# Applications of Law of Sines and Cosines

BAM!!! Mr. Tarrou. Ok, I am going to do three
examples of involving triangles that require the use of Law of Sine or Cosine. And uh,
these are written on the chalk board, so I am not going to write out an entire word problem.
But, I am going to give you some setups that you might recognize as you read the word problems
in your trig or precalculus book that involve some applications. So, for are first example
let’s say that we have something leaning like a tree, a pole, the leaning tower of Pisa.
We want to find the length of that item. So we have something leaning, whatever it is
and we would like to know how long that is. We will call this a. You are able to go up
to it, touch it, and take some measurements and then from those measurements estimate
or calculate the length of that object. So you come up to it and you put a protractor
on it or some kind of instrument that allows you to measure the angle of incline because
this is not plum or square. That incline let’s just say comes out to be seven degrees. So
from there, you come out and you walk a certain length, or measure out a certain length, and
let’s just say that it comes out to be thirty feet. Single mark is for feet. You take tool
that you are using to measure these with and the incline or the degrees from this point
up to the top of this item you are interested in measuring comes out to be…trying to keep
my lines a little straight here…comes out to be 68 degrees. Now, from just those measurements
we would like to find the length of this item. Well, what do we have? Not enough to just
go, eh..it is 50 feet. We don’t have enough measurements in the triangle to calculate
this length at this point. But, if the ground is level and from that level ground we have
measured a perpendicular line, and the angle from perpendicular is seven degrees, then
if this again is perpendicular then this is 90 degrees. That ninety degrees is made up
of two separate angles. One being seven degrees that you were able to measure. Or, if you
could measure seven degrees, you could probably measure this degree but hey it is a textbook!
It is what was given. So if this is ninety, this is seven, then this has to be 83 degrees.
We have either the Las of Sine or the Law of Cosine at our disposal to try and solve
this problem. Well don’t forget to use the Law of Cosine, you have to have two sides
and an included angle. For the Law of Sine you need to have a matching pair of opposite
side and angle setup. So, since we know that this is… Do we have enough? Let’s see here.
We don’t. We don’t know two sides and and an included angle. We have 83 degrees and
we don’t know it’s opposing side. We have 68 degrees and we don’t know the length of
that side. That is what the problem is about. Now we have thirty feet. That is the only
side measurement that we have. So, to be able to use that 30 feet and be able to use the
Law of Sine to try and figure this out… Again the Law of Cosine needs two sides and
an included angle and we only know one side so there is no way for us set up the Law of
Cosine for this problem. We need to know the opposing angle and now that we have two angles
that are inside the triangle, we can find the third angle up here. That is going to
be 180 minus 68 minus 83 degrees. What is that? Let’s see here. 80 and 60 makes 140.
Now we have 143 and now we have 151 right? So 180 minus 151 is going to be 29 degrees.
Ok, so now we have that side angle pairing that we need to be able to set up the Law
of Sine. So, we have thirty over the sine of twenty-nine degrees is equal, to looking
for this length of a, so that is going to be a over the sine of 68 degrees. We want
to get a alone so we know that length, right. That means that we want to undo this division
of the sine of sixty-eight, so we are going to multiply both sides by the sine of sixty-eight.
We get thirty times the sine of sixty-eight over the sine of twenty-nine degrees and that
is equal to a. Because I cannot do that off the top of my head, that is going to be approximately
57.4 feet or units long. Just double checking to make sure that I have got my numbers right
here. BAM! There you go. So, we had a leaning object of a tree, tower, pole, leaning tower
of Pisa. We have an angle of incline. We also had a length that we measured away from the
base of the object of thirty feet. We were also given the sixty-eight degree angle of
incline from this point up to the top of our item. From all of that information we were
able to figure out what the length of that was through the Law of Sine. So, BAM there
you go. Let’s do another example. We are going to do another measurement of item here. But
this time we are going to have less contact with whatever that item is that we are measuring.
So, there is some object here. Maybe it is a mountain, maybe it is a tree…I don’t know
why you couldn’t walk up to a tree and start taking some measurements like the previous
problem, but there are some item here that we want to measure the height of and we cannot
make any physical contact with that item to take any measurements. Like if was a mountain,
you can’t dig your way to the base of the mountain and measure straight up or something
like that. This item we want the height of. So you are going to come out here. Just some
random point and you are going to look up at it and you take a measurement. That angle
of incline that you are looking at is forty-two degrees. That is the only measurement that
you are able to take now because you cannot get to the base and do any kind of leaning
angles or whatever. Ok, so you step back some distance and we are going to step back 100
