I didn't jump thru the LOC functions - I just subtracted the LOS angle from 180 and got 113.8. Why do people need to make things harder than needs to be...
My teacher always warned us about this when using the sine rule, which is why they encouraged us to use the cosine rule even though its longer and harder to remember.
An alternative is to indirectly use the law of sines on this one, to find the angle that wasn't asked for. Then use the 180 degree total interior angle rule, to find the angle we want to know. A triangle always has at least two acute angles, for which the sine rule will work without needing to think about the multivalued nature of inverse sine. Once you find both acute angles, it's trivial to find the remaining one.
As a teacher and tutor, I'd have to say I don't think that is a good idea. You can't always use the Law of Cosines. Sometimes, you have no choice but to use Law of Sines (the opposite is also true). For example, if you are given only two angles and one side, you can't use the Law of Cosines. If you are given only three sides, you can not use the Law of Sines. If you aren't given a pair of an angle and the side opposite that angle (or enough information to find such a pair), you can't use the Laws of Sines. Also, if the problem specifically asks you to find ALL possible triangles, you really need to understand, and actually USE the fact, that the inverse functions are multivalued. This is easier to do with sine, because the Law of Sines only gives you positive numbers and sine is positive in BOTH Q1 and Q2. The cosine is not, so you have to be very careful to not drop signs in those calculations, and a lot of students, in my experience, do just that.
Love the video! I realized at 3:24 that it would be 180-66.2 because inverse sine is constrained. I really like the main point you made, I'm still watching though!
If you append a line segment CD parallel to AC such that it forms the base of an isoceles triangle BDC, we see that the Law of Sines is for the angle between the new far side and the base of the isoceles. Given that base angles for the I.T are the same, we can subtract that from 180 by vertical anglrs to get the initial angle This can be proven generally that law of sines for obtuse angles still holds up so basically c = 180° - (law of sines angle) in general as you said
I believe the absolute simplest way to understand what is going on is by scetching a single complete sine wave. Since you want the inverse sine, you mark the value you use to find the angle on the y-axis. Now draw a parallel through this value to the x-axis to figure out where it intersects the sine wave and suddenly you will see: there are 2 such intersection points. Thus you need to be carefull which one to use.
I replied this to another comment. But now I think about again, there’s actually no need. Since I gave them all three sides so we must have only one possible triangle and that is the obtuse one.
Sine rule is easier to compute than cosine rule, so it is best to use and calculate the angle, with an awareness that sin(180-A) = sin(A) so you have to choose between A or 180-A. Obtuse angles are only applicable to those opposite to the longest side and one that is greater in length than the hypotenuse of the smaller sides. If this extra attention is too much to ask for relative to using cosine rule, than as the video suggests, just use the cosine rule.
I drew a line perpendicular to the 5.8 line that intersects C, and used basic trig equations to get the angles of both triangles which arrive at the same answer(~52+~61.8 = ~113.8). Would you take points off because I didn't use LOS or LOC? :)
@@blackpenredpen It actually depends on what type of calculator you have. I know on my calculator (Casio-991EX) the brackets aren't needed for either half unless you type it in cackhandedly (and then you only require the opening bracket on the numerator and it will add the closing one when forming the fraction). (You can activate a mode that requires all the brackets although I haven't seen anyone actually use it as it makes typing in the calculations illegible for half the functions including fractions)
When it comes to measurements, never trust the picture! In this case, see that c is too big for the Pythagorean theorem, so angle C is more than 90 degrees.
If the question says Angle C > 90 degree (In short Obtuse Angle) and we know Range of Inverse Sin Rule is [pi/2 , -pi/2], then we can easily find the Angle B as We know Angle B < Angle C, Therefore Angle B can be applicable to find it in LOS, after that then Do 180 - Angle B - Angle A = Angle C, You will get your answer still. (Alternative Way)
Very nice video. I personally had never learnt about using sine and cosine outside of right triangles, so this definitely was an eye opener for me. Thanks, keep making great math content!
In Ireland we have booklets for the sine and cosine rules thankfully, we’d still be inclined to use sine rule over cosine here though, we only use cosine rule 100 percent when we only have 3 sides and no angles
I got 113.765 degrees, and I did it in a visual programming language so it was actually pretty fun. First time trying to do proper algebra in that format, was a bit strange at first but became intuitive very quickly.
Everytime we use the law of sines to calculate an angle, we have to verify the possibility of a second answer. It's mandatory. We then must determine if this second angle is valid or not. Say we are given angle A and sides a and b so we can calculate angle B. Afer proceeding with the law of sines, this answer, say B1, has to be used to find B2 = 180º - B1. Then we have to calculate B2 + A. If the result is less than 180º, this second angle B2 is valid and we will get two possible triangles. However, if this sum is > 180º, the second triangle does not exist and B2 is discarted.
thanks for this!! I've been trying to think of sin, cos, ln as multivalued functions, to be more careful in my math. This is a good case. Don't put this on a test please
I can be wrong but I do remember a rule from my student past. If we solve an equation x^3 + p*x + q = 0 we must take some specific values of cbrt(-q/2 + sqrt(...)) and cbrt(-q/2 - sqrt(...)) that satisfy the condition cbrt(-q/2 + sqrt(...)) * cbrt(-q/2 - sqrt(...)) = -p/3. It will be obvious if we put x = cbrt(-q/2 + sqrt(...)) + cbrt(-q/2 - sqrt(...)) into x^3 + p*x + q = 0. This product of the cube roots should make some parts "annihilate". This one of the ways to prove this formula. We were shown this method on the physics lessons.
