal to the sum of the angles ABC, BAC, or which is the same thing, is the supplement of the angle ACB, the circle passes through the points A, C, B; but then the angle ADC, BDC, unless they have been erroneously |