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.
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.
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 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.
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.
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!
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
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 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
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.
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)
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.
@@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)
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.
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.
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.
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
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.
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
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
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 ☺️🌹
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.
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? :)
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).
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.
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!
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.
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 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°
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.
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…
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
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 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.
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.
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
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).
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)
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 😊
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 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?
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
@@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 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 ??
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
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 ❤🎉
That's true, and as someone who saw similarly non-existent geometry & trig questions, I can relate to getting inclined to check since I do so quite a lot with these sort of questions :) That said, I promise that I've seen WAY worse questions from BuzzFeed [may not be surprising] and in one of my former school modules [probably more surprising] 😬 I can definitely understand why BPRP would write 28.25° instead of the more precise 28.2540053...° since assuming the other values as exact, 28.25° would be close enough. Also, writing the more precise value will likely make the students less inclined to use the Law of Sines 🤷 BuzzFeed, on the other hand, had less excuses. One of the triangles they asked to find the area of can be shown to not exist via Pythagoras which was ironically what they asked about in the very next question from the same quiz! 😒 Also, one of my school modules asked me to solve for x in a perpendicular bisector problem involving congruent triangles. They really expected me to set the corresponding lengths 5x + 12 and 4x equal without realizing that their supposed answer of x = -12 would make both segments have a NEGATIVE LENGTH [not displacement] of -48 😭 So, compared to those triangular atrocities, BPRP's triangle is way more reasonable! But if you still want to use a more exact triangle question with the same tricky-ness, you could use the 3-7-8 triangle labelling the angle opposite the side length 7 as 60° and ask the students for the angle opposite the side length 8 👍