I feel proud of myself, that I got the explaination completely correct (by myself) by imagining the unit circle But instead of using arccos, I just subtracted 66.2 from 90 and added 90
Quick reminder that I tell my students, longer side is always opposite to bigger angle, smaller side to smaller angle, etc. They would have realized that 66, can’t be correct because then, angle B is bigger than 66. Of course, that still ask them to have the knowledge that ambiguous cases exists. But, whenever possible, use the cosine law to find the largest angle (using the side/angle relationship) and you will avoid the problem, if there is only a unique solution.
OK this is really cool. I was thinking about how two different angles can give the same y coordinate on the unit circle, but that ambiguous triangle thing is something I've either never seen before or completely forgotten
Am I the only one who went with a basic strategy of splitting the triangle into right-angle triangles and calculate the angle from there? 😅 Given the triangle ABC whose AB = 3.9, BC = 3, CA = 5.8, m∠A = 28.25, I've solved for m∠B. I've first calculated the height of it using: - sin(28.25) = 3.9 / h - h = (3.9)(sin(28.25) - h ≈ 1.846 I've then calculated the angle on the right-side triangle and left-side triangle by considering height as the adjacent side and both of AB and BC as the tangent sides, and solving for θ using cosine: - Right-side -> θ = arccos(h / AB) = arccos(h / 3.9) = 61.75° - Left-side -> θ = arccos(h / BC) = arccos(h / 3) = 52.025 ≈ 52.03° Summing angles of both sides would then return m∠B, which is going to be equal to 113.77514644°, or 113.78° for short.
Solving the equation sin(x) = 0.915 isn't as simple as taking the arcsine of both sides to be left with x equaling some number. Similarly, solving x^2 = 4 is not as simple as square rooting both sides to be left with x equaling some number.
What I did was law of sines for A and B to get B then I used 180=A+B+C to get C because: 1. I know that all the inverse trig fns are good for acute angles. 2. There are at least 2 acute angles in a triangle 3. The acute angles are opposite the smallest sides of the triangle 4. I’m too lazy to do an unnecessary law of cos calculation
The problem is that taking arcsin or arccos from both sides of the equation is not equivalent. For instance, if sin(x) = k then x≠arcsin(k). Instead x = (-1)^n * arcsin(k) + pi * n (n is any integer) (In this particular case we choose n = 1, resulting pi - arcsin(k) ) As well: cos(x) = k then x = ±arccos(k) + 2pi * n (n is any integer) In russian schools we are taught to be as general as possible
You could have used the law of sines, just remember if there is a equation sinx = a then x = a + 2kpi or x = pi-a + 2kpi where k is an integer. That means the next solition is 180 degrees - 66.2 degrees which gives us 113.8 degrees. Now we have to choose one of these and obviously since the C angle is bigger than 90 degrees we choose 113.8 degrees.
Also notice that the wrong answer and the correct answer adds up to 180 degrees. This reflects one of the sine identity, where sin A = sin(180-A). I guess we must consider two cases: sin A and sin (180-A).
I literally just had a math test on this today. There was one problem that had two solutions because of this. They were both correct, I needed both of them for full credit.
I am very very happy that i found this channel 😊. Thank you soo much i just found your channel when i searched for integration of sin⁴x. Love from India ☺️🌹
No need to give that C° > 90° in the question. We can understand if an angle is bigger or smaller than 90° by using law of cosine. In an ABC triangle, where edges are a-b-c; a² + b² - 2abcosC° = c². In this equation; If C° = 90° ⇒ cosC° = 0 ⇒ a² + b² = c²; If C° < 90° ⇒ cosC° > 0 ⇒ a² + b² > c²; If C° > 90° ⇒ cosC° < 0 ⇒ a² + b² < c². In this case, let's calculate the values and analyze. Firstly, [3²+(3,9)²] ? (5,8)² then, (9+15,21) ? 33.64 finally, 24,21 < 33.64. Which means C° > 90. We can solve this question with that information right now. We know that C° > 90° even if arcsin gives us 66.2°. And since we know that sina° = sinb° if a°+b° = 180°, we can subtract 66.2° from 180° and get our answer 113.8°. These are the basic informations and reasonings a precalculus student should be capable.