feet and take another measurement of incline. Back a hundred feet, of course you are back
farther so you are not having to look so sharply up, you take another angle measurement and
that comes out to be thirty-four degrees. Now from just those three measurement, the
angle of elevation, back a hundred feet, and another angle of incline we are supposed to
be able figure out the height of that that object. Are you serious? It seems so obtuse,
or sort of just out there. How are we going to find that measurement of x? It is going
to take some work, I can tell you that! hmm.. Ok, well let’s see. If this is forty-two degrees,
then we know that this angle in here is equal to… Well, they are two angles forming a
straight line so they are supplementary angles or a linear pair. 180 minus 42, 180 minus
forty is now 140, now it is 138 degrees. So that first angle that I found is that angle
that is adjacent to the first angle of incline measurement of 42 degrees and we get 138 degrees.
Now what I am trying to do is, we have this one obtuse triangle. I am trying to fill in
the parts enough to where I can find/calculate some more measurements. So this is 42 and
this is 138 degrees. Now I have got this triangle which only has three angles and now we know
two of them. The 34 degrees that we took a physical measurement of and the 138 based
off of the 42 degrees. So this angle up here is 180 minus 34 minus 138. What do we have
here? We have got 138, now it is 168, and now it is 172. So, 180 minus 172 is eight
degrees. Just because I am paranoid because I don’t want to make a mistake in another
video, yes that is eight degrees. Ok, so we found this angle of 138 and this angle of
eight degrees. Now what can we find? We are not going to get x directly. We don’t have
enough information yet. But, we have this side of a hundred and we know the angle that
is opposite of that 100 feet of eight degrees. So if you have that side and angle pairing,
you have enough information to set up the Law of Sine. There ambiguous cases of the
Law of Sine where you might have more than one possible triangle from the information
given. But we have an angle-side-angle and that is one of the congruency theorems or
similarity theorems from Geometry, so when we get our answer from our calculator that
is going to be the one valid answer that there is so we will be safe. So a hundred over the
sine of eight degrees is equal to…. What we are going to find is this side right here
that is opposite 34 degrees, let’s call that a. I don’t really necessarily care about the
length of a for my final answer, but I need that value to build up enough information
to find out what x is. So, we have a over the sine of 34 degrees. Ok, well it is a divided
by the sine of 34 and I want a alone. I want to undo that division, so I am going to multiply
both sides by the sine of thirty-four. Get that typed into your calculator correctly
and you will get the answer of 401.8 feet. Ok, now I am going to just round these values
off to the nearest tenths just for convenience and what comes out of the calculator. If you
are in a science class talking about significant digits you will know more than me at this
point because I miss-speak a bit! My measurement here of 100 feet and there is no decimal at
the end of that implies that I have rounded off my measurements, my actual physical measurements
to the nearest ones place or whole number value. You cannot really have a calculated
value that represents a measurement more accurate than your real physical measurements. So,
you could argue that I really need to round this off to 402 feet, round this off to the
ones place like I did the 100 here as it was presented. But, I am going to leave it at
401.8 just as a math exercise. Your science teacher may have another opinion about that,
ok, and physics teacher and such. So a comes out to be 401.8. Now we finally have enough
information to find out the value of x, the height of that object whatever it is in your
word problem that you are reading in your application. What are we looking at here?
We are looking at a right triangle. When you have a right triangle, I guess I should have
said that we are going to be measuring vertically or perpendicular from the base. When you have
a right triangle you can go back to SOHCAHTOA. Sine is opposite over hypotenuse, cosine is
adjacent over hypotenuse, and tangent is opposite over adjacent. So, from the forty-two we want
the opposite side from 42 and we have the hypotenuse. When you want the opposite and
you have the hypotenuse that is going to require the sine function. So, the sine of forty-two
degrees is equal to x over 401.8. Again, we want to undo that division, that fraction
bar by multiplying both sides by 401.8. And x comes out to be 268.9 units. I have already
marked off feet, so BAM there you go:D That is the height of this item that we were able
to calculator without ever physically approaching it, touching it, taking any angle measures
off of it. We just stood out here and we looked up, we stepped back and looked up again, and
that was enough information to find the height of whatever that item may be…mountain, tree,
whatever your application problem is about. One more example. ALRIGHTY THEN!!! Next page
of notes. You travel east thirty miles, you then head south twenty-five degrees east
for seventeen miles. Ok, the first question
I want to ask based on that information is how far are you from where you started? That
is a pretty basic question. Well, we need to take these words and actually make some
kind of diagram out of them. So, you travel east thirty miles from where? I don’t know,
there! So we travel east
