# Trigonometry

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Topics:

• An application of plane trigonometry
• Early applied mathematics in the United States
• History of trigonometry
• An application of spherical trigonometry

The Fermi model we developed previously for the size of the earth was simple and easy to use, but very crude and inaccurate. Better methods have been known for centuries. This is excellently illustrated by an episode from the colonial period of United States history.

In the winter of February, 1768, a party of men began a 3-month march accross the Delmarva peninsula, measuring their progress each step of the way. The Delmarva peninsula is a spit of land on eastern coast of North America between the Delaware and Chesapeake bays. The men leading the expedition were Charles Mason and Jeremiah Dixon. Their names and the boundaries they had surveyed over the last five years would eventually symbolize the American divide between the northern and southern states. Mason and Dixon themselves worked in the culture of enlightenment science, and their survey was probably the most sophisticated astronomical and mathematical activity yet performed in the English colonies soon to become the United States of America. Their goal now was to take advantage of Delmarva's flat terrain to complete the Royal Society's first scientific surveying of a degree of latitude. There results would contribute to the fierce ongoing debate over the shape of the earth.

Mason and Dixon's surveying work for their measurement of a degree of latitude, as well as the establishment of the Maryland-Pennsylvania-Delaware boundaries, was based on one of the oldest branches of applied mathematics -- trigonometry. The practical use of triangles in geometry traces back atleast as far as the Babylonian empire, circa 1800 BC. The subsequent history of trigonometry is especially complicated, winding through several continents and empires over four thousand years before reaching the form we use today.

Mason and Dixon were experts in two important generalizations of trigonometry developed during the Renaissance for modelling the earth -- triangulation and spherical trigonometry. These were the most valuable applied mathematics techniques of the time and have remained essential tools in cartography and navigation ever since. We will give a brief overview of both and illustrate their importance by redesigning Mason and Dixon's survey plan for the Maryland--Pennsylvania boundary.

## Triangulation

Dutchman Gemma Frisius (1508-155) was a mathematician, astronomer, and geographer by inclination despite being a doctor by profession. His 1533 second edition of Cosmographicus liber Petri Apiani added an appendix that is the first known description of triangulation. Frisius proposed that by using a carefully measured baseline, together with angle sightings to landmarks recorded using astronomical instruments, an entire map with accurate distance scaling could be drawn. Jacob van Deventer and Gerardus Mercator quickly applied Frisius's ideas, and Dane nobleman Tycho Brahe (1546-1601) further developed triangulation in his attempts to map Denmark in the 1570's. But it was Willebord Snel van Royen (1580-1626), Brahe's student, who was first to publish the "successful" large-scale application of triangulation (1617). Snell was interested in one of the same questions as Mason and Dixon -- estimating the earth's size. And after lengthy observation and calculation, he succeeded. Snell estimated there were 28,500 Rhineland rods (at 3.77 meters per rod) in one degree of latitude. That's $$107$$ km per degree, or $$6.15$$ megameters for the earth's radius if the earth is a sphere, much better than our Fermi estimate, but not meeting Snel's own expectations for precision. The task of completing such map calculations was subsequently simplified through the use of Napier's and Brigg's logarithms (circa 1624).

Suppose we have a set of landmarks. Any three of these landmarks $$i$$, $$j$$, and $$k$$ defining a triangle, and $$\theta _ {ijk}$$ is the observed angle from $$i$$ to $$k$$ centered at $$j$$. We adopt a sign convention so that angles that are oriented to rotate counter-clockwise are positive while clockwise angles are negative, so complementary angles (represented by reversing the start and end indices) cancel out and $\theta_{ijk} + \theta_{kji} = 0.$ Under the assumptions of Euclidean geometry, the sum of interior angles of a triangle $\theta _ {ijk} + \theta _ {jki} + \theta _ {kij} = \pi.$ As long as all angle arithmetic is done modulo $$2 \pi$$ so every angle is reduced to fall between $$-\pi$$ and $$\pi$$, this formula holds for both the interior and exterior angles of the triangle. If all angles are known, the distances $$L _ {ij} = L _ {ji}$$ between landmarks $$i$$ and $$j$$ can be determined by the Law of sines, which can be written in indexed form $\begin{gather} \frac{L_{ik}}{\sin \theta_{ijk}} = \frac{L_{ji}}{\sin \theta_{jki}} = \frac{L_{kj}}{\sin \theta_{kij}}. \end{gather}$ When the angles and distances are known, location coordinates $$\vec{z} _ j$$ for each landmark $$j$$ relative to the first two landmarks $$z_1$$ and $$z_2$$ can be calculated using the vector equation $\begin{gather} \vec{z}_j = \vec{z}_i + \frac{L_{ij}}{L_{ik}} \begin{bmatrix} \cos \theta_{jik} & -\sin \theta_{jik} \\ \sin \theta_{jik} & \cos \theta_{jik} \end{bmatrix} (\vec{z}_k-\vec{z}_i). \end{gather}$

Once the map was drawn, distances between all landmarks can be quickly inferred without need for field observation.