why does angle A equels to 28.25 degrees? We must chek it using ToC. And... it's wrong. Then, we can't use this or one of sides. Or the problem is incorrect
@@blackpenredpen can you do a proof of descartes rule of signs. I would do anything to see the proof. As a matter of fact, how much do i pay you? I'm dead serious 😊
we get this a lot in physics (i dunno how to name it in english but we call the chapter "sinusoïdal") and our teacher says find the angle C and then calculate cosC, if you find it >0 then you take that angle. But if you find the cosC
And if anyone wondering if someone do go 66.25° then some people gonna realize that angle between 3 and 5.8 is 84.93° which should not be possible cause angle between 2 smaller side of triangle should be the biggest one Okay also the picture too if I wanna add
I'll ask you a question What is -ln(-1) =? A . iπ B . -iπ C. A ans B both D. Can't possible E. None of the above I am not testing you but by watching just your videos i got that question in my mind
My first thought was that you were trolling your students by giving them the wrong value for angle A :). But I checked with WolframAlpha and it's legit: arccos((5.8^2+3.9^2-9)/(2*5.8*3.9)) = 28.25 degrees Weirdly, WolframAlpha has a minor bug here, that if you add "in degrees" to that input, it gives you an error saying a pure number can't be converted to degrees, but that's obviously incorrect because arccos doesn't return a pure number, it returns a value in radians (even WolframAlpha admits as much when you put the input above).
If you use the Law of Sines on angles A and B and subtract both from 180 degrees it also works. Surely this is a simpler method? Whoops! Just noticed that, unsurprisingly, I am not the only one to suggest this. I enjoy your videos, though, even if I can only understand some of them. P.S. The Law of Sines is a piece of cake on a slide-rule, as are all proportions.
To be honest as a math teacher and tutor I absolutely hate discussing the 'ambiguous case of sine law' as it's called in the Ontario curriculum. I get that it's important but it's just tedious and annoying :p I think it doesn't really make sense to students until they've learned how to find all solutions for sin(x) = something which usually occurs in a higher level course
i knew there would be a problem since sin^-1 varies from -pi to pi only , but heres a trick! Just find angle B and use the fact that A+B+C=180 ( B is acute so no need to worry), no need of cosine rule!
I wish blackpenredpen had been more explicit about the MISTAKE he did when applying the law of sines. It is simply NOT TRUE that sin x = y implies that x = arcsin y. sin x = y implies that x = 90° ± (90 - arcsin y) + n*360°, with n = any integer (which just adds more full turns left or right). Make n = 0 so you don't have repeated solutions for the same angle just rotated 360 degrees, and you get sin x = y implies that x = 90° ± (90 - arcsin y), which is EITHER arcsin y (if you pick the -) OR 180 - arcsin y (if you pick the -). Solving for x in sin x= y and coming up with x = arcsin y is the same type of MISTAKE than solving for x in x^2 = y and coming up with x = sqrt(y), instead of x = ±sqrt(y) So how to get the correct answer, without relying on the accuracy of the drawing to figure if ancle C is acute or obtuse? Apply law of sin on angles A and B to solve for B. Angle B is GUEARNTEED to be acute, the angles adjacent to the longest side of a triangle must be acute. Then, with angle A and B known, solve for C doing A+B+C=180. Finally, note that this problem has redundant information. Even more, I strongly suspect that this exact triangle doesn't exist. One of the sides or the given angle (A) must be only approximately correct. So how to solve it if you have only the 3 sides? Well, you can apply the law of cos 3 times to find the 3 angles, or 2 times and the 3rd angle is found by adding the 3 angles to 180. Bot for me, because the law of sine is simpler and more practical than the law of cos, the easiest and shortest way is to get any angle suing the law of cos, then using the law of sin to find one of the 2 angles that are known to be acute, and then get the 3rd angle by adding all three = 180.
What happens then the angle is very close to 90 degrees. It would be nearly impossible to use the picture to determine whether the angle was supposed to be acute or obtuse just by looking at it.
One of the properties of functions with inverses is that they have to be bijective, but sin is only bijective in a certain range (-pi/2 to pi/2), so as a result the inverse has the same range. Sin is also bijective at an infinite number of places but we chose between +-pi/2 cos it contains the origin- so is nice.
because if you scetch a complete sine wave and try to do the inverse operation by drawing a parallel to the x axis through the sine value and intersecting it with the curve you will see: there are 2 such intersection points. So which one is the correct one? Both of them are correct, as the sin of (in this case 66.2° and 113.8°) both those intersection points have the same value. ( sin(66.2) is equal to sin(113.8) ) In other words, as someone has already pointed out: the sin function IS a function, but the inverse is not. Edit: Something similar is true for the cos function. Scetch it and you will see that cos(x) is equal to cos(-x).
I think you just have to remember that, if you use the law of sine to find an obtuse angle, you have to subtract your answer from 180º. It's an extra step, but so is using the law of cosine to solve that problem.
