Working With Right Triangles
to Determine Heights and Distances
By understanding the relationship of particular sides in a right
triangle, along with some simple trigonometric functions and helpful
formulas, we can easily
solve for unknown angles, heights, or distances.
One of the earliest examples of a geometric application is that of
Thales, the ancient Greek philosopher and mathematician, who lived from
about 640 BC. to 546 BC. In trying to determine the height of a
pyramid, he placed an upright stick at the foot of the pyramid's shadow
and discovered that the relation between the pyramid's height and
was the same ratio as that of the stick's height to its shadow.
In other words, if a stick 3 feet high were
to cast a shadow 5 feet long, while at the same time a pyramid's shadow
measured 750 feet, we can then determine the height of the
The pyramid, as well as the stick, can be visualized in the
framework of a right triangle. And though viewing the stick this
way is a
much smaller scale than that of the pyramid, yet their tangent ratios
are identical, which means that their angles are also of the same
degree. We can, therefore, learn from the smaller what we need to
know for the larger. For in determining the tangent of an angle,
"opposite" to the angle is divided by the one "adjacent" to it.
And for now, we are thinking of the "opposite" as the height, and the
"adjacent" as the length of the shadow.
For instance, suppose that an angle has an "opposite" side
of 3 and
an "adjacent" side of 4. Dividing the opposite by the adjacent
give us a ratio of 0.75. For the formula for the tangent ratio of
an angle is "TAN = OPPOSITE / ADJACENT," and for our
example that means 0.75 = 3 <divided by> 4.
What I want you to also see in this is how helpful this formula can be
in finding unknowns. For since "tan = opposite / adjacent," then
"tan * adjacent = opposite" and "opposite / tan =
adjacent." So with this formula, and its re-arranging, we can
determine three different unknowns.
Let us now try this with the pyramid and stick.
The stick stands 3 feet and casts a shadow of 5 feet. We are
thinking of its height as the "opposite" to the angle, and the length
of the shadow as the "adjacent" to the angle; and since TAN = OPPOSITE
/ ADJACENT, then 3 / 5 = 0.6. So "0.6" is the tangent of the
Now if we only knew the angle and the adjacent -- and not the opposite
-- we could determine the "opposite" by multiplying the adjacent
by the tangent; and in
this case it would be 5 * 0.6, which would result in 3. Or
if we didn't know the adjacent, we could divide the opposite by the
tangent, which would be 3 / 0.6 to come up with 5.
Though the stick is much shorter than the pyramid, they still have the
same angle degree caused by the sun. Their heights and shadows
greatly differ in length, but their own ratio between their heights and
shadow lengths are the same. They both have the same tangent
ratio. So because of that, we can use the same tangent ratio of
the stick and its shadow to determine the height of the pyramid.
To prove that the tangent ratio remains the same for the pyramid and
stick, think of any acute angle. Now if you added several right
angles in it that were an inch or so apart, they would each still be
sharing the same angle A. And as the right triangle increases in
size as we move from one of those right angles to the next, still each
one maintains the same tangent ratio between their opposite and
adjacent. For instance, if the "opposite" side were 3 and the
"adjacent" side 4, the ratio would be 0.75; but if these sides were 6
8, respectively, we would still have a tangent ratio of 0.75. And
this continues wherever we would place a
right angle on the same acute angle, whether the vertex of the 90
degree angle would be 10 inches away from the vertex of angle A, or 12
inches away, or 14 inches away, etc.
How High is the
the pyramid has the same tangent ratio as the stick of 0.6, then
0.6 * 750 = 450
So the pyramid is 450 feet in height.
wasn't that much easier than having to awkwardly climb up a 450-foot
structure with an enormously long tape measure?
If Thales had a calculator in his day, there is also another way that
the pyramid's height could have been easily determined -- and without
using the stick. One would merely have to measure the
length of the pyramid's shadow and obtain the angle of elevation from
the foot of that shadow to the peak of the pyramid. The angle
would be 30.96 degrees. By inputting that number into a
calculator and pressing the TAN (tangent) button, we come up with
0.6. Then we can plug the value into our formula: "tan = opposite
/ adjacent," but
re-arrange to "tan * adjacent = opposite" to now determine the value of
"opposite." Therefore, 0.6 * 750 = 450.
