The distance of the highest point of a Vertical Curve is 70.59m from PC. If the length of the curve is 120m and the slope of the forward tangent is -4.4%. L1 is 80m and L2 is 40m. 1. Determine the slope of the back tangent in % 2. Determine the value of the maximum offset in meters. 3. Determine the elevation of the hp in meters is Elev. PC is 110m.

The offset distance of the simple curve from PT to the tangent line passing through the PC is equal to 120.20m. The stationing of PC is at 2 + 540.26. The simple curve has an angle of intersection of 50° a. Compute the degree of curve. b. Compute the external distance. c. Compute the length of long chord.

1. The tangents of a simple curve have bearings of N 20° E and N 80° E, respectively. The radius of the curve is 205m. a. Compute the external distance of the curve. b. Compute the middle ordinate of the curve. c. Compute the stationing of point A on the curve having a deflection angle of 6° from the P.C. which is 1 + 200.00

The offset distance of the simple curve from PT to the tangent line passing through the PC is equal to 120.20m. The stationing of PC is at 2 + 540.26. The simple curve has an angle of intersection of 500. a. Compute the degree of curve.b. Compute the external distance.c. Compute the length of the long chord.

4. A point X on a simple curve has a deflection angle of 18 degrees with the back tangent. Theradius of the curve is 200m and the central angle is 70 degrees. Determine the distance from pointX to the starting point of the curve.5. Determine the perpendicular distance from X to the long chord of the curve.6. Determine the perpendicular distance from X to the back tangent.

The azimuth of the tangents of the simple curve is 514 degrees and 759 degrees respectively. The radius of the curve is 213.2 meters. Station A is lying on the curve having a deflection angle of 8 degrees from PT. Determine the: Show Solution Distance of the middle ordinate of the simple curve in meters.

QUESTION 2 The tangent thru the PC has a direction due to North; and the tangent through the PT has a bearing of N 50° E. It has a radius of 200 m. Stationing of PC is 12+060. a. Compute the tangent distance of the curve. Answer: T= b. Compute the long chord of the curve. Answer: Lc = m m c. A line is drawn from the center of the curve (at the central angle) and intersects the curve at point B. This line, when extended, makes an angle of 62° with the tangent thru PC. What is the stationing of point B? Answer: Sta. B = Note: Unit has been provided after the blank space. No need to indicate units on your answer.

A simple curve have tangents AB and BC intersecting at a common point B. AB has anazimuth of 180° and BC has an azimuth 230°. The stationing of the point of curvature atA is 10 + 140.26. If the degree of curvature of the simple curve is 4°.a. Compute the length of the chord from A.b. Compute the distance AB of the curve.c. Compute the stationing of point “X” on the curve on which a line passing through thecenter of the curve making an angle of 58° with line AB, intersects the curve at point“X”.

The azimuth of the tangents of the simple curve is 514 degrees and 759 degrees respectively. The radius of the curve is 213.2 meters. Station A is lying on the curve having a deflection angle of 8 degrees from PT. Determine the: Show Solution Length of chord from PC to PT in meters.

2. A simple curve has a central angle of ( 23 ) and a degree of curve of 5°. a. What is the value of Tangent, T? b. What is the value of External Distance, E? c. What is the value of Middle Ordinate, M?