As mentioned earlier, why bother with the Law of sines is law of cosines can be used (and doesn't deal with the ambiguous case). The problem would have been more interesting if the 5.8 would have been left out. Then the student would have to use the law of sines (SSA case)
uuh let's call the height on c h (forming 2 right triangles) sinA = h / 3.9 => h = sinA * 3.9 = 1.85 sinB = h / 3 = 1.85 / 3 => B = arcsin(1.85 / 3) = 37.97° C = 180 - A - B = 180 - 28.25 - 37.97 = approx. 113.78 didn't need any law; neither sine nor cosine
I have a mysterious query in my mind that "We can make every negative value positive by adding it to 0." (like -2+0 = 2) because 0 is greater than any negative value in number line and its rule in addition that we take sign of greater value with answer .... am I right ??
By giving them an extra angle not only did you give them a AAS problem you also gave them a Donkey Theorem problem which had “two solutions” when you ignore the third side. Also you need to keep in mind sine is also positive in the 2nd quadrant…
@@blackpenredpen ... Good day to you, Thank you for showing the SSA (ambiguous) case and also in this case the difference in application between the sine and cosine rule (advantage -disadvantage) in your presentation; at last it has become clear to me the 2 possibilities of triangles (obtuse angle C or acute angle C; supplementary angles)! Thank you again for as always a great presentation and take care, Jan-W
I saw the catch with sine coming from a mile away about the time the sine law was written on the board... I was like oh no it doesn't work that simple for angles bigger than 90°
Law of sines gives us two possible angles C = 66.2 or C = 113.8 but one answer is wrong 6:35 Yes but you can get the other solution from reduction formula nevertheless one solution will be wrong
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An insidious multiple choice question would give the wrong angle for A. Then you list the one answer for the law of cosines, two answers for the law of sines (obtuse and acute) and 90 degrees as options, along with the correct answer none of these, because the triangle does not exist.
Learn more about the law of sine and cosine on Brilliant: brilliant.org/blackpenredpen/ (now with a 30-day free trial)
It doesn’t give me 30 days free trial :( but thank you so much! The app is nice.
I Have A Lovely Problem In Mathematics You Should Look It Atleast One Time.
The Problem Is
Given- (X)^1/2 + (Y )^1/2 = (5)^1/2
Then y=f(x)=?
Conditiion Is the The Graph should Not Be Changed
I didn't jump thru the LOC functions - I just subtracted the LOS angle from 180 and got 113.8. Why do people need to make things harder than needs to be...
My teacher always warned us about this when using the sine rule, which is why they encouraged us to use the cosine rule even though its longer and harder to remember.
An alternative is to indirectly use the law of sines on this one, to find the angle that wasn't asked for. Then use the 180 degree total interior angle rule, to find the angle we want to know.
A triangle always has at least two acute angles, for which the sine rule will work without needing to think about the multivalued nature of inverse sine. Once you find both acute angles, it's trivial to find the remaining one.
exactly what my teacher said
As a teacher and tutor, I'd have to say I don't think that is a good idea. You can't always use the Law of Cosines. Sometimes, you have no choice but to use Law of Sines (the opposite is also true). For example, if you are given only two angles and one side, you can't use the Law of Cosines. If you are given only three sides, you can not use the Law of Sines. If you aren't given a pair of an angle and the side opposite that angle (or enough information to find such a pair), you can't use the Laws of Sines. Also, if the problem specifically asks you to find ALL possible triangles, you really need to understand, and actually USE the fact, that the inverse functions are multivalued. This is easier to do with sine, because the Law of Sines only gives you positive numbers and sine is positive in BOTH Q1 and Q2. The cosine is not, so you have to be very careful to not drop signs in those calculations, and a lot of students, in my experience, do just that.
@@michaelmann8800 sorry, i originally meant more of use LOC whenever possible/whenever we have the choice
Did 5 years of engineering at university, no teacher has mentioned that before.
Love the video! I realized at 3:24 that it would be 180-66.2 because inverse sine is constrained. I really like the main point you made, I'm still watching though!
Because that cosine theorem more strong
Thank you!
If you append a line segment CD parallel to AC such that it forms the base of an isoceles triangle BDC, we see that the Law of Sines is for the angle between the new far side and the base of the isoceles. Given that base angles for the I.T are the same, we can subtract that from 180 by vertical anglrs to get the initial angle
This can be proven generally that law of sines for obtuse angles still holds up so basically c = 180° - (law of sines angle) in general as you said
I believe the absolute simplest way to understand what is going on is by scetching a single complete sine wave.
Since you want the inverse sine, you mark the value you use to find the angle on the y-axis. Now draw a parallel through this value to the x-axis to figure out where it intersects the sine wave and suddenly you will see: there are 2 such intersection points. Thus you need to be carefull which one to use.
You always inspired me during my calculus classes before I started studying health sciences. Thank you so much :)
Be sure to clearly state C>90°, otherwise the student can claim "Drawing not to scale." 😁
But I do appreciate this scenario.
I replied this to another comment. But now I think about again, there’s actually no need. Since I gave them all three sides so we must have only one possible triangle and that is the obtuse one.
@@blackpenredpen yep
2:52 plot twist of the year
Sine rule is easier to compute than cosine rule, so it is best to use and calculate the angle, with an awareness that sin(180-A) = sin(A) so you have to choose between A or 180-A. Obtuse angles are only applicable to those opposite to the longest side and one that is greater in length than the hypotenuse of the smaller sides. If this extra attention is too much to ask for relative to using cosine rule, than as the video suggests, just use the cosine rule.
