Angle of Elevation and Depression Right Triangle Trig

Angle of Elevation and Depression Right Triangle Trig


BAM! Mr. Tarrou. In this video we are going
to being doing some more right triangle trigonometry, or in other words we are going to be dealing
with sine, cosine, and tangent. SOHCAHTOA. And we are going to be doing these problems
specifically with angles of elevation and depression. And of course I am yelling SOHCAHTOA,
because SOH is sine which is opposite over hypotenuse, CAH which is adjacent over hypotenuse,
and TOA which the tangent of an angle is equal to the opposite side of a right triangle over
its adjacent side. SOH CAH TOA. Ok, so um… we have already been introduced to the basic
sine, cosine, and tangent functions and worked a little bit with this right triangle trigonometry.
We are just going to be doing 3 examples with varying levels of difficulty that deal with
angles of elevation and depression. Now I am making these videos for honors Geometry.
We come back to these videos in PreCalculus. So if you are in a PreCalculus level class,
these examples should be helpful but probably won’t take care of the questions that you
have near the end of your assignment. They can get a bit more complicated that we are
going to do in this video with our three examples. So I have got a couple of buildings here.
I know that one is 80 feet tall. I just made up these numbers. I don’t really know if that
is a good height of a building or not. But, it is what we have. From that building we
have… and we are going to assume that they are on horizontal or level ground in comparison
to each other. We have, you know, this horizontal line. And an angle of elevation is how much
you have to look up at something. Ok, you are elevating your point of view to see something
else. You are looking up. An angle of depression is simply looking down off of horizontal based
on wherever your eye level is at. Right now we are just going to say that we are at the
top of this 80 foot building. If the angle of depression is 32 degrees… so we are going
to focus in on this little angle here being 32 degrees. How far apart are the buildings?
So what I am going to do is
just draw a quick little right triangle here. We know that this building here is 80 feet
tall. Now I have an angle of depression. I am looking from here over to the other building
and I want to find out how far away the building is. Well, that means that I want to find out
how far apart the buildings are so I am looking at the horizontal line right here. That means
that this length is going to be our unknown value of x. Well, I am not going to be able
to solve for x unless I get some more information about this right triangle. Now, I am looking
at this right triangle right here coming from our point of view horizontal to the other
building and hopefully that building is perpendicular or straight up and down. So this is going
to be a 90 degree angle. And, you know, I am going to go ahead and take this 80′ that
really does not appear to be a part of the triangle and of course transpose it to the
other building. That is why I am talking about, as far as the purposes of this example, our
two building… the base of our building are on level ground as well… or the same height.
It gets more complicated if they are not. You might have to use Law of Sine or Law of
Cosine for those kinds of problems. So these two bases of these buildings are level across
each other. Angle of depression is looking down, or angle of elevation is looking up
is always off of horizontal. And that means that this side over here is 80′. Ok, if you
just found my videos I like to emphasis that we are going to be doing again right triangle
trigonometry which means we are going to be dealing with sine, cosine, and tangent. And
as you look at your diagram and you are trying to decide what trig function you are going
to use, I like to first identify what angle we are dealing with. We always name the sides
of our triangle again with one of the acute angles, never with the right angle in our
right triangle. You want to name the sides of the triangle based on the angle we are
going to use. So I want use this angle that is marked off as 32 degrees. So this length
that I have marked off as 80 feet is the opposite side. The side opposite 90 degrees is the
hypotenuse, and the leg that is helping to make the angle that you are going to use in
your problem is the adjacent leg. It is extremely important… It is extremely important for
you to name the sides of the triangle before you start the problem. Because if I were using
this angle here on the bottom of the right hand corner, well this would be the opposite
leg and not the adjacent leg. So the names on those legs are dependent one more time
on the angle that you are going to use in your problem. So now that we have the triangle
properly labelled, we want to decide sine, cosine, or tangent. What do you have and what
do you want? I am going to repeat that through the whole video. What trig function are you
going to use? Well what side do you have and what side do you want? I have the opposite
side and I want the adjacent side. Well opposite and adjacent… SOH, sine is opposite over
hypotenuse. CAH, cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So based
on what I have and what I want, the opposite and adjacent we are going to use the tangent
function. So the tangent of 32 degrees is equal to the opposite over the adjacent side
of 80 over x. You cannot solve for a variable when it is in the denominator of an equation,
never really basically. That is what I like to say when I am teaching. So we need to get
that x out of the denominator. We are going to multiply both sides by x. What I am doing
is this is 80 divided by x. I don’t want that division of x, I don’t want x in the denominator,
so I am doing the inverse of division which is multiplication. And of course that is fine
as long as you do this to both sides of your equation. So now we have x times the tangent
of 32 is equal to 80. I have x times the tangent of 32, so I want to undo that multiplication
now that I have that x out of the denominator. I have it up in the main line of the equation.