Together, the angle rules and the law of sines create a nonlinear system of equations for the angles and distances between landmarks. Let's consider an example of triangulation among landmarks when there are no overlapping edges, two angles are known in each triangle, and there is a single baseline to determine scale, location, and orientation.

[ Data : hide , shown as table , shown as CSV shown as QR ]

# right landmark, middle landmark, left landmark, angle (degrees)
#
# An example of a mapping problem based on triangulation,
# where triangle interior angles have been observed between 6
# landmarks.
#
# All angles are implicitly signed
# and give counterclockwise rotations
#
3,2,1,70
2,1,3,50
4,2,3,64
2,3,4,61
5,2,4,52
2,4,5,62
5,4,6,48
4,6,5,65


One approach to solving this system of equations is to apply the isolate-and-substitute method we use for linear systems. Assuming a baseline of length $$1$$ between landmarks 1 and 2, we can use the observed angle data to fill in the rest of the map.

[Show code]
#!/usr/bin/env python3

"""
This script solves a simple triangulation problem by brute-search.

https://en.wikipedia.org/wiki/Solution_of_triangles
"""

from scipy import *
from matplotlib.pyplot import *

def completeAngleCalcs(angles):
# Complete all triangle angles based on the # sum of interior
# angles = pi, and 2 independent interior angles of same sign given.
#
# For each triangle, there are 3 interior angles and each has
# a complement, so 6 total.
#
# This loop
for x in list(angles.keys()):
i,j,k = x
y = j,k,i
z = k,i,j
assert z in angles or y in angles, "some solution methods are missing: %s"%(str(z)+str(y))
if y in angles and z in angles:
print("\tclosure %s%s%s error = %f"%(i,j,k,pi-angles[x]-angles[y]-angles[z]))
elif y in angles and z not in angles:
angles[z] = (pi - angles[x] - angles[y])%(2*pi)
elif z in angles and y not in angles:
angles[y] = (pi - angles[x] - angles[z])%(2*pi)
# complementary angles
u = k,j,i
v = i,k,j
w = j,i,k
angles[u] = (-angles[x])%(2*pi)
angles[v] = (-angles[y])%(2*pi)
angles[w] = (-angles[z])%(2*pi)
assert x in angles
assert y in angles
assert z in angles
assert u in angles
assert v in angles
assert w in angles

def completeLengthCalcs(angles, lengths):
def calc(ej,Hj,Hk):
return lengths[ej]*sin(angles[Hk])/sin(angles[Hj])
changing = True
while changing:  # loop until no updating has occurred
changing = False
for Hj in angles:
i,j,k = Hj
Hk = j,k,i
Hi = k,i,j
# edges are always labelled with smaller number first
assert i != k
ej = ((i,k) if i < k else (k,i))
if ej not in lengths:
continue
assert j != k
ei = ((j,k) if j < k else (k,j))
assert i != j
ek = ((i,j) if i < j else (j,i))
if ei not in lengths:
lengths[ei] = calc(ej,Hj,Hi)
changing = True
if ek not in lengths:
lengths[ek] = calc(ej,Hj,Hk)
changing = True

def completePositionCalcs(angles, lengths, nodes, z):
# positions are represented with complex numbers
changing = True
while changing:
changing = False
for e in lengths:
for (i,j) in [e,(e[1],e[0])]:
if i in z and j not in z:
for k in nodes:
ai = (k,i,j)
if k in z and ai in angles:
A = angles[ai]
L = lengths[e]
M = lengths[((k,i) if k < i else (i,k))]
T = (L/M)*exp(1j*A)
z[j] = z[i] + T*(z[k]-z[i])
changing = True
# print(j, ai, z[i], z[k], A, L, M)