Because the height of the pyramid was the same ratio to its shadow as
the stick was to its, we could have also used this following formula
for determining the pyramid's height
--- = ---
Here we have plugged in all the values
except the pyramid's height:
We can now solve "x" in either of the following ways::
1) Divide the second denominator by the first denominator, and then
multiply the result by the first numerator.
2) Or, divide the first numerator by the first denominator,
and then multiply the result by the second denominator.
In either method, your answer should be 450.
This can also be double-checked by cross-multiplying: Just think of it
in the shape of an "X" and multiply the first numerator by the second
denominator. Then multiply the first denominator by the second
numerator. Both results should be the same. In this case,
Looking More Into the Right
We considered, above, an example of using a right triangle to determine
an unknown height; and there are also more examples to follow.
But let us first take a brief look at a right triangle; and with only
three sides, it shouldn't take us too long to learn them. These
sides are called the hypotenuse
(which is always the longest side), the adjacent
(to the angle), and the opposite
(to the angle).
The right angle is directly across from the hypotenuse, and it is
sometimes labeled with a capital “C,” which would then correspond to
the small “c”
that labels the hypotenuse. The small square box, which is sometimes
placed in the inside corner of the right angle, points to the
In working with the same right triangle, the sides of the opposite and
the adjacent will switch places when we move the focus from angle A to
angle B, but the hypotenuse will still remain the same for either
If the vertex of the right angle is in the bottom right of the
right triangle and the vertex of angle A is in the bottom left, then
at the top. The side across from angle A is called the “opposite” and
usually labeled “a,” and the side across from angle B is usually
labeled “b” and called the “adjacent” to angle A (but is also the
“opposite” to angle B).
The term “adjacent” signifies that that side is adjoining the angle.
This is why side b is the adjacent to angle A, while side a is the
adjacent to angle B.
The Pythagorean Theorem
There is a helpful formula called the Pythagorean
Theorem that helps us in figuring out the lengths of a
right triangle's sides. First, and moving clockwise with our above
right triangle, let us label the sides as “c” for the hypotenuse, “a”
for the opposite, and “b” for the adjacent. The Pythagorean Theorem
states that “c^2 = a^2 + b^2.” To see this demonstrated, let us suppose
that c = 5, a = 4, and b = 3. By plugging these values into the
formula, we then have c^2 = 25, a^2 = 16, and b^2 = 9. So squaring the
lengths of sides a and b and then adding them will total 25, the same
as the square of c.
We can also see in this that by knowing any two of the sides, we could
then easily determine an unknown side. We merely need to re-arrange the
formula, such as “c^2 - a^2 = b^2,” if we were wanting to determine the
length of b. By plugging in our above values for c and a, we then have
25 - 16 = b^2. So, b^2 = 9.
We then need to find the square root of that answer to show the actual
length. The square root of 9 is 3; so there you have it.
Using a calculator that has squaring and square root functions makes
this very easy.
To make this even easier, if you have an HP calculator with a “solver”
feature, try inputting this short program:
The calculator recognizes the “SQ” (square) function and displays the
labels of HYPOT, OPP, and ADJ on the screen. With it, you can plug in
any two values to determine the third with the press of a button. Just
use the actual lengths of whatever two sides you want to input, and
then press the button for the one you want to determine. The inputted
numbers, as well as the result, will all be displayed in actual
lengths, rather than in those numbers squared.
Another way you can determine side lengths – and by initially knowing
the length of only one side – is by using a trig function with the
angle and the known side. This is where we get into the “Soh-Cah-Toa,”
which is a helpful mnemonic for remembering which trig function to use
with which sides of the right triangle.
“Soh,” for instance, stands for
“SINE(of the angle) = OPPOSITE / HYPOTENUSE.” “Sine” is often
abbreviated as “sin,” but still pronounced to rhyme with “twine” even
then. This formula means that the opposite divided by the hypotenuse
will be the SINE of the angle. So if you inputted the angle into a
calculator and pressed the SINE button, it would give you the same
answer as when dividing the “opposite” by the “hypotenuse.”