I want to undo that multiplication. I am going to do that with division. So I am going to
divide both sides by the tangent of 32. That tan(32) divided by tan(32), anything divided
by itself is one. So this is just simply x. This is equal to… This is going to have
to come out of your calculator. I have to refer to my notes for that. Since this is
my first example and I am going kind of slow. I understand that, I am hopefully helping
you and not boring you here. Whenever you touch the sine, the cosine, or the tangent
button on your calculator you need to make sure that it is in the correct mode (degree
here). Now if you are just in Geometry. Like… That is who I am doing this video for, my
Geometry students. Your calculator is probably already going to be in degree mode. But if
you are in Precalculus or Trigonometry you switch back and forth between radian and degrees.
A radian is another way of measuring rotation. So you want to make sure that if you touch
the sine, cosine, or tangent button that your calculator is in the proper degree mode. Otherwise
it does not make any difference for the purposes of these questions. This is in degrees and
I want to make sure that my calculator is in degree mode. Enough of that talking. 80
divided by the tangent of 32 degrees is approximately 128 feet. Now I say approximate because the
tangent of 32, when you take the tangent of almost any… actually any trig function of
almost any angle you get an huge decimal. It just runs on forever… not repeating or
ending. So there is almost always some level of rounding off. Because my problem gave me
a measurement that is just simply 80 feet. It is not super specific. I don’t know if
it is really 80.02 or something like that in feet. It is just rounded off to 80 feet,
so I rounded this off two 128 feet. Ok, so that is how far apart the buildings are. Let’s
just put that right there. Now coming over to part to B. I will start speeding up these
examples. If the angle of elevation is 23 degrees, how tall is the building on the right?
Now what I want to do is focus on
the fact that we are looking up. Our angle of elevation is 23 degrees. I want to find
out how tall this building is on the right. Well I already know that this length is 80
feet from over here. All I have to do is find out how tall this piece of the building is.
Let’s just call that y. I don’t think I have used that variable yet. We are going to find
that length and simply add these two lengths and then we will know how tall the building
is. So I am going to have to use this information… the fact that the buildings are 128 feet apart.
That little hatch mark stands for feet. If there were two of them, it would be inches.
Now we are going to solve for y. Now again, what do you have and what do you want? Well
if I am using this angle of 23 degrees, my angle of elevation, then this is the adjacent
side, this is the opposite, opposite the 90 degree angle is always the hypotenuse. And
I have the adjacent, I want the opposite… adjacent… opposite… opposite over adjacent.
That is tangent again. So the tangent of 23 degrees is equal to y over 128. Now with this
one my variable is in the numerator, so it is going to be a little bit less work to figure
it out. All I have to do is multiply both sides by 128. Again I want to undo that division.
That multiplication and division cancels out. We get y is equal to 128 times the tangent
of 23. Now you are going to put it in the calculator…. depends the kind of calculator
you have. You want to do the tangent of 23 degrees, that is size of the angle. Find that
answer first and then multiply it by 128. If you have a two line scientific calculator,
or maybe a graphing calculator, where you can see the numbers and variables and all
that… well not variables… but the math functions as you type them into the calculator,
you can just do 128 times tangent of 23. It will give you the right answer. What you don’t
want to do, is do 128 times 23 and then take the tangent. That would be very very wrong.
The tangent is applied to just the angle measurement, and the angle measurement is 23 degrees. So
128 times tan(23) is approximately 54.3 feet. I have a feeling that… I only rounded it
off to the tenths. I have a feeling this must have been… I don’t remember exactly what
I got out of my calculator, but it must have been really close to 128 because I am just
using my notes here. If your teacher is really stressing the significant digit idea, really
you are not supposed to represent any calculated values to be any more accurate than your measured
values. So if we wanted to be super strict on this, your teacher might you want to be
rounding this off to 54 feet to have it appear to be no more accurate than the given measurement
of 80′. I am going to just keep the decimal though and not worry about that. The overall
building then, if this is 54.3 feet then of course we have the two parts of the building
and the overall building is going to be… well… 54.3 plus 80. And, 8 plus 5 is 13,
so 50 plus 80 is going to be 134.3. Excellent. Ok, now we are going to stop and read this.
