10-36.This bisects the interior angleat the point of intersection:1. A2. B3. C4. D10-37.For a degree of curvature of 1°,the radius is 5,729.58 ft. Whichof the following equations couldbe used to derive this value?1.2.3.4.Each of the aboveLearnning Objective: Recognizecorrect procedures and performmathematical computations to solvesimple horizontal curvesituations.10-38.To solve for the tangent distance,you must know what information?1.2.3.4.Point of tangencyPoint of curvatureCentral angle and radiusEach of the above10-39.You must know the degree of thecurve to solve for which of thefollowing information?1.Chord distance2.Curve distance3.Tangent distance4.External distance10-40.When calculating the length of thecurve using the chord definition,you obtain a value sightly lessthan the true length of the curve.1.True2.False10-41.What is the recommended procedurefor laying out a curve?1.Swing the arc with a tape2.Set up a transit at the PIand turn the interior angles3.Set up a transit at the PCand turn the interior angles4.Set up the transit at the PCand turn deflection angles10-42.The degree of curve required forthe layout of a road section is20°. When you lay out this curve,what chord length should you useto minimize the difference betweenarc and chord distances?1.10 ft2.25 ft3.50 ft4.100 ftIN ANSWERING QUESTIONS 10-43 THROUGH10-52, YOU ARE TO LAY OUT A HORIZONTALCURVE BY ARC DEFINITION, USING THEFOLLOWING DATA:PI = Sta. 16 + 24.60I = 60°D = 8°10-43.HOW long is the radius(R) for thecurve?1.708.20 ft2.716.20 ft3.720.20 ft4.728.20 ft10-44.What is the plus station at thePC?1.12 + 11.102.12 + 12.603.20 + 38.104.20 + 40.2010-45.What station should you markthe stake at the PT?1.23 + 74.602.23 + 46.103.19 + 61.104.19 + 51.1070
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