For more information, please see full course syllabus of Trigonometry
For more information, please see full course syllabus of Trigonometry
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Trigonometry in Right Angles
Main formulas:
Master formula for right triangles: SOHCAHTOA!

Example 1:
A right triangle has short sides of length 3 and 4. Find all the angles in the triangle.Example 2:
A right triangle has one angle measuring 40^{°} and opposite side of length 6. Find the lengths of all the sides.Example 3:
The lengths of the two short sides of a right triangle are in a 5:2 ratio. Find all angles of the triangle.Example 4:
A right triangle has short sides of length 3 and hypotenuse of length 7. Find all the angles in the triangle.Example 5:
A right triangle has one angle of 65^{°} and hypotenuse of length 3. Find the lengths of all the sides of the triangle.Trigonometry in Right Angles
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 tanθ = [5/12] ⇒θ = arctan([5/12]) ⇒θ ≈ 22.6^{°} (Make sure your calculator is in degree mode)
 tanϕ = [12/5] ⇒ϕ = arctan([12/5]) ⇒ϕ ≈ 67.4^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
sin33^{°} = [x/7] ⇒ x = 7sin(33^{°}) ⇒ x ≈ 3.8 (Make sure your calculator is in degree mode
cos33^{°} = [7/y] ⇒ y = [7/(cos33^{°})] ⇒ y ≈ 8.3
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 Missing side length: Use Pythagorean Theorem a^{2} + b^{2} = c^{2}
tanθ = [7/4] ⇒θ = arctan([7/4]) ⇒θ≈ 60.3^{°}
sinϕ = [4/(√{65} )] ⇒ϕ = arcsin([4/(√{65} )]) ⇒ ϕ ≈ 29.7^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 Missing side length: Use Pythagorean Theorem a^{2} + b^{2} = c^{2}
sinθ = [4/9] ⇒θ = arcsin([4/9]) ⇒θ ≈ 26.4^{°}
cosϕ = [4/9] ⇒ϕ = arccos([4/9]) ⇒ ϕ ≈ 63.6^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
sin56^{°} = [x/7] ⇒ x = 7sin56^{°} ⇒ x ≈ 5.8
cos56^{°} = [y/7] ⇒ y = 7cos56^{°} ⇒ y ≈ 3.9
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 tanθ = [12/11] ⇒θ = arctan([12/11]) ⇒ θ ≈ 47.5^{°} (Make sure your calculator is in degree mode)
 tanϕ = [11/12] ⇒ϕ = arctan([11/12]) ⇒ϕ ≈ 42.5^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
sin27^{°} = [4/y] ⇒ y = [4/(sin27^{°})] ⇒ y ≈ 8.8 (Make sure your calculator is in degree mode)
tan27^{°} = [4/x] ⇒ x = [4/(tan27^{°})] ⇒ x ≈ 7.9
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 Missing side length: Use Pythagorean Theorem a^{2} + b^{2} = c^{2}
tanθ = [6/3] ⇒θ = arctan([6/3]) ⇒θ ≈ 63.4^{°}
tanϕ = [3/6] ⇒ϕ = arctan([3/6]) ⇒ϕ ≈ 26.6^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
 Missing side length: Use Pythagorean Theorem a^{2} + b^{2} = c^{2}
sinθ = [7/11] ⇒θ = arcsin([7/11]) ⇒θ ≈ 39.5^{°}
cosϕ = [7/11] ⇒ϕ = arccos([7/11]) ⇒ϕ ≈ 50.5^{°}
 Start by drawing a right triangle with labeled side lengths and angles
 Recall SOHCAHTOA, sin x = [Opposite/Hypotenuse], cos x = [Adjacent/Hypotenuse], tan x = [Opposite/Adjacent]
sin68^{°} = [x/9] ⇒ x = 9sin68^{°} ⇒ x ≈ 8.3
cos68^{°} = [y/9] ⇒ y = 9cos68^{°} ⇒ y ≈ 3.4
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Trigonometry in Right Angles
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Master Formula for Right Angles 0:11
 SOHCAHTOA
 Only for Right Triangles
 Example 1: Find All Angles in a Triangle 2:19
 Example 2: Find Lengths of All Sides of Triangle 7:39
 Example 3: Find All Angles in a Triangle 11:00
 Extra Example 1: Find All Angles in a Triangle
 Extra Example 2: Find Lengths of All Sides of Triangle
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
Transcription: Trigonometry in Right Angles
We are trying some examples of right angle in trigonometry, we are finding lengths of sides and angles in right triangles.0000
The master formula we are using here is SOHCAHTOA, and remember that SOHCAHTOA only works in triangles where one of the angles is a right angle, so in right triangles.0012
SOHCAHTOA does not work in other kind of triangles, we are going to learn some complicated rules later on called the law of cosines and the law of sines, that will work in any kind of triangle.0026
Right now we are just looking at right triangles and we are using SOHCAHTOA to figure out the relationships between the length of the sides and the measure of the angles.0037
What we are given here is a right triangle with a short side of length = 3 and hypotenuse of length = 7.0046