We will set up the diagram then I am going to have to erase the chalk board because I
am running out of room. Somebody in a hot air balloon 1300′ in the air needs to look
down, that would be an angle of depression, 6.5 degrees to see you. How far are you from
the hot air balloon? Very common with application problems as you get into the harder ones,
they will not give you the diagram. They will just give you word description of the problem
and you have to come up with that picture yourself. So um… I think it is pretty obvious
that we have some ground. Ta da. You are down here… chilling… kind of skinny. And there
is a hot air balloon. A little basket. Not very well drawn… but eh. Good enough. And
that hot air balloon you are told is 1300 feet in the air. Now, the way that you measure
the distance between a point and a line, the line being the horizontal ground that we are
standing on, if we were going to be assumed to be standing on a hill ourselves and looking
up, or having someone in a hot air balloon looking down, that would not be a right triangle
and we would need to use Law of Sine or Law of Cosine to solve the problem. We are just
doing right triangles here. That person is looking down at you. So for them to see you
from the basket, they are having to look down… an angle of depression of 6.5 degrees… that
is 6.5 degrees must be them up in the hot air balloon like the problem says looking
down at you. And ignoring some problems with my drawing, we are assuming again because
we are doing right triangle trigonometry that we are standing on level ground. Angles of
elevation and depression are always off the horizontal as well, and what that means is…
We have 6.5 degrees in our diagram. Well that 6.5 degrees is not inside of our triangle
where we can use it. But, if this ground is to be assumed to be horizontal, angle of depression
is off of horizontal, that means that these two line segments are parallel. And once again,
if you have parallel lines being cut by a transversal that z shape… which we can often
find… is one way of identifying alternate interior angles. And alternate interior angles,
when two parallel lines are cut by a transversal… those alternate interior angles are going
to be equal. So that means that this 6.5 degrees can be brought down into the triangle and
we can use it. Ok, now before I erase the rest of this board, how far are you from the
hot air balloon? That is little bit vague on the wording, so let me clarify. I want
to use the… basically I just don’t want to use the tangent function again! So I want
to be directly… not if you were to walk over, how far would it be for you to be directly
underneath the balloon. But specifically, if you were to pull a string, how far are
you away from that person in the hot air balloon? So we are going to be finding out this value
here of x which is going to be the hypotenuse of the right triangle. So as I pause the video,
why don’t you get these sides named and even try to do the problem yourself. Because this
will be done very quickly once I get the board erased. BAM! Thanks for watching. Ta da. Whatever.
I got the chalk board erased and I have labelled the sides. So again, based on what do you
have and what do you want? Based on the 6.5 degree angle, we have the opposite leg and
we are looking for the hypotenuse. Opposite… hypotenuse… SOHCAHTOA… that would be the
sine function. So the sine of our angle measure which is 6.5 degrees is equal to the opposite
side of 1300 over your hypotenuse of x. Again you cannot solve for a variable that is in
the denominator. Now… um… I prefer to just multiply both sides by x. You can go
ahead and pop this over 1, it is going to do the same thing anyway, and show… You
can write this as a fraction and show that there is one fraction on each side of the
equation and cross multiply. Which is fine as well. But at any rate, the x is going to
come out of the denominator by doing the inverse of that math operation. So we are going to
get x times the sine of 6.5 degrees is equal to 1300 feet. We are going to divide both
sides by sin(6.5). We get x is equal to… right off the top of my head… if I can find
it… it is on the next page, 1300 divided by sin(6.5) is 11,483.8 feet. That is approximate.
I think we are pretty safe assuming that hot air balloon is not exactly 1300 feet, so it
is probably rounded off to the nearest foot. So we would say that this was approximately
11484 feet. You know, rounding that up of course because of this 8 which is 5 or bigger.
Alright, I have one more example before we are done with the video. WHOOOOOW! Last example.
Right here. Now if you are in Algebra 2 honors or PreCalculus you are going to have problems
with this similar looking diagram but based on… like what parts of this diagram you
are looking for can be much more work than what we are going to do in this particular
problem. I have videos or examples like that in my PreCalculus videos. What we are going
to do though is for the first time in this video, we will actually find an angle of a
right triangle. Then we are going to find this little piece here, a length x between
this angle and this angle. Ok, so the first thing we want to do is find the angle of elevation.
