to find G C; F G being determined as above directed. As < F B G (= 50′-10'-40′): G B C (10′) = :: FG (8 ft.): G C 2 ft. Given F B C 83'; G B C = 17' and F G = Ans. G C 8.5 ft. 33 ft. to find G C. if G is between F and C; but if not; G C = 5.61 ft. PROPOSITION 10. Dividing Land. An easy rule for finding the angles of a right angled triangle, the sides being given. To the hypothenuse add half the longer leg. Then, as that sum is to the shorter leg, so is 86° to the angle opposite the shorter leg. This rule, which is easily remembered, is very useful in many calculations in the field, where tables cannot be conveniently used. The greatest error does not exceed-4 minutes. The rule is therefore sufficiently exact for most purposes in surveying to which it may be applied. 76 A MANUAL FOR SURVEYORS. Example. Given the 3 sides of a right angled triangle, 30, 40 and 50, to find the angles: (50 +20): 30: 86°-36° 53′ the less angle. Example 2d. Given the hyp. and greater leg of a right angled triangle 50 and 40 angle opposite less leg 365°, to find the less leg. As 86° 36° 70: 30, as required. : This rule may be readily applied in cutting off a trapezoid from a given tract of land that shall contain a given number of acres, the angles being nearly right angles. D E A Let there be given A B south, B C west 40 per. and CD N. 4° W., to cut off a trapezoid A B C D containing 5 acres, by a line A D parallel to B C. First, 80040 20 CE approximate. In this case the leg and hypothenuse are nearly equal, we may use one for the other, therefore, As 86° : 4° : : (20 + 10): 1.4 = E D approxi mate. AD AE+ED=BC+ED 40 +1.4 41.4 approximate. Twice the area of a trapezoid, divided by the sum of the parallel sides, gives the perpendicular distance between them. 1600 ÷ (40 + 41.4) = 1600 ÷ 81.4 = 19.66 CE=AB. : As 86° 4° (19.66 + 9.83): 1.37 = D E. AD 40+ 1.37 41.37 sufficiently correct. — 1600 ÷ (40+ 41.37) 19.66 as before (the proof.) The angle B being a right angle, no correction is required on that side of the trapezoid. Another method is to find the area of the triangle D C E, and cut off a small trapezoid equal to it, either within or without, as the case may require. In the preceding example E C + D E = area of C DE 20+.7 = 14., this, divided by AD 41.4, gives .34, and this subtracted from E C, 20, because A D is greater than C B, gives the correct value of E C 20 = 20 — .34 19.66, as be fore. In most cases the use of the traverse table is more expeditious to find D E. Taking the ap proximate value of C E in a lat. column, under 4°, gives, in a departure column, 1.4 D E, from which a correct value of C E is found. With 19.66 in a lat. column, under 4° in a distance column, is found 19.71 = - CD. Example 2d. Given A B south, B C west 40 per., and C D N. 4° E. to find A B, C D and A D, when the trapezoid A B C D contains 5 acres. Ans. A B 20.36 perches. F D E H B When it is required to cut off a trapezoid from a trapezoidal piece of land, it may sometimes be done in the following manner: Given A B N. 40° E. 44.4 perches, B C S. 50 E. 60.8 perches, C D S. 40° W. 46 perches, and D A N. 49° W. 60.9 perches, to cut off 5 acres by a line E F parallel to A B. First.-800 44.4 18 E A approximate. = = Dg=DC-A B 46-44.4 = 1.6. : AD: AE::Dg: Eh or 60.9: 18: 1.6: .47. EF=Fh+hEAB+ Eh 44.4 + .47 44.87. Ah FB 1600 (44.4 +44.87) the perpendicular corrected. In a lat. column with 17.93 under 11° 17.93 <gAD in a distance column, is found 17.94 A E. When calculations are made by J. Gummere's rule, (the sides A D and B C, as in the above example, being nearly parallel,) great care must be used in extending the decimals to 3 places to ensure accuracy, as two decimals only may throw the line EF a pole from its true position in many cases that occur in practice. PROPOSITION 11. To determine the correct bearings of the lines of survey where local attraction deflects the needle from its usual direction or magnetic position. Let ABCD be several sides of B a survey on which the needle is disturbed by some extraneous mat ter. X A D |