We want to find all the angles in the triangle so let me try to graph this out.0056
We know that it is a right triangle, one angle is a right angle, short side has length = 3 and the hypotenuse has length = 7, we want to find all the angles in the triangle.0067
I label the angles as theta and phi, I think it will be probably useful to find the third side of the triangle.0079
we know that the third side satisfies if I call it x for the time being.0091
x^{2} + 3^{2} = 7^{2}, so x^{2} + 9 = 49.0096
X^{2} is 40 and x = square root of 40, we can factor 4 out of that so 2 square root of 10.0108
That one did not come out to be very neat as some of other triangles have then.0121
I want to find this angles now, theta if I use SOHCAHTOA, I know that sin (theta) is equal to the opposite / hypotenuse.0125
That one I deliberately picked one where I would not have to use the square root to make my life a little bit simpler.0137
So, opposite is 3, the hypotenuse is 7, theta itself is arcsin(3/7).0142
Now remember here, if you calculator is set to radian mode then you will get a very strange answer looking here.0154
If you are looking for degrees, you have to set your calculator to degree mode.0161
On my calculator that is a matter of pushing the mode button and then moving it over and selecting the degree option.0166
It has two options, degree and radians, you want to set the degree options if that is the kind of answer you are looking for.0172
In my case, I’m going to do the inverse sin of 3 / 7, and it tells me that it is 25.4 degrees.0180
That tells me what theta is, I would like to figure out now what phi is, if I use the sin = opposite / hypotenuse.0202
If I use that for phi then I’m going to have to look at this ugly number 2 square root of 10.0211
Instead, I’m going to use cos(phi) = adjacent/ hypotenuse, phi is right there so the adjacent is 3.0218
Cos(pi) is 3/7, phi is arcos(3/7) and if I work that out on my calculator I get 64.6 degrees approximately.0230
I’m rounding here to the nearest decimal place, so now I have phi.0252
That gives me all the angles of the triangle because the last angle is a right angle, but I want to check these by seeing whether these angles adds up to 180 degrees.0258
25.4 + 64.6 those add up to 90 and the last angle is a 90 degree angle, it does in fact give me a 180 degrees.0270
That is very satisfying to see that does checked, the key there was to remember the SOHCAHTOA formula which works in all right triangles.0286
And then when we figured out that we are given two sides of the triangle, we are able to figure out the angles using sin(theta) =( opposite / hypotenuse), cos(phi) = adjacent / hypotenuse.0294
For our last example of right triangle trigonometry, we are given a right triangle that has one angle of 65 degrees and a hypotenuse of length=3.0000
I want to find the lengths of all the sides of the triangle, let me draw this out.0012
We got one angle on the corner, I will call it (theta) and we are given that 65 degrees, we got a right angle and we know that the hypotenuse is length = 3.0022
The question is to find the lengths of the other sides of the triangle and I’m going to use my standard formula SOHCAHTOA, (Some Old Horse Caught Another Horse Taking Oats Away).0033
I’m going to use the sin portion of that formula, so sin(theta) = (opposite/hypotenuse).0048
I do not know what the opposite is but I know that the hypotenuse is 3 and I know that (theta) is 65 degrees.0064
The opposite is equal to 3 x sin(65 degrees), remember to set your calculator to degree mode.0076
If you have it set in radian you will get a very confusing looking answer here.0083
I typed in 3 × sin(65) and I get that, that is approximately equal to 2.7.0090
That tells me that the side opposite theta is approximately 2.7 units long.0100
Finally, I am going to use the cos part of SOHCAHTOA, cos(theta) = adjacent/hypotenuse, and the hypotenuse is still 3 units, the cos(theta) is cos(65 degrees).0107
If we solve that for the adjacent side, we get ((3 cos(65 degrees)) and the calculator tells me that it is approximately equal 1.3 units.0129
That tells me that the adjacent side to (theta) is 1.3.0137
Now again, I figured out each one of those sides using SOHCAHTOA, I’m going to check it using the Pythagorean theorem.0159
As I check here, I will do 1.3^{2} + 2.7^{2} and that should be approximately equal to 1.3^{2} + 2.7^{2}.0168