and we do that for thirty miles. We then we are going to stop here if you will and look
at our compass and we are going to turn to S 25 E. Well north, south, east, west. So
from due south we are going to turn twenty-five degrees east. From south we are going to favor
the eastern direction by 25 degrees. I think I drew that a little bit too small, but that
is that little part to the right of the vertical line. So we are to the right of south, or
east of south. We travel seventeen miles so my line is a little bit long here. Maybe we
will just make that longer. Ok, so thirty miles east, seventeen miles south twenty-five
degrees east. We want to know how far are we from where we started? So that is what
we are looking for. Do we have enough information to figure that out? Since the original measurement
was directly east and north-south is perpendicular to that this could be a worse problem, or
a more difficult problem, but traveling east like this doesn’t make it too terrible difficult.
If this is east and this is a north-south line then this part of the diagram is ninety
degrees. Ninety degrees and an additional twenty-five degrees, from south 25 degrees
east, well those add up to what? Ninety and twenty-five is 115 degrees. Ok, well if you
see what we have here? We have two sides and an included angle. So you might be thinking
hey, you say we are doing Law of Sine and Law of Cosine and I never did write the formulas
up, did I? Sorry about that. How far are we from where we started? Well with two sides
and an included angle, that is Law of Cosine. So x squared, the side that we don’t know,
is equal to the other two sides squared and added together
minus two times those other two sides times
the cosine of the included angle. I should have reminded you this in the previous two
examples, but if you are in precalc and trig you have been bouncing your calculators back
and forth between degrees and radians. Make sure that you are in degree mode. Otherwise,
you are going to get all of these answers incorrect. X squared is equal to, when you
get all of this typed into the calculator correctly 1620.04. You now square root both
sides of the equation and we get x is equal to approximately 40.25 miles. If you don’t
want a calculated measurement that is more accurate than your physical measurements,
then it is approximately 40 miles from where you started. Good. So that was not too bad.
Now I have another question. Remember you are here because you traveled from home, or
port, or whatever. I want the bearing from here to get you back home. Ok. What direction
would you take to get from this point back. I don’t want to go back in the same path,
I want to go straight there. I only want to travel 40 miles and not 47 miles. Well, if
I want a bearing. This is a north-south line. If I want a bearing from there back home,
let’s see. This part that I definitely should have drawn a bit bigger. Let me get the 17
out of the way. That little angle right there, how many degrees is that? Well, every time
you draw a north-south line when you talk about bearing or directions like when you
are reading a compass, every time you draw a north-south line those are all parallel
right. It is north-south. So, we have two parallel lines intersected by a transversal.
How long has it been since you heard those words, right? When you have alternate interior
angles, those angles are equal. So this little angle here that I have now drawn is also 25
degrees. So, if I can find this angle, then I can add it with twenty-five and be able
to say that to get back home my bearing is north blank, and then we are going to find
how many degrees there are to the west. So, what do we have? I don’t need this x in here
any more because I know what the measurement is. It is 40.25 miles. How are we going to
use that to now find this measurement in here of angle y? Well, I have got this side is
30 and the opposing angle is what I am looking for. I have this length of 40.25 and I know
it’s opposing angle. Again, any time you have a setup where you know an angle and it’s opposite
side, then you have the ability to set up the Law of Sine. Don’t forget, do not try
and find an angle opposite the largest side with the Law of Sine. If that angle ends up
being obtuse, you will be able to get the answer from the Law of Sine because your calculator
will not be able to give you that answer. Inverse sine with your calculator only gives
you answers between negative ninety and positive ninety degrees. Ok, so a little cautionary
note there. Let’s finish this problem and call it a day. So, we have 40.25 over the
sine of 115 degrees equals 30 over the sine of y. I hope I used the same setup as I did
in my notes here. Yep, I did. Our variable is in the denominator and we cannot solve
for a variable when it is in the denominator. I am going to cross multiply. There is only
one fraction on each side. We are going to have 40.25 times the sine of y equals 30 times
the sine of 115. Sine of y. We want to get the 40.25 away so we divide both sides by
40.25. The sine of y is equal to thirty times the sine of 115 over 40.25. That comes out
to be .676. Now we have the sine of y equals that .676. Y is equal to the inverse sine
of .676. That comes out to be 42.5 degrees. That is not your answer! That is this little
piece in here. Well it is not so tiny. This is y which is part of the bearing value, right?
So we are going to head, to get from here back home, it is going to be north 25 plus
42.5 west. So it is going to be north 67.5 degrees west. Let me just double check that.
Close enough. I am Mr. Tarrou. BAM!!! Go Do Your Homework:) Thank you for watching. I
do appreciate it very very much.