For instance, suppose we have a right triangle with an angle of 45
its “opposite” is 9 inches, and the “hypotenuse” is 12.73 (rounded to
the nearest hundredth). If we would input 45 into a calculator and
press the “SINE” button, the result would be 0.707 (rounded to the
nearest thousandth). But also if we divide the opposite by the
hypotenuse (9 <divided by> 12.73), we'll come up with the same
Now let us see a practical way we can use this:
Wire-Lengths for an Antennae
Two feet down from the top of a 15-foot antennae in a large field, you
want to run
four guy wires at a 45 degree angle. How long does each wire need
First of all, try to visualize the problem in the framework of a right
triangle. And be aware of what two sides you'll be giving your
attention to. One of the sides will be the side you are trying to
determine the length of, and the other side will be the one needful to
In our particular example, we have a 15-foot antennae, but the wires
will attach at the 13-foot mark; so we can think of the antennae
itself (up to the 13-foot mark) as being the side of the right
is “opposite” the angle. So the “opposite” is representing the height.
At the base of the “opposite” is the vertex for the right angle, and
running along the bottom from it is the “adjacent” (which we are not
given the length of for this problem, but we won't need to know that).
The adjacent continues to the vertex of our 45 degree angle; and the
line that angles up from that to where the wires will be connected is
the hypotenuse (which, therefore, represents the length of one of the
So since this problem will concern itself with the “opposite” and the “hypotenuse,” which of the formulas
in “Soh-Cah-Toa” will we need
for that? “Soh” is the only
one that deals with both of these sides; so we'll be using the SINE (sin) function – and not the
COSINE (cos) nor the TANGENT (tan) for this problem.
Remember, the “Soh” formula means
“SINE(of the angle) = opposite <divided by> hypotenuse.”
But since we don't know the length of the hypotenuse, how could we
re-arrange this formula, and what operation would we use, to determine
If you have trouble figuring that part out, try comparing it to the
following simple equation:
6 / 2
In this formula, the “2” is in the same location as the hypotenuse; so
how could we re-arrange the 3 and the 6, and what operation would we
use, to come up with 2 for the answer? That's pretty simple:
6 / 3
So we can then re-arrange our formula
the same way:
= Opposite / Sin(angle)
But better still, you might also find
this helpful. What observation can you make from the following
that will be helpful for you in
remembering the re-arrangement and operation to use for a formula?
First of all, whether we are dealing with SOH, CAH, or TOA, if we want
to determine the value of the second letter in any of these formulas,
we simply swap places with that and the first letter and change the
operation to multiplication. We can also see that it is only when
determining the value for the second letter that we will multiply
We also note that if we want to determine the value of the third letter
in SOH, CAH, or TOA, we also swap that with the first letter and let
the operation of division remain. So that's all there is to
So in re-arranging our “Soh” formula to solve for the
hypotenuse, we then have...
= opposite / sin(angle)
hypotenuse = 13 / sin(45)
hypotenuse = 13 / 0.707106781187
hypotenuse = 18.38
Therefore, four times that (for 4 wires) will give you 73.52 feet. Plus, in addition to
that, you'll want to include whatever extra inches you'll need for
tying the wires.
Here's another problem:
High Up on the Building is the Ladder Leaning?
If a 20-foot ladder, at an angle of 60 degrees, is leaning against a
building, how high up on the building is the tip of the ladder?
In putting this problem into the framework of a right triangle, we have
the 20-foot ladder represented by the hypotenuse; and the unknown
height up the building represented by the “opposite.” So this will also
use the “Soh” formula; and, therefore, the “SINE” function.
sin(angle) = opposite / hypotenuse
In our previous problem, we used this formula to solve for the unknown
hypotenuse. Now, however, we will be solving for the unknown “opposite.”
sin(60) = opposite / 20
* 20 = opposite
0.866025403784 * 20 = opposite
opposite = 17.32
So a 20-foot ladder, leaning at a 60-degree angle, will reach up to 17.32 feet on the side of a building.
with 30-60-90 and 45-45-90 Degree Angles
The 30-60-90 Degree Triangle
In the above examples, we used one right triangle that had a 60-degree
angle, and another with a 45-degree angle. These triangles have
ratios that can be easy to remember. For in a 30-60-90 degree
triangle, the hypotenuse will always be twice the length of the
shortest side; and the middle-length side will be "(the square root of
3) times the length of the smallest side." So if the hypotenuse
is 10, what would the other sides be? The shortest side would be
5; and, using the square root function of a calculator, the
middle-length side would be 8.66025403785. So if we know
the length of just one of the sides, we can then easily figure out the
length for the others.
in Using the 30-60-90 Degree Triangle
So let's try this. Suppose, for instance, that a kite is being
flown, and the 300-foot string is at a 60-degree angle. How high
is the kite?
