Problem 5. A long chord from the P.C. to P.T. of a compound curve is 150 m. long and the angle it makes with the longer and shorter tangents are 6° and 10° respectively. Find the larger radius of the compound curve so that the common.

Two straights AB and BC meet in an inaccessible point B and are to be connected by a simple circular curve of 600 m radius. Two points P and Q were selected on AB and BC respectively, and the following details were obtained. Angle APQ = 150 deg, Angle CQP = 160 deg and PQ = 150 m %3D %| A. Distance AP. B. Distance QC. C. Sta. C if Sta. P is at 10+178.50

The angle of intersection of T.S and S.T is 70 degrees. The radius of the circular curve is 260 m and the spiral curve length is 110 m. Determine the element of the spiral curve.   a. Determine the external distance of the curve.  b. Determine the central angle of the circular curve.  c. Determine the deflection angle at the end point of the spiral.  d. Determine the offset from tangent at the endpoint of the spiral.  e. Determine the distance along the tangent at the midpoint of the spiral.

The deflection angles of two intermediate points A and B of a simple curve are 3° 15' and 8° 15' respectively. If the chord distance between A and B is 40m. Show the illustration. a. Find the radius of the curve. b. Find the length of curve from PC to A. Find the length of chord from PC to B.

The line connecting the PC and PT of the simple curve is 120.4m. The radius is 213.6m. If the stationing of the PC is 9+346.1, determine the deflection angle of PT.

SURVEYING 4D

Problem 52 The slope of the curve y- xy - 3x 1 at the point (0, -1) is а. 1 с. -1 b. 2 d. -2