## 100 thoughts on “Applications of Law of Sines and Cosines”

• ### ProfRobBob Post author

That is what I like to hear:D/

• ### MongTran Tran Post author

When it turn south.Is it possible if 180-25 ?.Sometimes i confused.Thanks teacher.

• ### ProfRobBob Post author

? I am not sure what you mean…sorry. 180-25 is 165 degrees. For bearings that would be 25 degrees East of South.

• ### ProfRobBob Post author

Thank you so much!!! I hope continue to be helpful for you for a long time:D

thank you so much, you made it much easier for me to do my homework! subscribed! 😀

• ### ProfRobBob Post author

Thank you all the way from Canada for subscribing and watching my channel! I hope I can continue to assist in making your homework easier to complete. Keep watching and please share my channel with your friends:)

hey, i'm from canada too! 🙂 please include more application videos as those are the ones the trouble me and i will gladly share your channel to my friends 🙂

• ### ProfRobBob Post author

Sorry for the confusion…I was thanking YOU in Canada…I am in Florida:) Thank you again for all the support!

oh.. my bad haha.. and thanks to you, i got 100% on my math test! you're the best! 😀

• ### ProfRobBob Post author

BAM!!!!!….that's what I'M TALKIN about!!! Giant CONGRATS to you:D

and sorry I don't have any other notes on application ready to film:( But please keep watching!

thanks 😀 and aww 🙁 it's alright, ill watch the rest of your videos cause i have a math exam coming up haha and it's kinda weird that there is an add for a math tutor IN a math tutorial video… haha

it sucks how they can't choose which ads to put in for what videos because some people may use that site instead of watching your AWESOME videos, but i guess it is their loss 😛 hahaha i'll try 😉

• ### ProfRobBob Post author

I think there is a way to block individual advertisers but I have not done any real studying into that yet. I am learning as I go:) I am so excited you like my videos so much. The biggest problem for me is at times I find it difficult to talk, write, and think at the same time. I learned today I need to make a couple of small corrections:( it is frustrating and a bit embarrassing to try and help but make mistakes myself for the whole world to see! This is a work in progress:D

