APPLICATION TO HEIGHTS AND DISTANCES. 1. A HORIZONTAL PLANE is one which is parallel to the water level. 2. A plane which is perpendicular to a horizontal plane, is called a vertical plane. 3. All lines parallel to the water level, are called horizontal lines. 4. All lines which are perpendicular to a horizontal plane, are called vertical lines ; and all lines which are inclined to it, are called oblique lines. õ. A HORIZONTAL ANGLE is one whose sides are hori. zontal. 6. A VERTICAL ANGLE is one, the plane of whose sides is vertical. 7. An angle of elevation, is a vertical angle having one of its sides horizontal, and the inclined side above the horizontal side. 8. An angle of depression, is a vertical angle having one of its sides horizontal, and the inclined side under the horizontal side. I. To delermine the horizontal distance to a point which is in. accessible by reason of an intervening river. 35. Let C be the point. Measure along the bank of the river a horizontal base line AB, and select the stations A and B, in such a manner that each can be seen from the other, and the point C from both of them. Then measure the hori. A B zontal angles CAB and CBA with an instrument adapted to that purpose. Let is suppose that we have found AB= 600 yards, CAB= 57° 35', and CBA= 64° 51'. The angle C=180° — (A + B)= 57° 31'. To find the distance BC. sin C 57° 34' ar. comp. 0.073619 sin A 57° 35' 9.926431 AB 600 2.778151 BC 600.11 yards 2.778231 IL. To determine the altitude of an inaccessible object above a given horizontal plane. FIRST METHOD. 36. Suppose D to be the inaccessible object, and BC the horizontal plane from which the alti Bs tude is to be estimated : then, if we suppose DC to be a vertical line, it will represent the required altitude. А Measure any horizontal base line, as BA; and at the extremities B and A, measure the horizontal angles CBA and. CAB. Measure also the angle of elevation DBC. Then in the triangle CBA there will be known, two angles and the side AB; the side BO can therefore be determined. Having found BC, we shall have, in the right-angled triangle DBC, the base BC and the angle at the base, to find the perpendicular DC, which measures the altitude of the point D above the horizontal plane BC. Let us suppose that we have found BA= 780 yards, the horizontal angle CBA = 41° 24'; the horizontal angle CAB=96° 28', and the angle of eleva tion DBC=10°43'. In the triangle BCA, to find the horizontal distance BC. The angle BCA = 180° -(41° 24' +96° 28') = 42° 08'= C. In the right-angled triangle DBC, to find DC. 0.000000 tan DBC 10° 43' 9.277043 BC 1155.29 3.062692 REMARK I. It night, at first, appear, that the solution which we have given, requires that the points B and A should be in the same horizontal plane; but it is entirely independent of such a supposition. For, the horizontal distance, which is represented by BA, is the same, whether the station A is on the same level with B, above it, or below it. The horizontal angles CAB and CBA are also the same, so long as the point C is in the vertical line DC. Therefore, if the horizontal line through A should cut the vertical line DC, at any point, as E, above or below C, AB would still be the hori. zontal distance between B and A, and AE, which is equal to AC, would be the horizontal distance between A and C. If at A, we measure the angle of elevation of the point D, we shall know in the right-angled triangle DAE, the base AE, and the angle at the base; from which the perpendicular DE can be determined. 37. Let us suppose that we had measured the angle of elevation DAE, and found it equal to 20° 15'. First: In the triangle BAC, to find AC or its equal AE. sin c 42° 08' ar. comp. 0.173369 : sin B 41° 24 9.820406 AB 780 2.892095 : In the right-angled triangle DAE, to find DE. 0.000000 tan A 20° 15' 9.566932 AE 768.9 2.885870 DE 283.66 2.452802. : D Now, since DC is less than DE, it follows that the station B is above the station A. That is, DE- DC=283.66 - 218.64 = 65.02= EC which expresses the vertical distance that the station B is above the station A. BF A REMARK II. It should be remembered, that the vertical distance which is obtained by the calculation, is estimated from a horizontal line passing through the eye at the time of observation. Hence, the height of the instrument is to be added, in order to obtain the true result. SECOND METHOD. 38. When the nature of the ground will admit of it, measure a base line AB in the direction of the object D. Then measure with the instrument the angles of elevation at A and B. Then, since the exterior angle DBC is equal to the sum of the angles A and ADB, it follows that the an BY gle ADB is equal to the difference of the angles of elevation at A and B. Hence, we can find all the parts of the triangle ADB. Having found DB, and knowing the angle DBC, we can find the altitude DC. This method supposes that the stations A and B are on the same horizontal plane; and therefore it can only be used when the line AB is nearly horizontal. Let us suppose that we have measured the base line. and the two angles of elevation, and AB = 975 yards, found A =15° 36', DBC=27° 29'; required the altitude DC. First: ADB= DBC-A=27° 29' -15° 36'=11° 53'. III. To determine the perpendicular distance of an object below a given horizontal plane. 39. Suppose C to be directly over the given object, and A the point through which the horizon. tal plane is supposed to pass. Measure a horizontal base line AB, and at the stations A and B conceive the two horizontal lines AC, BC, to be drawn. The A oblique lines from A and B to the object are the hy. pothenuses of two right-angled triangles, of which AC, BC, are the bases. The perpendiculars of these triangles are the distances from the horizontal lines AC, BC, to the object. If we turn the triangles about their bases AC, BC, until they become horizontal, the object, in the first case, will fall at C', and in the second at ('". Measure the horizontal angles CAB, CBA, and also the angles of depression CAC, C"BC. |