Solution: COGO Intersections and Curves

Problem (2)

Part (a)

Partial solution: coordinates of remaining points.

Solution logic

(1) Start at B and compute clockwise as far as possible computing thru the arc centers.

Because these are tangent curves:

line-curve intersections are 90°
curve-curve intersections are colinear

Forward comp from B to C - use 137.00 ft & Az = 102°32'50".
Forward comp from C to arc center O₁ - use 200.00 ft  & Az=102°32'50"+180°+90°=375°-360°=12°32'50"
Forward comp from O₁ to D - use 200.00 ft & Az=12°32'50"+180°-38°30'00"=154°02'50"
Forward comp from D to arc center O₂ - use 300.00 ft and Az=154°02'50"
Forward comp from O₂ to E - use 300.00 ft & Az=154°02'50"+180°+30°00'00"==364°02'50"-360°=04°02'50"
Forward comp E to O₃ - use 30.00 ft & Az=04°02'50"+180°=184°02'50"
Forward comp O₃ to F - use 30.00 ft  & Az=184°02'50"+180°+75°00'00"=473°02'50"-360°=79°02'50"
Forward comp F to O₄ = use 400.00 ft & Az=79°02'50"+180°=259°02'50"
Forward comp O₄ to G - use 400.00 ft & Az=259°02'50"+180+15°=454°02'50"-360°=94°02'50"

(2) Start at B and go counterclockwise as far as possible.

Forward comp from B to A - use 273.84 ft & Az=102°32'50"+74°25'10"=176°58'00

(3) Use COGO dir-dir intersection to compute H.

From A - Az = 176°58'00"+180°+98°52'25"=455°50'25"-360°=95°50'25"
From G - Az=94°02'50"+180°00'00"-90°00'00"=184°02'50"

Partial answers

Part (b)

Partial solution: Lot 17's area to nearest 10 st ft: 124,740 sq ft