• ### ProfRobBob Post author

Since the ads are what support my channel, I allow them to run…but as you can see, they have a captive math audience and those are the ads they run…and yes, I don't like seeing his face on MY channel any more that you do but what they choose is out of my hands. Anyway, keep up﻿ the GREAT grades and keep watching me…haha:)

if you want, ill do it for you 😉 hahaha and don't worry about mistakes, pratice makes perfect right? 😀 but to be honest, i hope your channel does well in the future and ill continue to watch your videos even when im not in the mood to do math hahaha

• ### Yu Jeff Post author

if only u were my teacher 🙂

• ### Yu Jeff Post author

im just having trouble with making word problems(3D) into a picture

• ### ProfRobBob Post author

I wish I was your teacher too:) I hope I can be your YouTube teacher for as many topics as possible. I also hope you do great in your class!!!

• ### ProfRobBob Post author

I don't do 3D Problems in my PreCalc class. I hope you can figure out the drawings…they are a challenge even in 2D.

• ### Ezra Grant Post author

The lines, they are so strait!

• ### Yu Jeff Post author

i have a quetion: From an airplane,the angles of depression to two forest fires measure 18 and 35 degrees. One fire is on a heading of N15degreesW. the other fire is on a heading of S70degreesE. what are headings?

• ### ProfRobBob Post author

HAHA… Thanks. My drawing usually impresses my Geometry students this year too:) Thank you for watching as well.

• ### ProfRobBob Post author

In the level of textbook I teach, I have done problems that sound similar as far as the headings are concerned but not with the added complexity of the angle of depression from the plane. Sorry…I will not be much help here.

• ### ProfRobBob Post author

I can't draw just anything, but I have gotten good at geometry diagrams through my years of teaching:D Thanks and thank you for watching.

• ### Yu Jeff Post author

@profrobbob i understand i am in foundations 11 of course

• ### Yu Jeff Post author

@profrobbob but could you explain to me what a heading looks like?

• ### Lissette Sosa Post author

Save the drama for your mama xD
P.S. This was really helpful… Thanks!

• ### Alorie Post author

i have a trig test tomorrow and this just saved my life! THANK YOU

• ### ProfRobBob Post author

Sometimes my students at school need reminding:)…LOL
Your'e welcome…and thanks for choosing my channel to watch!

• ### ProfRobBob Post author

Your'e welcome, thanks for watching:)….and let me know how you did on that trig test!

• ### ProfRobBob Post author

I used a heading…or a bearing…in my third example. Unless your book means headings can only be a degree measure. In that case a heading of 30 degrees would be like North 30 degree East, or a heading of 170 degrees would be like South 10 degree East, or a heading of 330 degrees would be like North 30 degrees West. Heading or Bearing is measured from due North with a clockwise rotation. I hope this is what you needed.

• ### ProfRobBob Post author

Together…Lets make math FUN!

i already thought that math was fun! hahaha 🙂

• ### ProfRobBob Post author

Yipppeeeee….another believer!

there are so few of us though 🙁

• ### Davy Lowman Post author

Thank you for the refresher!!!

• ### ProfRobBob Post author

You're welcome…anytime…& thanks for watching!

• ### ProfRobBob Post author

Thank you for that:)

• ### shane pander Post author

how do i use law of sines with: 44degrees34', and 101degrees46'. please help. thanks. profrobbob is awesome.

• ### ProfRobBob Post author

I'm glad you liked my videos and found them to be so helpful. Thanks for subscribing too, I really appreciate the support and please share with you friends! (sorry for the reply delay, but I have my channel set up to approve all comments:)

• ### ProfRobBob Post author

Are you asking about converting minutes into a decimal to make them easier to enter into your calculator? You can divide the minutes by 60 to find the decimal. You need more that just degrees to do law of sines, you need a length opposite one of the angles. Thank you very much for the support:)

• ### Ian Murphy Post author

You were a lot of fun watching, great at explaining, and proved to be quite helpful in reminding myself of this information. Should help with my pre-calc exam tomorrow!

• ### ProfRobBob Post author

Thank you for saying so:)) Now BAM!…go pass that exam with all your new found knowledge!!!