Specifically of a. I have two angles of elevation here. I have… If I am way back here, I can
see my angle of elevation from the level ground to 207 feet above is 41 degrees. But I want
to find this angle of elevation here, angle ‘a’. So if I am going to do that, I am going
to have to be able to determine what type of trig function I want to use. That means
that I want to with this inside triangle, identify that this is my adjacent leg, my
opposite leg, and my hypotenuse. So again, our examples here having us using tangent.
From ‘a’ I have the opposite side, I also have the adjacent side, but unlike the previous
problem where I had the angle measure this time I am going to be looking for it. So I
am going to use a tangent function to find angle ‘a’ because of the two sides that I
have in this particular triangle. If I gave you this length of of 207 and I have you this
length up here, where you had the opposite and had the hypotenuse… well then you would
be using the sine function because the sine of an angle is opposite over hypotenuse. We
have opposite and adjacent, so the tangent of ‘a’ is going to be equal to… Remember
that you put an angle measure into a trig function and you get out the sides of the
triangle. That is going to be opposite over adjacent which is 207 over 98. Now, I can’t
do all of this in my head. So, 207 divided by 98 is 2.112. So the tan(a)=2.112. Now
I want to get that tangent function away from that ‘a’ so I can solve for that unknown value.
Well, it is a tangent function, right. So whenever you want to undo a function, you
must apply its inverse function. In a previous scene where I was trying to get say x out
of the denominator, it was something divided by x so I had to undo that division operation…
math operation… by doing the inverse function which is multiplying. If I want to get the
tangent away from ‘a’, I can’t divide both sides by tangent. That would not make any
sense because tangent itself is a math function. I want to do the inverse math function. That
means ‘a’ is going to be equal to the inverse tangent of 2.112. So this little negative
one on your calculator, if you studied… I am sure you have, negative exponents you
know that something like 3^-2 is equal to 1/3^2. That becomes 1/9. Well when you talk
about functions, this little -1 in the exponent does not mean that I want to find the reciprocal.
It is an inverse math operation. So where as a regular trig function you put in an angle
measure and get out the sides of a triangle, with an inverse trig function you are going
to put in the sides of the triangle and get out the angle measure… instead of what you
normally do which is you put in the angle and get out the sides of the triangle. So
the inverse tangent of 2.112 is 64.7 degrees… approximately. Ok, now that makes sense from
our diagram because if I am here and I have to look up 61 or 64.7 degrees… you know…
as I get closer to a building I have to look up and up and up more and more and more. But
as I a move back the building… it appears to be smaller right. Because, it is going
to seem smaller as you move away. So, if I step farther back to look up at this top which
is 207 feet above the ground, I am not going to have to physically strain my neck as much
to look up. So this angle elevation should be less than this angle of elevation here,
and indeed it is. Of course 41 is less than 64.7. So I am just trying to give you some
way of… uh… you know identifying if this number is at least reasonable. Now I want
to find out what x is. I am not going to need this 64.7. These problems are kind of independent
of each other. I want to find out what x is, well x is only a piece of this bottom side.
It is only a piece of this leg. Now, that means that… you know… we do have this
obtuse triangle out here on the left. In trigonometry and precalculus you will learn something call
law of sine and law and cosine that allows you to deal with non-right triangles. But
we are only in Geometry Honors, or at least that is what I am doing this for, so we don’t
know that yet. What that means is if I am going to identify what the measurement or
the length of x is, I am going still have to work with this large right triangle. We
are just going to find the length of this entire leg and we will call that y. We are
going to find that entire length and then simply subtract out the 98 feet. Well, it
is …. the same design here. From the 41 this is the opposite side and this is the
adjacent side. So again, what trig function do you use? I guess I made too many examples
using tangent, but what do you have and what do you want? You have… you want… opposite…
adjacent… tangent. So the tangent of our angle measure of 41 degrees is equal to opposite
which is 207 over the adjacent side of y. Multiply both sides by y to get it out of
the denominator again. Divide both sides by the tangent of 41. And y is equal to 207 divided
by tangent of 41 is 238.1 feet. That again is approximate. So if the entire leg is 238.1,
then x… Make sure you answer the question. We are only looking for this small segment
here. So x is going to be equal to 238.1 minus 98. Ok, so the subtraction thing without being
completely dependent on our calculator. Well clearly the 8’s are going to cancel out. 8-8
is 0. 230 minus 90, is going to be 140. So we have 140.1 units. Whether this is feet,
miles, inches… I don’t know. But I do know that this is the end of my last example. So
I am Mr. Tarrou. BAM! Go Do Your Homework:D

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