If you work that out on the calculator that actually gives you 8.98.0187
There is a little bit of rounding when I found those values that is very close to 9 which is 3 squared.0192
That tells us that the Pythagorean theorem is satisfied by these lengths which means that we almost certainly did those right.0200
Again, that came back to writing down all the information we had in the triangle and labeling one of the angles.0206
We labeled the angle we were given as (theta=65 degrees) and then using the SOHCAHTOA relationships to set up some equations.0212
And then solving for the length of the sides that we did not know, the sin gave us the length of the opposite side.0221
The cos SOHCAHTOA relationship gave us the length of the adjacent side.0230
And then it was an easy matter to check that those actually satisfy the Pythagorean theorem.0235
That is the end of the lecture on trigonometry in right triangles.0240
We will come back later and talk about trigonometry in triangles that do not necessarily have a right angle.0245
We will be using the law of sines and law of cosines to analyze those.0251
Thanks for watching the trigonometry lectures on www.educator.com.0255
Hi, these are the trigonometry lectures for educator.com.0000
Today, we're going to talk about trigonometry in triangles that have a right angle.0005
These are called right triangles.0009
The master formula for right triangles, we've seen it before it's the SOH CAH TOA.0011
That's the word that I remember to know that the sin(θ) ...0017
Let me draw a right triangle here.0022
If θ is one of the small angles, not the right angle, then the sin(θ) is equal to the length of the opposite side over the length of the hypotenuse.0026
Cos(θ) is equal to the adjacent side over the hypotenuse.0042
The tan(θ) is equal to the opposite side over the adjacent side.0050
For shorthand, sine is equal to opposite over hypotenuse, cosine is equal to adjacent over hypotenuse, tangent is equal to opposite over adjacent.0053
It's probably worth saying SOH CAH TOA often enough until it sticks into your memory, because it really is useful for remembering these things.0063
If you have a hard time remembering that, the little mnemonic that is also helpful for some students is Some Old Horse Caught Another Horse Taking Oats Away.0071
That spells out SOH CAH TOA for you.0083
One key thing to remember here is that SOH CAH TOA only works in right triangles.0085
You have to have one angle being a right angle.0093
If you have a triangle that is not a right triangle, if you don't have a right angle, don't use SOH CAH TOA because it's not valid in triangles that don't have right angles.0096
That's if you have no right angle.0116
We're going to learn in the next lectures on educator.com, we'll learn about the law of sines and the law of cosines.0122
Those work in any triangle where you don't need a right angle, but when you have a right angle, it's definitely easier and better and quicker to use SOH CAH TOA.0128
Let's try that out with some actual triangles.0138
On the first example, we have a right triangle with short sides of length 3 and 4, and we want to find all the angles in the triangle.0140
Let me draw a triangle here.0149
We're told that the short sides have length 3 and 4.0154
Of course, one angle is a right angle, so I don't need to worry about that.0158
I'll call these angles θ, and I'll call this one φ.0162
First thing we're going to need to know is what the hypotenuse of this triangle is.0166
h^{2}=3^{2}+4^{2}, Pythagorean theorem there, which is 9+16, which is 25.0172
So, h is the square root of 25, h is 5.0184
Let me draw that in there.0187
Now, I want to figure out what θ and φ are.0191
I'm going to use SOH CAH TOA.0197
Let me write that down there for reference, SOH CAH TOA.0200
I'm going to figure out what θ is by using the SOH part of the SOH CAH TOA.0205
Sin(θ) is equal to the opposite over the hypotenuse.0214
The opposite angle to θ is 3 and the hypotenuse is 5.0225
So, θ=arcsin(3/5).0231
I'm going to work that out on the calculator.0238
My calculator has an arcsine button, it actually writes it as sin^{1}, which I don't like that notation because makes it seem like a power.0241
In any case, I'm going to use inverse sine of 3/5.0251