• ### ProfRobBob Post author

Happy I could help:)

• ### ProfRobBob Post author

And THANK YOU for watching, taking time to comment and liking my video! I hope you will continue to use my videos for extra help and share my channel with your friends:D

• ### Watto Watt Post author

Been following your series, Good job and excellent presentation. One Question though shouldn't your bearing be taken clockwise from the position your at. If so I believe the last answer should be 292.5 degs from the point "y". Just a thought. Thanks again.

• ### ProfRobBob Post author

A bearing of 292.5 is equivalent to N 67.5 W. Thank you very much for watching. I appreciate the support very much and I am glad you like my videos:D

• ### pato spamerr Post author

Thank you so much this helped me more than you can imagine!

• ### ProfRobBob Post author

Happy to help anytime!

• ### onelove2937 Post author

drama mama lol
i still payed attention tho! thanks

• ### ProfRobBob Post author

Somedays at school, I have to remind them…lol
Thanks for paying attention…that is a practice that all teachers appreciate:)

• ### Jonathan Harvey Post author

my teacher said that it was sinA/a=sinB/b=sinC/c but you have it as a/sinA is she incorrect?

• ### ProfRobBob Post author

No…that is just another form of saying the same thing:) Thank you very much for watching and asking for clarification

• ### nmuzaic Post author

you are an exceptional teacher. thank you.

• ### ProfRobBob Post author

You're welcome and THANK YOU for choosing my videos to learn from:)

• ### kkawesome1234 Post author

Really nice job man! Very clear, and loved the enthusiasm! Hopefully i'll go in with confidence and kill that test tomorrow

• ### ProfRobBob Post author

Thanks for liking and supporting by subscribing! And I'll repeat your last line preceded by BAM!!!…go take that test with confidence!!! Then let me hear how you afterwards how you "killed it"!

• ### Mihir Borkar Post author

the power of calculators!!! bless the calculator!!

• ### ProfRobBob Post author

YES…sure does beat using a slide-rule (not that I know how to use one…lol:)

13:06 HAHA!

• ### ProfRobBob Post author

Thanks for laughing all the way from Spain!

• ### yurilegs1 Post author

I wish you were my math prof. You explain waaaaaayy better than her lol

• ### ProfRobBob Post author

THANKS…I'm here 24/7 whenever you want to watch another one of my 435 videos:)

• ### Draco-X-RPG Post author

It's really sad I have to pay my Trig prof when I come to your channel to actually learn how to do and understand the ideas behind Trig. She is that bad and I get more from reading the Trig book than going to her class and if you've ever read a math/trig book you know that is saying something. I would much rather pay you to run my class.

• ### ProfRobBob Post author

How about YOU start a study group with the rest of your class…you can all watch my videos together, work on homework together, then pass the hat and "Tip the Teacher" after you ALL pass the class together:)
Thanks for watching and supporting!

Great vid

• ### Emmanuel Carballido Post author

goddd, I would have loved to have him as my trigonometry professor

• ### Kazim Post author

Could you just break it into vector components and add the x component to the 30 km east and then use the y component and use the pythagorean theorem to solve for x since it becomes a right angle triangle problem then ﻿

• ### rizvif Post author

Hi, great video…I know you talked about sig figs and the 100 has 1 sig fig so you would have to round the 401.8 down to 400 not 402.  402 has 3 sig figs…..

• ### dima71387 Post author

I was reading this trig book with application problems. What it said to do is:
1. tan 42 = b / x  and  tan 34 = 100 + b / x where b is the unknown side
2. Solve for b so, b = tan 42 x  and b = (x / tan 34) – 100
3. Set them equal to each other and solve for x, with little algebra manipulation they come up with x = 100/(cot34 – tan42) or 268'.
Your videos been helpful for me transitioning from military back to college, yet very challenging because there us always another way to solve a problem. I'm almost done with your trig playlist and school didn't start yet. Thank you for your videos and tricks of the trade.