# Triangle interior angles
#
# all angles are implicitly signed
# counterclockwise rotation is positive
#
# angles measured in degrees
anglestr = """\
3,2,1,70
2,1,3,50
4,2,3,64
2,3,4,61
5,2,4,52
2,4,5,62
5,4,6,48
4,6,5,65"""
# the triangles and edge set can be inferred from the angle set

def main():
nodes = set()
angles = {}
F = lambda x : (x[0], x[1], x[2], float(x[3]))
for x in anglestr.split('\n'):
i,j,k,a = F(x.split(','))
A = a*pi/180
angles[(i,j,k)] = A
print(nodes)
completeAngleCalcs(angles)

lengths = {}
lengths[('1','2')] = 1.
completeLengthCalcs(angles, lengths)

positions = {}
positions['1'] = 0.+0j
positions['2'] = lengths[('1','2')]+positions['1']
completePositionCalcs(angles, lengths, nodes, positions)

print('# lengths of all edges')
for e in lengths:
print(e,lengths[e])

print('# Point positions')
for i in positions:
a = positions[i]
print(i,a.real,a.imag)

Z = array([ positions[i] for i in nodes])
x = Z.real
y = Z.imag

for e in lengths:
u = (positions[e[0]].real, positions[e[1]].real)
v = (positions[e[0]].imag, positions[e[1]].imag)
plot(u, v, 'g-')
plot(x, y, 'ko')
xlabel('x-coordinate', fontsize=18)
ylabel('y-coordinate', fontsize=18)
# shift landmark labels to make them easier to see
positions['1']+= -0.07j-0.04
positions['2']+= -0.07j-0.04
positions['3']+= -0.02j-0.12
positions['4']+=  0.02j
positions['5']+= -0.03j+0.04
positions['6']+=  0.02j
for i in nodes:
text(positions[i].real, positions[i].imag, i, fontsize=14)
savefig('triangulationEX.png', bbox_inches='tight', transparent=True)
savefig('triangulationEX.pdf', bbox_inches='tight')
show()

main()

## Spherical Trigonometry

While some tasks like determination of latitude were relatively simple for surveyors of the day, other tasks such as the calculation of longitude were beyond the skills of all but the most capable technicians of the day. Colonial surveyors had already made two attempts to mark off the trans-peninsular line dividing Delaware from Maryland, both of which had failed. In addition to technological issues for making accurate measurements, the distances involved were so large that the earth's curvature could introduce significant errors in surveying. Experts in surveying, including spherical trigonometry were needed, and so, on the advice of the British astronomer royal, Mason and Dixon had gotten the job.

Spherical trigonometry is the study of angles, distances, and areas, on the surface of a sphere. It is an ancient field as well, though not nearly so ancient as planar trigonometry. Menelaus of Alexandria wrote Sphaerica on spherical geometry and spherical trigonometry, circa 100 AD. This early origin is due in part to the desire of ancient Greek astronomers to reconcile measurement differences between the plane of the earth's orbit and the earth's equitorial plane. Even though we can never see more than half of the night sky at a time, early astronomers recognized that the night sky is conveniently mapped as a sphere. From Alexandria, spherical trigonometry was further developed in India and then the Islamic empire, before reaching Europe. The theory was effectively complete by the 1600's. The classic textbook for the subject is Spherical Trigonometry by Todhunter (1886).

### Great circles

Before we can discuss the concept of a triangle on a sphere, we must first agree on some concept of a line for the sides of the triangle. In Euclidean geometry of the plane, a line is straight and traverses the shortest distance between two points. On a sphere, none of the lines between two points will ever be straight. The curvature of the sphere itself means that all lines will curve. But we can still ask which curving line between two points is the shortest curve.

Suppose you are standing at the south pole of the earth, and you want to walk to the north pole as directly as possible. The most direct paths will be to follow a single line of longitude up the globe. It doesn't matter which one you pick, as long as you stick to the one with which you start. Along the way, the lines of latitude mark your progress, and are perpendicular to the lines of longitude. The equator marks the half-way point, and the other lines of latitude mark similar progress. If you stop anyplace along the line of longitude, you will have traversed the shortest path from the south pole to that point.

This idea can be adapted to finding the shortest path between any two points on a sphere. To find the shortest path across the sphere, draw the plane through the start point, the end point, and the center of the sphere. The intersection of this plane and the sphere is called a "great circle" because no other circles drawn on the sphere can have larger radius. The arc of this great circle between the start and end points will be the shortest path. Moving allong the shortest path between the start point and end point corresponds to following this greath circle. If you keep following this great circle, you will eventually reach the antipole of your start point, exactly opposite the point on the sphere where you started. For brevity, we can loosen our definition of "straight", and define a straight line on a sphere as any line corresponding to an arc from a great circle.