There's a very important step here that many students get confused about which is that, if you're looking for an answer in terms of degrees, which in real world measurement, it sometimes easier to use degrees than radians.0257
You have to set your calculator to degree mode.0270
Most calculators have a degree mode and a radian mode.0274
In fact, all calculators that do trigonometric functions have a degree mode and a radian mode.0278
The default is probably radian mode.0284
If your calculator is set in radian mode and you try to do something like arcsin(3/5), you'll get an answer that doesn't look right.0287
You may be confused if you're checking your answers somewhere, it may not agree with what the correct answer it.0297
What you have to do is set up your calculator in degree mode, if you want an answer in degrees.0304
My calculator is a Texas Instruments.0311
It's got a mode button, I just scroll down, it say's RAD and DEG to convert it from radians to degrees.0315
I'm going to convert it into degree mode and then I'll get an answer in terms of degrees.0323
That's a step that many students forget and they get kind of confused when they get an answer which is in terms of radians, but it doesn't agree with the degree answer they were looking for.0330
Now, I've got my calculator set in degree mode.0338
I'll do the arcsin(3/5) which is 0.6.0342
It tells me that that is approximately equal to 36.9 degrees.0348
I found one of the angles in the triangle.0358
Of course, another one is a right angle, so it's 90 degrees.0359
For φ, I think for φ, I'm going to practice the cosine part of the SOH CAH TOA.0362
I know that cos(φ) is equal to adjacent over hypotenuse.0369
That's the adjacent side to φ, so φ is over here, it's adjacent side is 3, hypotenuse is still 5.0378
So, φ=arccos(3/5).0390
Again, I'll do that on my calculator.0396
The calculator tells me that that's 53.1 degrees, approximately equal to 53.1 degrees.0404
Now, there's a little check here you can do to make sure that you work this out correctly.0412
We know that the angles of a triangle add up to 180 degrees.0416
If we check here, 36.9+53.1 plus the last angle was a 90degree angle, a right angle, if you add those up, 36.9+53.1 is 90 plus 90, you get 180 degrees.0421
That tells me that my answers are right.0439
The key formula here to remember is SOH CAH TOA.0443
Everything comes down to drawing the angles in the triangle, and just using sine equals opposite over hypotenuse, cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent.0446
Now, we're given a right triangle and we're told that one angle measures 40 degrees.0460
Let me call that angle right here, that would be 40 degrees.0472
The opposite side has length 6.0480
I want to find the lengths of all the sides in the triangle.0486
Of course, finding the angles is no big deal because one side is a right angle, we're told that it is a right triangle.0491
I can find the other angle just by subtracting from 180.0498
In fact, 18040 is 140, minus 90, is 50.0502
I know that other angle is 50.0511
The challenge here is to find the lengths.0513
We're going to use SOH CAH TOA.0516
I'm going to apply SOH CAH TOA to 40.0521
I know that sin(40) is equal to 6 over the hypotenuse, because that's opposite over hypotenuse.0526
The hypotenuse, if I solve this, that's equal to 6/sin(40).0535
Remember to convert your calculator to degree mode before you do this kind of calculation.0545
6/sin(40) is approximately equal to 9.3.0551
That tells us the length of the hypotenuse, 9.3.0561
Now I'd like to find the length of the other sides.0567
I'm going to use the cosine part of SOH CAH TOA.0572
I know that cos(40) is equal to the adjacent over the hypotenuse.0574
If I solve that for the adjacent side, that's equal to hypotenuse times the cos(40).0585
I know the hypotenuse now.0592
If I multiply that by cos(40).0596
What I get is approximately 7.2 for my adjacent side.0601
The sides of my triangle are 6, 7.2 and 9.3.0611
Again, there's an easy way to check that.0618
We'll check that using the Pythagorean theorem.0620
I want to check that 6^{2}+7.2^{2} gives me 87.8, which is approximately equal to 9.3^{2}.0623
I know that I got those side lengths right.0656
The third example here, we have the lengths of the two short sides of a right triangle, are in a 5:2 ratio.0662
Let me draw that out.0668
