# 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

is the sound in the beginning your dog? 😛 great video by the way 🙂

Yes it was…nice catch! If it had been around christmas time I would have said it was santa paws:))

Thank you by the way too!

hahaha funny 😉 and no problem! 😀

I wish you were my teacher! When I did this in class last week I didn't understand any of it but then I watch this and it was so simple and made total sense! THANK YOU SO MUCH!

You are so welcome! Now don't keep my channel a secret…go out there and share it with all those other students in your school!…then if there is a big group of you, you can all get together in one room and watch my videos and it will be just like I "was" your teacher:D

really loved the way you explained all of it, even though I'm a Filipino, I understand it crystal clear… it helped me very much

From now on every time I don't get what my teacher is teaching, I'll watch it from you =)))

That's awesome to hear! I'm glad my videos are able to pick up where your teacher might not have enough time to explain:) Please share my channel with your other classmates and friends, cause that's how I groW!

i wish my teacher would be like you proff!!love the way you explained!! c:)

You just keep watching my videos and sharing with your friends and before you know it…it WILL be just like I was your teacher:D Glad to see you re still watching too:)

You are the most AWESOME teacher i ever had…. Good Job Prof.!

Thanks Jeff, I think that's the first time I've heard that from a YouTube student…usually they say "I wish I had you as my teacher" I know there is no live interaction but I do try to teach in my videos the same way I teach in my classroom:) Thanks for being such a great student! Please be sure to share my channel with others and help my continued efforts to groW!

Thank you ProfRobBob. I have to take a test for Precal today and my professor said this was going to be on the test even though he didn't teach the class this. This was very helpful and easy to understand.

You're welcome candysell! I hope you went into class and passed that test like BAM!!!

my trig teacher likes to wait until the last five minutes of class to try and explain to us how to do things, this helped me out SO MUCH! thank you!

You're welcome…thanks for watching:)

very helful!

Even though I go to a high school – I felt like homeschooled when I watched your video. It's more understanding because no distractions.. it's just yourself 😀

My Geometry Teacher isn't even half as good as you!

hummm… where the 23 came from for part b?? i understand that angles of depression and elevation are equal… confused!!!

Great Video it really helped

=)

Can't thank you enough. So grateful for your videos, trig just became easier 🙂

thanks so mush for this tutorial you brought me to the next

thank you so much u made the complex thing in math so simple. my teacher cant explain. You remind of my uncle, he has his master in math and explain everything in the simplest formation.

This guy is amazing!

you are so great than my prof. in trigo.. hahaha thank you very much Godbless =)

You are such an amazing teacher!! Thank you so much for all the helpful videos!

wait i'm in high school but can i still watch this??? is this for college level>?

YOU DA' MAN!

Nice shirt!

I always look forward to that Bam in the beginning.

ProfRobBob , you saved my teacher's life , I almost killed her. THanks a lot

thank you

Thank you so much mr. robdob, I clearly understand your tutorial than my Professor even though hes using our language.

THANK YOU VERY MUCH. I CLEARLY UNDERSTAND 🙂

I followed how you did it, but im still messing up somewhere 🙁

thanks this really helped for my finals in IB

Thank you so much!!! We have a test tomorrow in my Honors Geometry class and I was so lost before I came here! I will definitely be sharing this link with my friends so they can feel as good as I do about tomorrow's test 🙂

You only helped my right ear

AWESOME!! now i know how to solve.. you made it easier… good job! 😀

Hi.. this video was so helpful…You are a genius! omg Your handwriting <3 . I have my mock igcse exam tomorrow and a question in my book has annoyed me a lot. It is very hard. Could you please help me!!! So here it goes 🙂

Consider that u have been fortunate enough to be able to get into a perfect position to observe a full solar eclipse. Using the information provided below, and identifying any assumptions that you need to make, calculate how long it will take for the moon to pass fully across the sun?

Diameter of sun -1.392 x 10^6

Diameter of the moon-3.5 x 10^3

Earth to sun-1,5 x 10^8

Earth to moon- 3.84 x 10^5

Hint: Begin by drawing a diagram to represent the problem

I love your classes and it is helping me tremendously in my Math class! I have a good teacher at college but he does not explain things in detail as you do. I enjoy your seeing you are full of energy and funny as well!

You are a lifesaver. This'll definitely help me on my test tomorrow. Thanks!

pro ! keep it up! awesome!!!!!!!!!!!!!!!

u're welcomed

Finally I understand this thank you so much you are a life saver 😅

This greatly helped me understand this lesson. I hope I'll be able to answer the problems for our graded recitation. 🙂

Great vids! Way to go!!

cool professor!

Thank you 😀

THANK YOU SO MUCH!! you're a great teacher.

thanks :)) your video included different views mostly encountered in a problem :))) great help! ^^

Really helpful!

thank you!

i learn more in youtube than in school 😂 thank you sir!

very helpful thank you

Dude thank you so much I'm taking a geometry EOI tomorrow a randomly found your channel and it's really helping me out, keep it up!!

also side note instead of multiplying each side by X to get rid of it could I cross multiply just put lets say sin(40) over 1 and 70 over X?

THANKS DID EXAM

A+

best guy for finals week

u sir just saved my ass from my exam

In the last problem at 18:57, you mentioned you would do a more in dept video on those types of problems. Forgive me for not looking too thoroughly, but do you have a video link ? Thank you very much for the video. It helped clear up things from my class 😀

Edit; Nevermind, I found it in the related video tab !

You are the best, using your videos to help my sister understand trig. Thanks so much.

Your videos are SO helpful. you cover every single concept perfectly in a short amount of time, but also give examples and the solutions to them. Just about an hour ago I felt unconfident about my math test tomorrow but now I feel like I'm ready! Thank you I can finally go to sleep