We can think of the string as being represented by the hypotenuse of a
right triangle, and the side "opposite" from the angle as representing
the height, which we want to determine. That height we can
visualize as an imaginary line coming straight down from the kite to
the earth, and then over at a right angle to where the kite-flyer is
standing. The distance from the vertex of that right angle to the
kite-flyer is represented by the adjacent side of the angle.
Now since the hypotenuse is going to be twice the length of the
shortest side, and the middle-length side will be "(the square root
of 3) times the smallest side," we can come up with the following:
length of string (hypotenuse) = 300 feet
distance from vertex of right angle to
kite-flyer = 150 feet
height of kite = 259.81 feet (rounded to the nearest
You can also double-check that by multiplying the adjacent (150) with
the tangent of angle A. (The tangent of 60 is
1.73205080757.) The result will be the same.
Suppose, though, that the string had been 200 feet long, and at the
same angle. How high would the kite be then?
of string (hypotenuse) = 200 feet
distance from vertex of right angle to
kite-flyer = 100
height of kite = 173.21 feet
(rounded to the nearest hundredth)
45-45-90 Degree Triangle
For this particular triangle, you need
to only remember that the hypotenuse is going to be "(the square root
of 2) times either of the "legs" of the right angle (for they are both
the same length). So to find the length of the other sides,
simply divide the length of the hypotenuse by the square root of
2. For example, if the hypotenuse is 12, then each of the other
sides will be 8.48528137426.
And so, if you didn't know the length of the hypotenuse, but did of the
other sides, multiplying that length by the square root of 2 would give
you the hypotenuse. If, for instance, the adjacent and opposite
sides were each 8, then the hypotenuse would be 11.313708499.
And since the opposite and the adjacent sides are the same length, then
if we were to walk out from the base of a building until we could see
its top at a 45 degree angle, the height of the building would be the
same distance we walked.
Or if we didn't know the length of a long beam that is leaning against
a structure at a 45 degree angle,
but we did know that the base of that beam was 12 feet out from the
building, or that it reached 12 feet up to the side of it, then we
could easily figure that the beam would be 16.97 feet long. For
we merely need to multiply 12
by the square root of 2 for that answer,. which can be done very easily
with a calculator.
Here's one more problem for the “Soh” formula that deals with
ratio of an angle. The hypotenuse is 15, and the opposite is 12.
What is the
degree of the angle?
Remember: “Sin(angle) = Opposite / Hypotenuse.”
12 / 15 = 0.8
If you then input that number (0.8) into a calculator and press the arcsine button, it will give you “53.13” as the angle degree.
More Antennae Wires
(with a different formula)
If in a field, you had an antennae and didn't know the height of it,
but you wanted to place 4 guy wires that would be grounded 10 feet out
around its base and that would have an angle of elevation of 50
toward the antennae, how long would the wires have to be?
Again, let us put this in the framework of a right triangle. This time,
however, we don't know the height (which would be represented by the
“opposite” side of the right triangle), but we do know the distance
base of the antennae to where the wires will be grounded. That distance
is represented by the “adjacent” side
of the right triangle. And we also know the angle of elevation that the
are to have, but we don't know how long those wires need to be. How
could we find the solution for this?
Since we don't know the wire-length (represented by the hypotenuse) nor
the height of the antennae (represented by the “opposite”), then we
can't use the “Soh” formula. But in looking at the “Soh-Cah-Toa” mnemonic, we see that
“Cah” will use in its formula
both the “adjacent” (which is
representing the distance from the base of the antennae to where the
wire is to be grounded) and the
Cos(angle) = Adjacent / Hypotenuse
“Cos” is an abbreviation for “cosine,” another helpful
We can now input what we know into this formula:
(cos)50 = 10 / hypotenuse
Since we need to determine the hypotenuse, we will have to re-arrange this formula and choose
the right operation; and since “cos(angle) = adjacent / hypotenuse,”
then “adjacent / cos(angle) = hypotenuse”:
Adjacent / Cos(angle) = Hypotenuse
10 / cos(50) = hypotenuse
10 / 0.642787609687 = hypotenuse
hypotenuse = 15.56 (rounded to
will, therefore, need 62.24 feet of wire (for the total of 4), plus
whatever extra inches
needed for fastening them.