We're given that it's a right triangle.0676
We've got lengths in a 5:2 ratio.0679
I don't actually know that these lengths are 5 and 2, but here's the thing, if I expand this triangle proportionately, it won't change the angles.0683
If I blow this up to a similar triangle that actually has side lengths of 5 and 2, I'll get a similar triangle with the same angles.0695
I can just assume that this triangle has actually side lengths of 5 and 2.0706
I want to find all the angles in the triangle.0711
I'll label that one as θ, that one is φ.0715
I know that sooner or later, I'm going to need the hypotenuse of the triangle, I'll go ahead and find that now.0719
h^{2}=2^{2}+5^{2}, that's equal to 4+25 which is 29.0722
My hypotenuse is the square root of 29.0738
I'd like to practice all parts of SOH CAH TOA, and we've used sine and cosine in the previous problems.0747
I'm going to use the tangent part.0757
Remember, tangent is opposite over adjacent.0760
Tan(θ) there, the opposite side is 5, and the adjacent side has length 2.0765
I'm going to find arctan(5/2), θ is arctan(5/2).0772
Remember that you want to have your calculator in degree mode here, because if you have your calculator in radian mode, you'll get an answer in radian which would look very different from any answer in degrees that you were expecting.0783
I calculate arctan(5/2), and it tells me that that is approximately equal to 68.2 degrees.0800
Now, I found θ, I've got to find φ now.0815
I could find φ just by subtracting θ from 90 degrees, because I know θ+φ adds to 90 degrees.0819
Remember, one angle of the triangle is already 90 degrees, the other two must add up to 90 degrees.0826
I could just find the other angle by subtracting but I want to avoid that.0833
I want to practice using my SOH CAH TOA rules.0839
The other reason is if I find it using some other method, then I can add them together at the end and use that to check my work.0841
I'm going to try using a SOH CAH TOA rule.0848
I'm going to find it using angle φ.0853
Looking at φ, I know that sin(φ) is equal to the opposite over hypotenuse.0854
Sin(φ) is equal to the opposite over hypotenuse.0866
The opposite side to φ is 2, and the hypotenuse is the square root of 29.0872
φ is equal to the arcsine of 2 over the square root of 29.0881
That's definitely something I want to put into my calculator.0886
I'll figure out inverse sine of 2 divided by square root of 29 ...0891
The inverse sine of 2 divided by square root of 29 ...0911
It tells me that that's approximately equal to 21.8 degrees.0917
That's what I got using SOH CAH TOA.0925
Again, I'm going to check it by checking that the three angles of this triangle add up to a 180 degrees.0928
I've got 68.2 degrees plus 21.8 degrees, those add up to 90, plus the last 90degree angle, the right angle.0937
Those do add up to 180 degrees.0951
That tells me that my work must probably be right.0954
That came back to looking at SOH CAH TOA, and figuring out what the angles were based on the SOH CAH TOA.0958
We know that tangent is opposite over adjacent, and we also used that sine is opposite over hypotenuse.0968
We'll try some more examples later.0974
1 answer
Last reply by: Dr. William Murray
Wed Oct 14, 2015 4:44 PM
Post by Vance Bower on October 14, 2015
Can you use sin, cosin, or tangent in any order for angle theta or c to get the angle measurements? In other words, you used the tan of theta and the sin of c in the third example. Could I have used the tangent of theta and the cosign or tangent of c, instead of the sin of c?
2 answers
Last reply by: Dr. William Murray
Tue Aug 5, 2014 11:50 AM
Post by Joshua Jacob on July 19, 2014
Sorry if this is a little unrelated but is it possible to do arcsin, arccos, and arctan on a graphing calculator? Is it the same thing as sin^1...etc.
Thank you for making these amazing lectures!
1 answer
Last reply by: Dr. William Murray
Thu Jul 18, 2013 8:25 AM
Post by Manfred Berger on July 7, 2013
Unless I'm missing anything crucial here, the fact that SOCAHTOA only works for right triangles stems directly from the basic definition of the the unit circle, which is essentially the pythagorian theorem.
1 answer
Last reply by: Dr. William Murray
Mon May 6, 2013 8:54 PM
Post by Emily Engle on May 5, 2013
How would you solve for the inverse ratios without a calculator?