425 By Practical Geometry is commonly understood, the mensuration of heights and distances of terrestrial objects, by means of plane trigonometry. EXAMPLE L In Fig. 1, of Plate 4, is represented a tower AB, of which it is desired to ascertain the heighth above the surface of the ground at B. Suppose the observer placed at the point C, at the distance CB equal to 130 feet, measured horizontally, or on a level from the foot of the tower at B; his eye being situated at E, elevated 5 feet above the line CB. Then with a common quadrant, or other instrument for measuring angles, let the angle AED be measured equal to 29 degrees, 59 minutes, which is formed by EA the line of sight from the observer's eye to the top of the tower, and the level or horizontal line from his eye to a point D elevated 5 feet above the bottom of the tower at B, corresponding to the elevation of the eye above the ground at C. This line ED being horizontal, and the face of the tower D being supposed perpendicular to the horizon, the angle ADE will be a right angle; consequently we have a rightangled triangle AED, in which are known the base ED = 130 feet, and the angle at E = 29°, 59′, when by Case 3, of right-angled Trigonometry, the perpendicular DE may found in the following way: As radius = 90°, 00' 10,00000 To the tangent of AED = 29°, 59′ = 9,76115 be = 130 = 2,11394 To log. of perpend. DA == 75 = 1,87509 But But as the point D is elevated 5 feet above the bottom of the tower, this quantity, added to the perpendicular just found = 75, will give 80 feet for the whole elevation of the tower, which was required to be known. Again, by reversing this problem, and supposing that an observer at E found the angle of elevation of the tower AED to be 29°, 59', and that he knew the height of the tower above the horizontal line ED to be 75 feet, but wished to know the distance between his station and the tower, or the measure of the line ED CB; by the same Case 3, of right-angled Frigonometry, he would say; As radius = 90°, 00' 10,00000 To the tang. of EAD == 60, 01 = 10,23885 The angle EAD being the complement of the observed angle AEB to 90 degrees, its measure is 60°, 01′, by means of which the horizontal distance of the observer from the tower is found to be 130 feet, agreeably to the statement in the first part of this case, which is applicable to the measurement of all accessible and perpendicular elevations. EXAMPLE II. Fig. 2, Plate 4. Suppose it be required to find the height of the point of a pyramid, or obelisk AB, situated on the top of a hill, at the same time that the observer can approach no nearer than the point D, at the foot of the hill. From D let the level, or horizontal line DC, be measured off equal to 360 feet, in such a direction that the obelisk when observed from C may be seen precisely over and in the direction of D. Then, with a proper instrument, let the angle ACD be measured equal to 30°, 00′, as also the angle ADE equal to 46°, 00′, When When these things are ascertained, we have an obliqueangled triangle CAD, of which we know the side CD = 360 feet, the angle ACD = 30 ̊, 00′, and the angle ADC = 134°, 00′ (being the supplement of the measured angle, ADE, to two right angles, or 180 degrees) and, consequently, the remaining angle formed by lines supposed to be drawn from the top of the obelisk to the two stations of the observer, that is, the angle CAD = 16', 00'. It was shown in Prop. 2 of Trigonometry, that in any plane triangle the sides are in proportion to the sines of the respectively opposite angles; we may therefore by this proportion discover the length of the sides AC or AD. Let AD then be found as follows: As the sine of CAD 16°, 00' = 9,44034 Again, in the triangle ADE, having a right angle at E formed by the imaginary lines DE, which is horizontal, and AE a perpendicular let fall from the point of the obelisk at A, we have the hypothenuse AD just found to be 653 feet, and the angle ADE observed to be 46°, 00'; hence the perpendicular EA may be found as follows, (Trig. Prop. 5.) : 90°,00′ = 10,00000 As radius |