I already love maths, because of your channel I feel it really interesting
Great to hear!
I remember getting this exact point wrong in the math competition 😂😂😅😅
Thanks for the clear explanation :))
I plan on being a physicist/ mathematician when i grow up lol i am in 5th grade but i learn calculus from ur vids! Math is awesome :)
👁️👄👁️
Great, I remember this from school. For some students it might help to draw the graphs of sine and cosine.
I’m just noticing how many markers this dude has in the background
I drew a line perpendicular to the 5.8 line that intersects C, and used basic trig equations to get the angles of both triangles which arrive at the same answer(~52+~61.8 = ~113.8). Would you take points off because I didn't use LOS or LOC? :)
oh wow great solution
5:03 it’s a fraction, why is the parentheses needed? The fraction bar always indicates you do top and bottom first separately, then divide
When you enter that expression on a calculator, those ( ) are needed.
@@blackpenredpen It actually depends on what type of calculator you have. I know on my calculator (Casio-991EX) the brackets aren't needed for either half unless you type it in cackhandedly (and then you only require the opening bracket on the numerator and it will add the closing one when forming the fraction).
(You can activate a mode that requires all the brackets although I haven't seen anyone actually use it as it makes typing in the calculations illegible for half the functions including fractions)
Finally a blackpenredpen video I actually understand😃
When it comes to measurements, never trust the picture! In this case, see that c is too big for the Pythagorean theorem, so angle C is more than 90 degrees.
If the question says Angle C > 90 degree (In short Obtuse Angle) and we know Range of Inverse Sin Rule is [pi/2 , -pi/2], then we can easily find the Angle B as We know Angle B < Angle C, Therefore Angle B can be applicable to find it in LOS, after that then Do 180 - Angle B - Angle A = Angle C, You will get your answer still. (Alternative Way)
Hi BPRD!
Super nifty! I'm definitely stealing this.
Be my guest. 😃
Very nice video. I personally had never learnt about using sine and cosine outside of right triangles, so this definitely was an eye opener for me. Thanks, keep making great math content!
Just list 2 values for the sine (0
Thanks Blackpenredpen. Still love your videos. You are such a good teacher.
Thanks!!
In Ireland we have booklets for the sine and cosine rules thankfully, we’d still be inclined to use sine rule over cosine here though, we only use cosine rule 100 percent when we only have 3 sides and no angles
I got 113.765 degrees, and I did it in a visual programming language so it was actually pretty fun. First time trying to do proper algebra in that format, was a bit strange at first but became intuitive very quickly.
Great enthusiasm but even my HS trig teacher instructed us in this from early on in the semester. So, ... this is old news but presented well.
He said it’s a trig class problem.
Everytime we use the law of sines to calculate an angle, we have to verify the possibility of a second answer. It's mandatory. We then must determine if this second angle is valid or not. Say we are given angle A and sides a and b so we can calculate angle B. Afer proceeding with the law of sines, this answer, say B1, has to be used to find B2 = 180º - B1. Then we have to calculate B2 + A. If the result is less than 180º, this second angle B2 is valid and we will get two possible triangles. However, if this sum is > 180º, the second triangle does not exist and B2 is discarted.
thanks for this!! I've been trying to think of sin, cos, ln as multivalued functions, to be more careful in my math. This is a good case.
Don't put this on a test please
I can be wrong but I do remember a rule from my student past.
If we solve an equation x^3 + p*x + q = 0 we must take some specific values of cbrt(-q/2 + sqrt(...)) and cbrt(-q/2 - sqrt(...)) that satisfy the condition cbrt(-q/2 + sqrt(...)) * cbrt(-q/2 - sqrt(...)) = -p/3.
It will be obvious if we put x = cbrt(-q/2 + sqrt(...)) + cbrt(-q/2 - sqrt(...)) into x^3 + p*x + q = 0. This product of the cube roots should make some parts "annihilate". This one of the ways to prove this formula. We were shown this method on the physics lessons.
Let
a = cbrt(-q/2 + sqrt(...))
b = cbrt(-q/2 - sqrt(...))
and
a*b = -p/3
then
x = a + b
so
x^3 + p*x + q = 0
=>
(a + b)^3 + p*(a + b) + q = 0
=>
a^3 + 3*a*b*(a + b) + b^3 + p*(a + b) + q = 0
=>
a^3 - p*(a + b) + b^3 + p*(a + b) + q = 0
=>
a^3+ b^3 +q = 0
=>
(-q/2 + sqrt(...)) + (-q/2 - sqrt(...)) + q = 0
=>
-q + q = 0
=>
0 = 0
We can reverse these steps so we get the aforementioned formula.
Speechless sir
Love from Hindustan ❤️
I feel proud of myself, that I got the explaination completely correct (by myself) by imagining the unit circle
But instead of using arccos, I just subtracted 66.2 from 90 and added 90
Awesome!
