Download Homework 1 Solutions

Modern Geometries
Due: Friday, January 27
Name:
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Problem 1 Use your knowledge of spherical trigonometry. Calculations should be correct
to four significant digits.
(a) Berlin is located at 52.52◦ North by 13.41◦ East. Chicago is at 41.88◦ North by 87.63◦ West.
Assuming the earth is a sphere of radius 3959 miles, find the distance from Chicago to Berlin.
Hint: use the north pole.
(b) The distance from Chicago to Honolulu is 4251 miles. Chicago to Lima, Peru is 3768
miles, and Lima to Honolulu is 5943 miles. If you were in Honolulu and had signs pointing
exactly toward Chicago and Lima, what would be the angle between the signs?
(c) A triangle is drawn on the earth with one vertex at the north pole. The other two
vertices are on the same line of latitude, one on the prime meridian and one on the longitude
90◦ West. The triangle has an area of 41 million square miles. What is the latitude of the
other two vertices?
(a) Together with the north pole, we are given side-angle-side for a triangle. From
Chicago to the north pole is a distance of 48.12◦ (remember we measure distance by
angles on a sphere!). The distance from Berlin to the north pole is 37.48◦ . And the
angle between the longitudes is 87.63◦ + 13.41◦ = 101.04◦ . Applying the spherical law
of cosines, with A the north pole and a the great circle distance from Chicago to Berlin,
we get cos(a) = cos(101.04◦ ) sin(48.12◦ ) sin(37.48◦ ) + cos(48.12◦ ) cos(37.48◦ ) = 0.4430.
So a = 63.70◦ = 1.1112. Multiplying by the radius of 3959 miles gives a distance of
4402 miles . (Actual distance is 4400.5.)
(b) This time we are given side-side-side. So this time we use the spherical law of cosines
to solve for the angle. If we put angle A at Honolulu, and convert the distances to angles
by dividing by the radius we get sides a = 3768/3959 = 0.9518, b = 1.074, c = 1.501. We
cos(a) − cos(b) cos(c)
cos(0.9518) − cos(1.074) cos(1.501)
get cos(A) =
=
= 0.6237. So
sin(b) sin(c)
sin(1.074) sin(1.501)
the angle is the inverse cosine of this, or 0.8972 = 51.41◦ .
(c) In terms of angles, the area of the triangle is 41, 000, 000/39592 = 2.616. So from
the spherical triangle area formula, we know the three angles of the triangle must add
to 2.616 + π = 5.752. Since one of the angles is 90◦ = π/2, the other two angles
(which are equal to each other because the two points are on the same latitude so the
triangle is isosceles) are 2.093. Now that we know all three angles, we can find the
sides using the supplemental law of cosines. Since one of the angles is 90◦ , the formula
simplifies to just cos(a) = cos(A)/ sin(C) where A and C are the two equal angles. So
cos(a) = cos(2.093)/ sin(2.093) = −0.5755. The inverse cosine of this is 2.184 = 125.1◦ .
So the two vertices are at 35.1◦ South .
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Problem 2 In plane geometry, there is no SSA law because of the ambiguous case of the
law of sines. Instead, given two sides and a non-included angle, you might get zero, one, or
two tringles depending on the relative sizes of the given sides and the angle.
(a) We will show that this is true in spherical geometry also, using examples. (i) If 4ABC
has A = 30◦ , b = 1, and a = 0.4, show no such triangle exists. (ii) If instead a = 0.7 show
that two triangles exist, while (iii) if a = 1.3 exactly one triangle exists.
(i) Since the spherical law of sines tells us sin(b)/ sin(B) = sin(a)/ sin(A) we find that
sin(B) = sin(1) sin(30◦ )/ sin(0.4) ≈ 1.08. Of course, sine cannot be this large, so no such
triangle can exist.
(ii) This time when we solve we find sin(B) = sin(1) sin(30◦ )/ sin(0.7) ≈ 0.653. This
allows for two angles for B, about 40.8◦ and 139.2◦ . This is akin to the ambiguous case
in the regular law of sines.
(iii) Again, we solve for B and find two angles, about 25.9◦ and 154.1◦ . This time,
though, the second angle won’t work, because it is opposite a side of length 1 while the
smaller angle of 30◦ is opposite a side of length 1.3. The longer side must be opposite
the larger angle, so this second triangle cannot actually exist.