High Up the Antennae Were the Wires Attached?
By the way, as you probably figured, we did not need to know the height
of the antennae for the solution because our only concern was that the
wires would be grounded 10 feet out from the base of the antennae and
then have an angle of elevation of 50 degrees to the antennae. But can
you figure out how high up on the antennae these wires were attached?
Which of the “Soh-Cah-Toa” formulas will you need to use for that?
With the information we now have for the hypotenuse (15.56), the adjacent (10), and the angle (50), we can use either the “Soh” formula or the “Toa” in order to find the height to
where the wires fastened. For since they both involve the “opposite” (which is representing
that height), we have the choice of using either that one that also
includes the hypotenuse or the
one that includes the adjacent.
If we chose the “Soh” formula,
so that we can determine the “opposite”
by inputting the value of the hypotenuse,
the formula will look like this:
sin(50) = opposite / 15.56
We then re-arrange this formula to determine the value of the
and since “sin(angle) = opposite / hypotenuse” then “sin(angle) *
hypotenuse = opposite.”
sin(50) * 15.56 = opposite
0.766044443119 * 15.56 = opposite
opposite = 11.92
(for how many feet up the
antennae the wires are attached,
and rounded to the nearest hundredth)
If, however, we used the “Toa” formula,
so that we could determine the
“opposite” by inputting the value of the adjacent, it could look something
tan(angle) = opposite / adjacent
tan(50) = opposite / 10
1.19175359259 = opposite / 10
Then we re-arrange the formula to the following:
* 10 = opposite
opposite = 11.92 (rounded to the
Far is the Base of the Ladder from the Building?
Here's another: If a 25-foot ladder were leaning on a 65-degree angle
against a building, how far would the base of it be from that building?
When we put this in the framework of a right triangle, the leaning
is represented by the hypotenuse, and we know the angle; but we need to
find out the distance the base of the ladder is from the building, and
this distance will be represented by the adjacent. So, again, we will
need the formula that will deal with the hypotenuse and the adjacent;
and that is “Cah.”
adjacent / hypotenuse
cos(65) = adjacent / 25
Since this above formula is so, then the following also is:
cos(65) * 25 = adjacent
0.422618261741 * 25 = adjacent
adjacent = 10.56
So the base of the ladder is 10.56
feet away from the building.
And if we divide the adjacent by the hypotenuse (10.56 / 25), we will
come up with “0.422,” which is the same as inputting 65 (our angle)
into the calculator and pressing the COS button. And inputting 0.422
into the calculator and pressing the arccosine button will give you the
angle of 65 degrees.
Tall is the Lighthouse?
Imagine that you were visiting a lighthouse and wondered about its
height. You took 12 steps out from the base of it, and figured that to
be about 36 feet. With an eye-level of 5.5 feet, you looked up at an
angle of 70 degrees to the very top of it. How tall is the lighthouse?
As we put this problem in the framework of a right triangle, we can
of the “opposite” side of the right triangle running up the height of
lighthouse, and with the vertex of the 90-degree angle 5.5 feet up from
its base (at our eye-level). The 36 feet that were walked from the base
of the lighthouse is represented by the “adjacent” side of the right
triangle. We also know that the angle from our eye-level to the top of
the building is 70 degrees. However, we don't know the length of the
hypotenuse, but that is not necessary for this problem.
So with this information, which of the three formulas will we need from
Since we know the adjacent and want to solve for the opposite, then we
will use the “Toa” formula:
tan(angle) = opposite / adjacent
Tan is an abbreviation for tangent, another trig
By plugging in the information we have so far, we come up with this:
tan(70) = opposite / 36
We then re-arrange the formula to solve for the opposite:
tan(70) x 36 = opposite
2.74747741945 x 36 = opposite
opposite = 98.91 (rounded to the
Since the vertex of angle A is at our eye level, then we need to add
that additional height of 5.5 feet to our answer.
So the actual height of the lighthouse
would be 104.41 feet (rounded to the nearest hundredth).
Far Away is that Building?
How about this one? You are walking down a very level road. In the
not-too-far-away distance, you can see a building that you know to have
a height of 100 feet. If from where you are, you can see at a 20-degree
angle to the top of the building, how far are you from the building?