That's the same what i did
This video is the umpteenth time I'm reminded how lucky I am to be subbed to your channel!
Also do you speak Chinese? 你也会说中文吗 I do and I am fluent! You are a brilllaint mathematician and I love learning from ur vids!
I do! I also have a Chinese channel 黑筆紅筆 😄。Thank you very much for your kind words! I want to wish you all the best for your future!
Quick reminder that I tell my students, longer side is always opposite to bigger angle, smaller side to smaller angle, etc. They would have realized that 66, can’t be correct because then, angle B is bigger than 66. Of course, that still ask them to have the knowledge that ambiguous cases exists. But, whenever possible, use the cosine law to find the largest angle (using the side/angle relationship) and you will avoid the problem, if there is only a unique solution.
OK this is really cool. I was thinking about how two different angles can give the same y coordinate on the unit circle, but that ambiguous triangle thing is something I've either never seen before or completely forgotten
I'm surprised you didn't give dimensions for a non-existent triangle.....
hey bprp when's the sin of 10 degrees with the cubic formula coming? I am so excited for that video!!
Am I the only one who went with a basic strategy of splitting the triangle into right-angle triangles and calculate the angle from there? 😅
Given the triangle ABC whose AB = 3.9, BC = 3, CA = 5.8, m∠A = 28.25, I've solved for m∠B.
I've first calculated the height of it using:
- sin(28.25) = 3.9 / h
- h = (3.9)(sin(28.25)
- h ≈ 1.846
I've then calculated the angle on the right-side triangle and left-side triangle by considering height as the adjacent side and both of AB and BC as the tangent sides, and solving for θ using cosine:
- Right-side -> θ = arccos(h / AB) = arccos(h / 3.9) = 61.75°
- Left-side -> θ = arccos(h / BC) = arccos(h / 3) = 52.025 ≈ 52.03°
Summing angles of both sides would then return m∠B, which is going to be equal to 113.77514644°, or 113.78° for short.
Solving the equation sin(x) = 0.915 isn't as simple as taking the arcsine of both sides to be left with x equaling some number. Similarly, solving x^2 = 4 is not as simple as square rooting both sides to be left with x equaling some number.
What I did was law of sines for A and B to get B then I used 180=A+B+C to get C because:
1. I know that all the inverse trig fns are good for acute angles.
2. There are at least 2 acute angles in a triangle
3. The acute angles are opposite the smallest sides of the triangle
4. I’m too lazy to do an unnecessary law of cos calculation
That's an excellent tricky test question!
The problem is that taking arcsin or arccos from both sides of the equation is not equivalent.
For instance, if sin(x) = k then x≠arcsin(k).
Instead
x = (-1)^n * arcsin(k) + pi * n (n is any integer)
(In this particular case we choose
n = 1, resulting pi - arcsin(k) )
As well:
cos(x) = k then
x = ±arccos(k) + 2pi * n (n is any integer)
In russian schools we are taught to be as general as possible
You could have used the law of sines, just remember if there is a equation sinx = a then x = a + 2kpi or x = pi-a + 2kpi where k is an integer. That means the next solition is 180 degrees - 66.2 degrees which gives us 113.8 degrees. Now we have to choose one of these and obviously since the C angle is bigger than 90 degrees we choose 113.8 degrees.
Thank you for sharing. Congratulations on over a million subscribers. 🎉
Also notice that the wrong answer and the correct answer adds up to 180 degrees. This reflects one of the sine identity, where sin A = sin(180-A). I guess we must consider two cases: sin A and sin (180-A).
Interesting problem and discussion!
Thanks!!!
would it be possible if this question is provided without diagram? because it seems that you do that because its not logical if C < 90
How we find the value of π^(ie)
I literally just had a math test on this today. There was one problem that had two solutions because of this. They were both correct, I needed both of them for full credit.
I am very very happy that i found this channel 😊. Thank you soo much i just found your channel when i searched for integration of sin⁴x.
Love from India ☺️🌹
No need to give that C° > 90° in the question. We can understand if an angle is bigger or smaller than 90° by using law of cosine. In an ABC triangle, where edges are a-b-c; a² + b² - 2abcosC° = c². In this equation;
If C° = 90° ⇒ cosC° = 0 ⇒ a² + b² = c²;
If C° < 90° ⇒ cosC° > 0 ⇒ a² + b² > c²;
If C° > 90° ⇒ cosC° < 0 ⇒ a² + b² < c².
In this case, let's calculate the values and analyze.
Firstly, [3²+(3,9)²] ? (5,8)²
then, (9+15,21) ? 33.64
finally, 24,21 < 33.64. Which means C° > 90. We can solve this question with that information right now. We know that C° > 90° even if arcsin gives us 66.2°. And since we know that sina° = sinb° if a°+b° = 180°, we can subtract 66.2° from 180° and get our answer 113.8°. These are the basic informations and reasonings a precalculus student should be capable.
why does angle A equels to 28.25 degrees? We must chek it using ToC. And... it's wrong. Then, we can't use this or one of sides. Or the problem is incorrect
?