(b) (Bonus) For angle A = 30◦ and b = 1 find the length of a that is the cutoff between
where there are no triangles and where there are two, and the cutoff between where there
are two triangles and where there is only one.
From part (i) above, the triangle doesn’t exist as long as sin(B) has to be larger than one.
If sin(B) = 1, then we can solve for a: sin(a) = sin(A) sin(b)/ sin(B) = sin(1)/2. In this
case, exactly one triangle exists, but then for slightly larger a two triangles exist. And
just like in the planar case, as we lengthen a two triangles will exist until the triangle
becomes isosceles—when a = 1. Beyond that there is only one triangle.
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(c) Show that AAS is not true in spherical either. Do this by giving examples of two angles
and a non-included side where there are (i) no triangles that work, (ii) two triangles that
work, (iii) exactly one triangle works.
We can swap between sides and angles by using the polar triangle. So no triangle exists
for a = 0.4, b = 1, A = 30◦ . Thus it’s polar triangle can’t exist either, which would have
A0 = π − 0.4, B 0 = π − 1, and a = 150◦ . Two triangle would exist if angle A0 is reduced to
π − 0.7 and the second disappears leaving just one when A0 is further reduced to π − 1.3.
Problem 3 In class, we used the small angle approximation on the spherical law of sines,
and it turned into the regular law of sines. Similarly for the spherical law of cosines. What do
you get when you apply the small angle approximations to the supplemental law of cosines?
(It may help to remember that in a plane triangle the angles have to add up to 180◦ !)
in this case, only a is small, since the other angles involved are angles of the triangle. So
we get cos(A) ≈ (1 − a2 /2) sin(B) sin(C) − cos(B) cos(C). Since a is small, the a2 term is
really small so we ignore it. Also, A + B + C = 180◦ so cos(A) = cos(180◦ − (B + C)) =
− cos(B + C). So we arrive at cos(B + C) = cos(B) cos(C) − sin(B) sin(C), which is the
angle addition formula for cosine.
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Problem 4 In class we saw Eratosthenes’ method of finding the radius of the earth. Come
up with your own method for doing so. You may assume the earth is a perfect sphere, and
that there is nothing on it, like houses or mountains or people to get in your way, unless you
want there to be somethere there to measure. You can also assume that you can make very
precise measurements if necessary, see very very far, hear very very well, etc. But you may
not travel more than, say, 20 miles. Bonus points for the most practical method!
Here are a number of methods, together with an estimate of their practicality.
• Make a loud noise, then see how long it takes to hear the echo from the sound
going all the way around the world. This is extremely impractical, as the sound
spreads out it will get quieter and quieter. (Well, until it gets halfway around
the world, and then the wavefront from the sound will come from all directions
and converge at the antipode!). In fact, (if I did the arithmetic correctly!) you
would need to create a sound of about 420 decibels (at ten feet from the source)
or else by the time it spreads out to halfway around the world the strength of the
wave wouldn’t be any stronger than the random currents of air molecules moving
around. Unfortunately, sounds over 195 decibels become shockwaves which tear
the air apart, and do not travel normally. So even if you could make a sound loud
enough to hear all the way around the world, the computations you would have to
do to figure out the proper speed of sound and speed of shockwaves are just too
difficult.
• Plant two sticks exactly vertical at some distance from each other. Precisely measure the distance between the bottoms of the sticks and the tops. Use these to figure
out the angle between the sticks, and then the radius is the distance between the
sticks divided by the angle. Practicality is not good, but doable. If your sticks were
about 3 meters long, planted about 30 km apart, the difference in distance between
the tops and bottoms would be about 1.5 cm. You would need to be extremely
precise in planting the sticks exactly vertically, but this is not impossible.
• Climb a tree, say, to a height of about 20 feet. Measure the distance to the horizon,
by having someone put something on the ground and moving it farther and farther
away until it just barely disappears behind the horizon. Use geometry to figure
out the radius of the earth. (Hint: your line of sight to the object just as it moves
over the horizon is a tangent to the sphere, so is at right angles to the radius,
forming two sides of a right triangle. You, up in your tree, form the hypotenuse.
So (r + 20)2 − h2 = r2 where h is the distance to the horizon and the 20 is because
you are 20 feet higher than the radius of the earth.) The practicality of this is
medium. If you are 20 feet up a tree, the horizon on the earth is about 5 1/2 miles
away, so you would have to spot your test object at this distance. This is actually
pretty reasonable if your test object is a light, and it is dark outside.
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