### The Spherical Law of Sines

Once we have agreed on a generalization of the concept of a straight line to a sphere, other geometric properties of a triangle can be developed. From three intersecting great circles, we can draw spherical triangles, and the theory of spherical trigonometry is the study of these triangles. We first note that in spherical trigonometry, it's convenient to measure distances in units of arc rather than standard units of distance. The angle of an arc is $$\ell/r$$ where $$\ell$$ is the length of the segment of the great circle and $$r$$ is the sphere's radius. For geography problems, we usually use the Earth's radius as $$r$$.

One important identity of spherical trigonometry is the spherical law of sines. For a spherical triangle, where $$A$$, $$B$$, and $$C$$ are the angles of the triangle, and $$a$$, $$b$$, and $$c$$ are the arc lengths of the sides opposite the given angles (measured in the same units as the triangle angles), the law of sines states $\begin{gather} \frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}. \end{gather}$

This law is tedious to derive by hand, but surprisingly easy to understand intuitively. The starting point is similar to the derivation of the plane-geometry law of sines. The volume of the parallel-piped rectangular prism created by the 3 position vectors of the triangle is the same no matter how we calculate it. By using 3 different trigonometric representations of these sides, we can get 3 different formulas for the volume. Setting the 3 formulas equal and simplifying yields the Law of Sines.

Suppose we have a spherical triangle $$ijk$$ where $$A$$ is the triangle angle $$kij$$, $$B$$ is the triangle angle $$ijk$$, and $$C$$ is the triangle angle $$jki$$. Let $$a$$ be the arc from $$j$$ to $$k$$, $$b$$ be the arc from $$k$$ to $$k$$, and $$c$$ be the arc from $$i$$ to $$j$$.

Now, think in terms of polar coordinates. Let $$i$$ be the z-axis. Since the angle $$A$$ lies in the plane orthogonal to $$i$$, $$A$$ is the angle between the plane containing $$0ij$$ and the plane containing $$0ik$$. Now express the vectors $$j$$ and $$k$$ in polar coordinates. Constructing a 3 by 3 matrix $$M _ i$$ out of these 3 vectors, use the determinate formula $$V^2 = \det (M^T M)$$ to find the volume $$V$$.

Repeat taking sequentially $$j$$ and $$k$$ as the z-axis. Then set these 3 volume formulas equal to each other and simplify. The computational part of the derivation can be performed using symbolic computing.

[Show code]
 1 2 3 4 5 6 7  from sympy.abc import a,b,c,A,B,C Mi = Matrix([[0,0,1], [sin(b),0,cos(b)], [sin(c)*cos(A),sin(c)*sin(A), cos(c)]]).T Mj = Matrix([[0,0,1], [sin(c),0,cos(c)], [sin(a)*cos(B),sin(a)*sin(B), cos(a)]]).T Mk = Matrix([[0,0,1], [sin(a),0,cos(a)], [sin(b)*cos(C),sin(b)*sin(C), cos(b)]]).T k = (sin(a)*sin(b)*sin(c)) ** 2 [ (M.T * M).det().factor()/k for M in [Mi,Mj,Mk]]
$M_i = \begin{bmatrix} 0 & \sin{\left (b \right )} & \sin{\left (c \right )} \cos{\left (A \right )}\\ 0 & 0 & \sin{\left (c \right ) \sin{\left (A \right )}}\\ 1 & \cos{\left (b \right )} & \cos{\left (c \right )} \end{bmatrix}$ $V_i^2 = \det (M_i^T M_i) = \sin^{2}(A) \sin^{2}(b) \sin^{2}(c)$ By symmetry, $\begin{gather*} V_j^2 = \det (M_j^T M_j) = \sin^{2}(B) \sin^{2}(a) \sin^{2}(c) \\ V_k^2 = \det (M_k^T M_k) = \sin^{2}(C) \sin^{2}(a) \sin^{2}(b) \end{gather*}$

Since all the angles are less than 180 degrees, all of the sin-functions are positive, and we can take the positive square roots without any loss of generality in our results.