@@blackpenredpen we can define triangle knowing 3 sides. Then, we can find all angles. But one were given. Maybe, incorrect
@@blackpenredpen can you do a proof of descartes rule of signs. I would do anything to see the proof. As a matter of fact, how much do i pay you? I'm dead serious 😊
can u plz make a video on the integral e^x secx (tanx)^2 dx
Outstanding ❤love from Bangladesh ❤
we get this a lot in physics (i dunno how to name it in english but we call the chapter "sinusoïdal") and our teacher says find the angle C and then calculate cosC, if you find it >0 then you take that angle. But if you find the cosC
And if anyone wondering if someone do go 66.25° then some people gonna realize that angle between 3 and 5.8 is 84.93° which should not be possible cause angle between 2 smaller side of triangle should be the biggest one
Okay also the picture too if I wanna add
I never knew that. Thank you.
Sir please tell me board dimensions and yt setup
It's because sin(90°+x)=sin(90°-x), and cos(x)=cos(-x). It means cos is symmetric around 0° and sin is symmetric around 90°
I'll ask you a question
What is -ln(-1) =?
A . iπ
B . -iπ
C. A ans B both
D. Can't possible
E. None of the above
I am not testing you but by watching just your videos i got that question in my mind
My first thought was that you were trolling your students by giving them the wrong value for angle A :). But I checked with WolframAlpha and it's legit:
arccos((5.8^2+3.9^2-9)/(2*5.8*3.9)) = 28.25 degrees
Weirdly, WolframAlpha has a minor bug here, that if you add "in degrees" to that input, it gives you an error saying a pure number can't be converted to degrees, but that's obviously incorrect because arccos doesn't return a pure number, it returns a value in radians (even WolframAlpha admits as much when you put the input above).
Lol. By giving them angle A, many would attempt to use LOS instead of LOC and might just answer 66.2 deg at the end. 😃
If you use the Law of Sines on angles A and B and subtract both from 180 degrees it also works. Surely this is a simpler method?
Whoops! Just noticed that, unsurprisingly, I am not the only one to suggest this.
I enjoy your videos, though, even if I can only understand some of them.
P.S. The Law of Sines is a piece of cake on a slide-rule, as are all proportions.
2:46 if you already expect the angle to be obtuse, you could just substract it from 180° to get the real answer
To be honest as a math teacher and tutor I absolutely hate discussing the 'ambiguous case of sine law' as it's called in the Ontario curriculum. I get that it's important but it's just tedious and annoying :p
I think it doesn't really make sense to students until they've learned how to find all solutions for sin(x) = something which usually occurs in a higher level course
i knew there would be a problem since sin^-1 varies from -pi to pi only , but heres a trick! Just find angle B and use the fact that A+B+C=180 ( B is acute so no need to worry), no need of cosine rule!
you are my hero
For sine law you can just take 180-66.2 because as you mentioned C > 90º
I wish blackpenredpen had been more explicit about the MISTAKE he did when applying the law of sines.
It is simply NOT TRUE that sin x = y implies that x = arcsin y.
sin x = y implies that x = 90° ± (90 - arcsin y) + n*360°, with n = any integer (which just adds more full turns left or right).
Make n = 0 so you don't have repeated solutions for the same angle just rotated 360 degrees, and you get
sin x = y implies that x = 90° ± (90 - arcsin y), which is EITHER arcsin y (if you pick the -) OR 180 - arcsin y (if you pick the -).
Solving for x in sin x= y and coming up with x = arcsin y is the same type of MISTAKE than
solving for x in x^2 = y and coming up with x = sqrt(y), instead of x = ±sqrt(y)
So how to get the correct answer, without relying on the accuracy of the drawing to figure if ancle C is acute or obtuse?
Apply law of sin on angles A and B to solve for B. Angle B is GUEARNTEED to be acute, the angles adjacent to the longest side of a triangle must be acute. Then, with angle A and B known, solve for C doing A+B+C=180.
Finally, note that this problem has redundant information. Even more, I strongly suspect that this exact triangle doesn't exist. One of the sides or the given angle (A) must be only approximately correct. So how to solve it if you have only the 3 sides? Well, you can apply the law of cos 3 times to find the 3 angles, or 2 times and the 3rd angle is found by adding the 3 angles to 180.
Bot for me, because the law of sine is simpler and more practical than the law of cos, the easiest and shortest way is to get any angle suing the law of cos, then using the law of sin to find one of the 2 angles that are known to be acute, and then get the 3rd angle by adding all three = 180.
What happens then the angle is very close to 90 degrees. It would be nearly impossible to use the picture to determine whether the angle was supposed to be acute or obtuse just by looking at it.