• ### tbrown2963 Post author

Every subscriber should give this man at least \$5 for all his hard work.  Prof Bob, is there a way you can put up a link to tip the teacher?  I appreciate all the hard work you put into these videos.  You are a very talented teacher.  It's one thing to know how to do math, but it is a totally separate gift to be able to explain it as well as you do.

• ### Robert Murphy Post author

Excellent tutorial.  Excellent set up.  Good energy to keep interest.

Solid video.  Helped me a lot!

• ### claire g. Post author

Why is my school making us solve this without a calculator… -_-

• ### santiago pedrini Post author

This guy might be the most helpful source when studying for Pre calculus tests. thanks for the help!

• ### mispetitefashionista Post author

Y can't u be my teacher!

• ### Dylan Geick Post author

Wish you were my teacher ahhhhhhh

• ### BoomPnaes Post author

Sir ProfRobBob why you use cosine 115 it cant be sine? i got confused

• ### JeepJason Post author

this helped me refresh after the long weekend before my final

• ### Monica Duran Post author

Thank you so much. Helped a lot

• ### Jacob Adamczyk Post author

For most questions like these to be valid, wouldn't the terminal side have to be supplementary to the "ground" or the base of the triangle?  Basically, wouldn't you have to be a height of 0 for your angle measures to be accurate?

• ### ProfRobBob Post author

Now Closed Captioned #math

• ### bonesjuones melo Post author

Mr tarou makes math fun!! BAM!! Go do your homework. I wish I was a student in your class sir! You are awesome!!

• ### Flyer Aerospace Post author

I just can't get enough of your truly educational and awesome videos… I am in awe … in a good way 😉

• ### Student Biology Post author

How do you get 68 degrees?

AWESOME!

• ### Toaster Boy Post author

Thank you Mr. Tarrou! I really appreciate your love for math and helping the general people! You have really made a big impact in my studies!

baaaaaaam

• ### aditya vala Post author

why is this so amazing

• ### Soumya Sharma Post author

Professor RobBob, sometimes my trouble is not understanding what the problem is asking- it's worded in such a way to make it seem like you have to solve for one thing, but really they want the measurement of something else. What can I do during exams to make sure I don't solve for the wrong thing?

• ### Brian Golden Post author

Couldn't you also find the opposite side of 138 degree and find x using the Sin90/480.8=sin34/h? Gave me the same answer.

• ### Exodus Post author

I'm not sure if you still check the messages on your videos anymore but I wanted to let you know that as an older student 40+ and without a high school diploma, and being out of school for 20+ years I was able to use most of your videos in math to help me learn and obtain a degree in Computer Forensics. If not for you passion for math and your ability to teach in a way that is easily understood I wouldn't be where I am today. So from someone who was completely lost when I started this journey, to someone who I believe has a decent graphs on his mathematical ability's I want to thank you from the bottom of my heart, you sir rock !! (If only all professors had the ability to educate like you)

• ### Lizzie Lee Post author

On #3, why did you do cos 115? don't you have to do cos 140 bc if you do 115 then the 17 feet side isn't at the right place (it isn't 25 degrees southeast or whatever)?

• ### Aly Cat Post author

Helping me every day, thanks so much!

• ### Shilpa Mishra Post author

Youtube is lucky to have Professor Tarrou as one of their eclectic mathematics patron. The chalk talks are awesome! Thank you for sharing 🙂
Regarding the quote "Save the drama for your mama." I do and so do many like me, having the ultimate assurance that mama is the only who loves and accepts us despite our whimsical dramas 😉

thank you so much for explaining driving problem with the north and south i was have trouble setting those up. i was putting the degree on the wrong side lol! cant wait for tomorrows test!

• ### Silvana Segura Post author

This is so helpful. i love watching his videos, they always make studying not so bad 🙂

• ### Tammi E Post author

It feels like I took algebra 100 years ago! Now I'm back in school & taking trig & I'm totally lost whenever we get to a new chapter. Your videos help tremendously and you make it a fun process! I love loving math! It's really fun when you actually get it:)
Thank you SO much!!!

Thanks