Two intersecting streets are shown on the right. The centerline direction for each is known as are their widths.

Given the coordinates of a point on each centerline, what are the coordinates of the sideline intersection at point P?

Since bearings are given, this is a bearing-bearing intersection.

Because point P is not on the centerlines it should not be connected directly to the two given coordinate points.

Instead, create two new points, G and H, on each sideline. These will serve as the baseline for the intersection triangle.

To create point G, go perpendicular to the centerline at the given coordinate:

Do the same to create point H:

Inverse to determine the length and direction of the baseline GH:

Sidelines have the same directions as their respective centerlines. Using those, compute the three angles:

Then using the Law of Sines, determine the two missing sides:

Compute point P from point G:

Math check - compute point P from point H:

check

Coordinates of point A are 2000.00' N and 500.00' E. Using the curve geometry shown, compute coordinates of the labeled points.

Because sine and cosine of Azimuths will always return the correct mathematical sign in coordinate computations, they will be used instead of Bearings.

Starting at point A, compute azimuths to the side points and all the way to point C

Side points G and D at A.

Side points H and E at B;

Continue to center of arc then cul-de-sac center.

Point F cannot be directly computed from a known point. Because its position is defined by the intersection of the southerly street sideline and the cul-de-sac, it is connected by radii to points O and C. Its position is computed from a distance-distance intersection

We have the length and direction of the baseline OC. We also have lengths of the other two sides of the intersection triangle: d_CF = 60.00' and d_OF = 300.00'-30.00' = 270.00'

Because a distance-distance intersection results in two possible point locations, we must take care to solve the correct triangle.

Use Law of Cosines to solve angle at C:

Then the direction from C to F:

And finally the coordinates from point C:

Similarly, point I is at the intersection of the northerly street sideline and the cul-de-sac.

Intersection traingle sides are 300.00', 60.00' and 330.00'. As before, there are two possible intersection locations. After determining correct triangle to solve:

Use Law of Cosines to solve angle at C:

Then the direction from C to I:

And finally the coordinates from point C:

Point T must be computed by intersecting the westerly lot line with the southerly street sideline.

Point T is connected to point S by a bearing and point O by the southerly curve radius.

Compute angle at S from directions of SO and ST:

Use Law of Sines to solve angle at T:

(since the angle at T is less than 90°, this is the correct angle).

Compute angle at O:

And length of line ST:

Segment area:

(Eqn F-7)

Central angle:

(Eqn F-3)

Set TSI at point B.

Sight point A with Az_B-A=255°42'45" set on the horizontal circle.

To stake 10+88.40 BC

Turn to horizontal angle reading = 238°25'40"
Stake point at 227.17 ft

To stake 11+00

Turn to horizontal angle reading = 235°42'10"
Stake point at 222.69 ft

To stake 12+00

Turn to horizontal angle reading = 209°17'17"
Stake point at 189.79 ft

Move the TSI to point C;

Sight point B with Az_C-B = 303°54'34" set on horizontal circle.

Turn to horizontal angle reading = 286°57'50"
Stake point at 226.66 ft

To stake 14+56.98 EC

Turn to horizontal angle reading = 300°08'26"
Stake point at 193.411 ft

Turn to horizontal angle reading = 310°48'54"
Stake point at 178.41 ft