I dont know how to calculate inverse of trigonometry please make a video on it
So why is the range of inverse sine limited like that? Nice video by the way 👍🏼
To avoid confusion i.e multiple answers
So they fixed a range of arcsin function from -pi/2 to pi/2
One of the properties of functions with inverses is that they have to be bijective, but sin is only bijective in a certain range (-pi/2 to pi/2), so as a result the inverse has the same range. Sin is also bijective at an infinite number of places but we chose between +-pi/2 cos it contains the origin- so is nice.
because if you scetch a complete sine wave and try to do the inverse operation by drawing a parallel to the x axis through the sine value and intersecting it with the curve you will see: there are 2 such intersection points. So which one is the correct one? Both of them are correct, as the sin of (in this case 66.2° and 113.8°) both those intersection points have the same value. ( sin(66.2) is equal to sin(113.8) )
In other words, as someone has already pointed out: the sin function IS a function, but the inverse is not.
Edit: Something similar is true for the cos function. Scetch it and you will see that cos(x) is equal to cos(-x).
Horizontal line test.
What do you think about √-6 × √-4=√(-1)²× 24 = -√ 24 ?
This is the first video I could solve from the thumbnail..
I think you just have to remember that, if you use the law of sine to find an obtuse angle, you have to subtract your answer from 180º. It's an extra step, but so is using the law of cosine to solve that problem.
I Have A Lovely Problem In Mathematics You Should Look It Atleast One Time.
The Problem Is
Given- (X)^1/2 + (Y )^1/2 = (5)^1/2
Then y=f(x)=?
Conditiion Is the That Graph should Not Be Changed.
How Is This?
sin pi-x =sin x, so use the more logical one
I love questions like this because they are so deceivingly simple. You don't even need the angle to solve it.
As mentioned earlier, why bother with the Law of sines is law of cosines can be used (and doesn't deal with the ambiguous case). The problem would have been more interesting if the 5.8 would have been left out. Then the student would have to use the law of sines (SSA case)
uuh
let's call the height on c h (forming 2 right triangles)
sinA = h / 3.9 => h = sinA * 3.9 = 1.85
sinB = h / 3 = 1.85 / 3 => B = arcsin(1.85 / 3) = 37.97°
C = 180 - A - B = 180 - 28.25 - 37.97 = approx. 113.78
didn't need any law; neither sine nor cosine
You are the best teacher
Hey bprp where is your 100 questions with lambert w function ?
More complete version of the Pythagorean Theorem has a name: Al-Kashi theorem. It’s important that we use proper name.
I knew law of sines wouldn't work but I couldn't remember why. Now it makes sense.
I have a mysterious query in my mind that "We can make every negative value positive by adding it to 0."
(like -2+0 = 2) because 0 is greater than any negative value in number line and its rule in addition that we take sign of greater value with answer .... am I right ??
3:52 no . 0 has no sign
3:55
By giving them an extra angle not only did you give them a AAS problem you also gave them a Donkey Theorem problem which had “two solutions” when you ignore the third side. Also you need to keep in mind sine is also positive in the 2nd quadrant…
you have always been a very big inspiraion for me thanks🥰
... A good day to you, At about time 5:02 you used c = 5.9 in the cos rule formula instead of c = 5.8 ...
Ah you are right. Thanks for pointing that out.
@@blackpenredpen ... Good day to you, Thank you for showing the SSA (ambiguous) case and also in this case the difference in application between the sine and cosine rule (advantage -disadvantage) in your presentation; at last it has become clear to me the 2 possibilities of triangles (obtuse angle C or acute angle C; supplementary angles)! Thank you again for as always a great presentation and take care, Jan-W
Aaand that's why it's a good idea to use Wolfram|Alpha as your everyday calculator when your calculation is more complicated than 2+2
I saw the catch with sine coming from a mile away about the time the sine law was written on the board... I was like oh no it doesn't work that simple for angles bigger than 90°
Integral (arcsinx/x) when :)
Al kashi 😁😁
?
If there is an angle obove 90 , it will be C. So use sin-theorm with 3 and 3.9 .....
cos∠C = -943/2340
Law of sines gives us two possible angles C = 66.2 or C = 113.8 but one answer is wrong
6:35 Yes but you can get the other solution from reduction formula nevertheless one solution will be wrong
Use law of sin on C, get 66.2˚, ”hmm..this doesn’t seem right”
Use law of din on the other unknown angle, get 37.97, C then must be 113.78 QED
wasnt the colour of the triangle in the thumbnail blue not red or am i going crazy lol
Hello teacher,, I watch your every class for many days and you are going to my favourite math teacher ❤. But hilarious is that I don't know your name😅😂.
I am going to be your big fan, So please tell me your name as soon as possible 😊
Btw best wishes for you from Bangladesh ❤🎉
An insidious multiple choice question would give the wrong angle for A. Then you list the one answer for the law of cosines, two answers for the law of sines (obtuse and acute) and 90 degrees as options, along with the correct answer none of these, because the triangle does not exist.
I used the law of cosine because I think it's easier. Got lucky :)
واحد من اهم مبرهنات الهندسة مبرهنة الكاشي التي أخدها الغرب و نسبوها لهم لأن مبرهنة فيتاغوس مأخودة منها اصلا
Can you please solve this 🙏
X³ - 3x +1 = 0