When thinking in terms of a right triangle, we'll let the building's
height be represented by the “opposite” of the right triangle, and the
distance we are wanting to determine will be the “adjacent.” So with
that in mind, it is the “Toa” formula that we are again going to use;
but with this problem, we will be solving for the “adjacent” rather
than the “opposite.”
Tan(angle) = Opposite / Adjacent
tan(20) = 100 / adjacent
0.36397 = 100 / adjacent
If “tan(angle) = opposite / adjacent,” then...
“opposite / tan(angle) = adjacent”
“100 / 0.363970234266 = adjacent”
adjacent = “274.75” (rounded to
So the distance to the building is
Far is it Across the River?
Here's another. Suppose you were walking along a peaceful river, and
you began to wonder about the width of it. If you could see on the
opposite bank of that river a large statue, how could you determine the
distance across? Which of the “Soh-Cah-Toa” formulas would be the one
to use for this?
To figure this, as you stand directly across from the statue, consider
where you are as being the vertex of the right angle, and the distance
from that point to the statue to be the “opposite” of the right
walk about 50 feet to form the “adjacent” part of the right triangle,
sure to keep on a straight course that won't mess up the right angle.
Now, where you stop will be the vertex of angle A. From there, figure
out the angle it is to the statue. That line of sight will be the
hypotenuse, but you don't need to know its length, only the degree of
angle A to the statue. For this example, we will say that the angle is
So you will once again need the “Toa” formula for this, since you know
the “adjacent” (50 feet), but want to determine the “opposite”
(distance across the river).
NOTE: Prior, we thought of the
“opposite” as representing the height; but now we have “placed the
right triangle on its side” in order to determine not the height, but
distance – or, in other words, not the vertical, but the horizontal.
opposite / adjacent
tan(83) = opposite / 50
8.14434642797 = opposite / 50
We then need to re-arrange our formula to solve for the “opposite.”
Since the above formula is so, then the following is also true:
8.14434642797 x 50 = opposite
opposite = 407.22 (rounded to the
So the river is 407.22 feet wide in
As we have seen, trig functions can be very helpful; and remembering
the three sides of a right triangle, along with what “Soh-Cah-Toa”
for, will help us in that.
Sin(angle) = opposite / hypotenuse
Cos(angle) = adjacent / hypotenuse
Tan(angle) = opposite / adjacent
I also included these formulas in the “solver” section of an HP
calculator, as three separate formulas. They look like this:
The “SOH:,” "CAH:," and "TOA:" are not necessary to run these.
They are merely titles, but they help me remember the sides these
certain trig functions work with. The
calculator recognizes SIN, COS, and TAN as trig functions. When I run
the one for "SOH," for example, the labels “A,” “OPP,” and “HYPOT” will
be displayed in the
screen. So by inputting values for any two of those displayed will
allow me to solve for the
unknown at the press of a button, and I don't have to do do any
re-arranging of the formula nor select an operation. So it works
very well. NOTE: The “/”
is not actually used,
but my keyboard doesn't show the division sign that the calculator uses
– a dash with a dot above and below it -- so I'm using this as a
High is the Kite?
Here's a final one: When I was about 7 years old, a teenage neighbor
was flying a kite and kept adding more string to it. There were several
of us kids there, and he would keep sending us down to the store to buy
more string -- and even some sandwiches later on. (This is a true
story.) Eventually, the kite was so high, we could no longer see it.
When he finally started reeling it in, he used a broom handle to wrap
the string around -- and it took him several hours. While it was at its
highest point, how could we have determined the actual height of the
Suppose that all the balls of string tied together equaled 3,000 feet,
and with an angle of elevation of 61 degrees. What would be the actual
height of the kite?
As we think of this in terms of a right triangle, we can picture the
feet of string as the length of the hypotenuse. Now we just need to
figure the height, which will be the side of the right triangle that is
opposite the angle. Which of the formulas in Soh-Cah-Toa do we need
that deals with the hypotenuse and opposite? We need the “Soh.”
sin(angle) = opposite / hypotenuse
sin(61) = opposite / 3,000
0.874619707139 = opposite / 3,000
Now we re-arrange the formula to determine the opposite:
0.874619707139 * 3,000 = opposite
opposite = 2,623.86 (rounded to the
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