Question

Trigonometric Leveling ZA=93*10'27" HT-5.96 SD4721.68 5 2 1 instrument, a zenith angle of 9310'27" is measured to a target whose verti cal distance above Point X is 4.92 feet. The slope distance from the instrument to the target is 3162.48 feet. The elevation of Point X is 326.98 feet. With the instrument at the same location, a vertical angle (clevation angle) of 6'37'37" is measured to a target set at 5.96 feet vertically above Point 0. The slope distance from the instrument to this target is 4721.68 feet. Ignoring the effects of curvature and refraction, compute the clevation of Point Q Pt.Q Horizontal Line ? IVA=6'3737" M ... ........... SD-3162.48

Answer 1

1 24 = 931027) TVA? 6 3 7'37) pt x = 326 Let us mart station observed ase _ @ У чS , А Т. ч. aim is to find alovahon of Point a fiast we have ho find clavation of 'horia Zontal line on a Zannith Angle 2980 10270 ich mop= = 93010²27)_900 consider DOAP, AP is perpendiculad to om on OA.

slope distance from For o I 3162.48' i sino = o.oss. on substituting this values, 0.055= AP Bilen: 48 AP = 0.0554 3162.483 = 175.1 ..Elevation of horizontal line oma elevation of point atxt 4.9 't ApS = 326.99 +4.9z+175.417 = So7.01 consider a OBT height TB perpendicular to horizontal is perpendiente

- от - (вет) - ir - - - 5 slope distanke on что from the calculator, sa sost) = . \\s чуцьк . от - чн . 64 6.5 = ѕчч. о. +1 = 1 + ga За = 3 , 4 best a) was 1-та . - 544. As s.aь = S . - (-)

BQ is distance of Point Q vertically above horizontal line Herlei Elevation of point Q Height BQ t elevation of horizontal line 538.94't 507.01 = 1045.95 fact is the final solution.