Question

THREE INDEPENDENT LINE OF LEVELS ARE RUN FROM BM1 TO BM2. ROUTE A IS 6KM LONG, ROUTE B IS 4 KM LONG, ROUTE C IS 8KM LONG. BY ROUTE A BM2 IS 82.27 ABOVE BM1, BY ROUTE B BM2 IS 82.4 ABOVE BM1 &BY ROUTE C BM2 IS 82.10 ABOVE BM1, THE ELEVATION OF BM1 IS 86.42. USING THE WEIGHTED MEAN VALUES WHAT IS THE WEIGHT OF ROUTE B? WHAT IS THE PROBABLE VALUE OF THE WEIGHTED MEAN? WHAT IS THE ELEVATION OF BM2? 4. 5. GIVEN: NORTHING 1000 950 EASTING 1000 1250 POINT NO. 2 <132 = 100。 d1-3-102.5m COMPUTE THE COORDINATE OF POINT 3 BY 2 POINT RESECTION

Answer 1

" Distance between 1 & 2 = squareroot (950 - 1000)^2 + (1250 - 1000)^2 so l_1 = squareroot 2500 + 62500 l_1 = 254.95 m applying sine rule we get sin alpha/l_2 = sin beta/l_3 = sin 100 degree/l_1 sin beta/102.5 = sin 100 degree/254.95 sin beta = 0.39 beta = 23.32 beta = 23 degree 19 26 .75 so by property of triangle alpha + beta + 100 = 180 degree alpha = 180 degree - 100 degree - 23 degree 19 27.75"" alpha - 56 degree 40 "" 33.25"" = 58.67 sin 56.67/l_2 = sin 100 degree/l_2 = sin 100 degree/254.95 by using the graphical approach we can know the coordinate of the point(3) or by the help & resection we can know the coordinates \r\n tan^-1 (50/250) = 11.30 x-cordinate of point 3 = x - coordinate of 1 + l_3 (0.70) x = 1000 + 102.5 (0.70) = 1071.75 y -coordinate pf point 3 = y - coordinate of 2 + 122.8 y = 950 + 122.88 = 1.72.88 "