There are a lots of other spherical-trigonometry identities's relating the six angles of a spherical triangle -- too many to list exhaustively. Perhaps the most important is the spherical laws of cosines, $\begin{gather} \cos c &= \sin a \sin b \cos C + \cos a \cos b,\\ \cos C &= \sin A \sin B \cos c - \cos A \cos B, \end{gather}$ which can be used to derive all other identities and which reduce to the familiar planar trigonometry identities $\begin{gather} c^2 = a^2 + b^2 - 2 a b \cos C, \\ \pi = A + B + C. \end{gather}$ For handy reference and flavor, here are the standard spherical trigonometry rules for right spherical triangles (where we assume $$C=90^{\circ}$$). \begin{alignat}{4} &\text{ }&\qquad \cos c&=\cos a\,\cos b, &\qquad\qquad &\text{ }&\qquad \tan b&=\cos A\,\tan c,\\ &\text{ }& \sin a&=\sin A\,\sin c, &&\text{ }& \tan a&=\cos B\,\tan c,\\ &\text{ }& \sin b&=\sin B\,\sin c, &&\text{ }& \cos A&=\sin B\,\cos a,\\ &\text{ }& \tan a&=\tan A\,\sin b, &&\text{ }& \cos B&=\sin A\,\cos b,\\ &\text{ }& \tan b&=\tan B\,\sin a, &&\text{ }& \cos c&=\cot A\,\cot B. \end{alignat}

## The Mason-Dixon line

Having discussed triangulation and spherical trigonometry, we now return to Mason and Dixon's task of surveying the boundary between Maryland and Pennsylvania.

By royal decree of the King of England, the southern boundary of Pennsylvania had been set at 40 degree's latitude. Such a simple rule seems at first pass like it would present no difficulties. Latitude (unlike longitude) can be estimated from relatively easy astronomical measurements and some star tables. However, latitude presents its own complications for surveying. For Mason and Dixon, the surveying of the southern border of Pennsylvania was a challenge because, unlike lines of longitude, lines of latitude do not follow the shortest paths between points -- they are actually curves on a sphere. As we discussed above, great circles pass through the antipole of their start point, but lines of latitude (other than the equator) do not. Because of of the great distance involved, Mason and Dixon had to make use of their knowledge of both triangulation and spherical trigonometry.

The technique Mason and Dixon chose to survey the boundary between Pennsylvania and Maryland was called the "secant method". The method is simple, but it required great care to execute. Starting at the corner of Pennsylvania, Maryland, and Delaware, face north, turn almost all the way (but not quite) west, and walk in a straight line for a fixed distance. Then, repeat as many times as necessary, facing north, turning west, and walking the same fixed distance. Starting at a point on the parallel, walking in a straight line is the same as proceeding on a great arc slightly above the parallel until it intersects the parallel a second time. Thus, the segment of the great arc is (roughly) a secant of part of the parallel. If turn angle and walking distance are pick right and everything is done accurately enough, the path followed will approximate a line of constant latitude.

Starting at the corner of Pennsylvania, Maryland, and Delaware, at 40° latitude (actually, everybody agreed to use a latitude slightly south of that, so the exact starting point was 39° $$43'$$ $$17.4''$$ North, 75° $$47'$$ $$18.9''$$ West longitude), Mason and Dixon chose to design their survey based on a secant arc of 10 minutes. If both ends of this 10 minute arc lie on the 40th parallel, then the two endpoints plus the north pole form a spherical isosceles triangle with the other two sides formed by lines of longitude. If we find the midpoint of $$10'$$ segment and connect the midpoint to the north pole, we then have two spherical right triangles.

We can now apply the identity's of spherical trigonometry to solve for the unknown angles and sides in the right spherical triangle in the diagram above. We start knowing $\begin{gather*} C = 90^{\circ}, \quad b = 5', \quad \text{and} \quad c = 90^{\circ} - 39^{\circ} 43' 17.4'' = 50^{\circ} 16' 42.6''. \end{gather*}$ Since this is a right-triangle, the identity $$\cos c = \cos a \cos b$$ implies $a = \arccos \left( \frac{ \cos c }{ \cos b} \right) = 50^{\circ} 16' 42.4''.$ And from the formula $$\tan b = \cos A \; \tan c,$$ we deduce $A = \arccos\left( \frac{\tan b}{\tan c}\right)= 89^{\circ} 55' 50.7''$ And for completeness, the identity $$\sin b = \sin B \sin c$$ implies $$B = 6'\, 30.0''$$. So, at each corner, the expedition would have to turn $$4'$$ $$9''$$ north of east and walk $$10'$$ in a straight line.

In May of 1764, Mason and Dixon made estimates equivalent to using $$r = 6.290$$ megameters as the earth's radius. Based on their estimate, a 10 minute arc would correspond to a distance of $$18.30$$ kilometers, though a modern estimate of the earth's radius would imply $$18.52$$ kilometers, an error of $$20$$ meters. Using standard interpolation methods, then, the rest of the parallel between the endpoints of the secant line can be filled in. At the midpoint of the secant arc, we can find the error between the arc and the parallel by subtracting the long leg from our hypotenuse, $$c-a = 0.2''$$, or: $\frac{(c-a)}{60 \times 60} \frac{2 \pi}{360} r \approx 6 \; \text{meters}.$

Precision calculation helped Mason and Dixon avoid wasteful back-tracking and correction during the survey. This had two important implications for Mason and Dixon's calculations. First, in order to get 1-meter precision in our solutions, they would sometimes have to carry around at least 7 decimal places or more in our calculations. This would make all multiplications -- which had to be done by hand -- very lengthy, if not for the use of logarithms. Second, imprecise knowledge of the relationship between celestial angles and terrestrial distances was an un-avoidable source of error as they proceeded, introducing a few meters of error for every arc minute of distance travelled. This would have to be corrected in the field. At night along the way, they would check their latitude using the stars and correct their position accordingly. Differences between observed and predicted times of eclipses and transits were used to calculate longitude.

Charles Mason and Jeremiah Dixon were able to use these calculations as as a basis for surveying of Pennsylvania's southern boundary. The expertise they gained allowed them to become the first British surveyors to accurately measure 1 degree of latitude for the Royal Society -- an important observation used in the great debate of the time over the earth's shape. Mason and Dixon's concluded with the collection of a number of important data values, including their best estimate as $$6.34$$ megameters for the earth's radius, a number slightly smaller than the modern excepted range, but accurate to about half a percent.

# Exercises

1. The philosopher Eratosthenes (276-194 BC) lived in North Africa and became chief librarian at the Library of Alexandria during the Ptolemaic empire. Legend has it that in Syene far south on the Nile river, at noon on the summer solstice, the sun shined all the way to the bottom of a deep well. The same was not true in Alexandria. Measuring the angle of the sun's shadow on the summer solstice in Alexandria, Eratosthenes found it differed from a vertical plumb bob by about 1/50th of the circumference of a circle. He also knew that camel caravans walking from Syene to Alexandria travelled about 5,000 stadia.

1. How many stadia did Eratosthenes estimate the circumference of earth to be?

2. How accurate was his estimate?
2. In the 1630's, Richard Norwood measured the altitudes of the sun as 51 degrees 30 minutes in London and 53 degrees 58 minutes in York. He then measured the distance from York to London as 367,200 feet (about 112 kilometers). What was his estimate for the radius of the earth (in kilometers)?

3. The Mason--Dixon boundary between Pennsylvania and Maryland came to be set at the parallel of 39° $$43'$$ $$17.4''$$ north, by mutual agreement of the states, rather than the 40° originally intended. If the southern boundary of Pennsylvania had been at exactly 40°, what should Mason and Dixon have used as their turn-off between successive arcs? (give an answer accurate to tenths of an arc second).

4. How would the survey design change if Mason and Dixon used 20' secants instead of 10' secants?

5. If the souther boundary of Pennsylvania had followed a great circle rather than a meridian, how much farther south would the westmost corner lie, assuming it started at the same easter corner with Delaware that it does today?

6. Page 88 of Haasbroek (1968) gives Snell's angle measurements in a table, with an accompanying map showing the angle labels.

1. Translate this angle data into 3-point form $$(a,b,c)$$ where $$a,b,c$$ are symbols for the citing locations, with $$b$$ the center of the angle, $$c$$ always counter-clockwise of $$a$$. Note that all angles are measured by inner arc so they are less than 180 degrees.

2. Using these data, and assuming the distance from Breda to Bergen op Zoom is 1 flab, how many flabs is it from Utrecht to Bergen op Zoom.

7. On July 3, 1764, Mason and Dixon used planar trigonometry to calculate the width of the Nanticoke river. Based on their radius of $$6.29 \times 10^6$$ meters, how many meters of error does the flat-earth model introduce into the determination of the Nanticoke river's width? (see page 59, and figure 73, page 214, of the transcription of their journal)

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