Here BPRP I thought I should let you know i found two other methods to solving this. You can check here: cdchester.co.uk/wp-content/uploads/2018/06/Area-of-an-Inscribed-Triangle-4-Methods.pdf
Because Math is the language of universe and nature !! İf you don't know English, You don't understant people speak english only ;) But You don't know Mats, You don't understant anything !!!
You can do this alternative way to make things faster : Step 1 : by using the law of sin. For any triangle, the area will be 0.5*b*c*sin(A) where A is the angle between side b and c Step 2 : by splitting the the isosceles triangle in two the base and the angle will be split in exact half. Simply 0.5*b = r*sin(B)
@@supertester23 RHS = 2sin(A)sin(B)(2sin(C)) (some manipulation of original RHS 4sin(A)sin(B)sin(C) where 2^2 = 4 = 2sin(C)(cos(A-B) - cos(A+B)) (product to sum, 2sin(A)sin(B) = cos(A-B) - cos(A+B)) = 2sin(C)cos((A-B)) - 2sin(C)cos((A+B)) (expansion) = sin(C+(A-B)) + sin(C - (A-B)) - sin(C+A+B) - sin(C-A-B) (product to sum once again) = sin(C + A - B) + sin(C - A + B) - sin(A + B + C) - sin(C - (A+B)) (expansion of input of sin()) = sin(pi - B - B) + sin(A + B + C - 2A) - sin(pi) - sin(C - (pi - C)) (using the fact A + B + C = pi) = sin(pi - 2B) + sin(pi - 2A) - 0 - sin(C - pi + C) (sin(pi) = 0 and using the fact A + B + C = pi and expansion) = sin(2B) + sin(2A) - sin(2C - pi) (sine is positive in the second quadrant) = sin(2A) + sin(2B) + sin(pi - 2C) (sine is an odd function) = sin(2A) + sin(2B) + sin(2C) (sin(pi-x) = sin(x)) = LHS Alternatively, use the fact that BPRP's method and the method Heliocentric used to find the area must give the same area as they are the same triangle... so (1/2)r^2[sin(2A) + Sin(2B) + sin(2C)] = 2r^2sin(A)sin(B)sin(C) which means sin(2A)+sin(2B)+sin(2C) = 4sin(A)sin(B)sin(C) of course, if A + B + C = 180 degrees since we are dealing with triangles.
@@supertester23 Since this formula using the double sines gives the area, we know it is equal to the area BPRP derived in the video. By cancelling r^2 and multiplying by 2 you arrange to get the equality OP derived, and we have this true only for A+B+C = 180 because this is only true when the angles are angles of a triangle, which sum to 180.
Wow, then, in this time we can use the law of sine, a/sin(A)=b/sin(B)=c/sin(C)=2r, now formula changes like S/abc=2r^2 sin(A)/a sin(B)/b sin(C)/c=2r^2 (1/2r)^3=1/4r, so we can know that S=abc/4r ! :)
In a circle of radius 1 , for any inscribed triangle with sides length a , b , c We have : (a^4) + (b^4) + (c^4) - 2 * ( (a^2)*(b^2) + (a^2)*(c^2) + (b^2)*(c^2) ) + (a^2)*(b^2)*(c^2) = 0
To solve b, you could also break triangle OAC into 2 equal pieces and calculate the area of one of the pieces(little right triangle) 2 ways. You will get 1/4 bh = 1/2 rhsinB, which is b = 2rsinB, a bit easier than law of cosine.
I did the exact same thing you are going to do in the future, finding the largest inscribed triangle for any circumference, and it was a hell of an exercise! Good to see you started up just like I did a while back, and I'm very eager to see if you do anything better than I did, and I sincerely hope you destroy my proof so that I can learn from my mistakes! Keep making these awesome videos!
Ruby Duby, you're not far away. Use the fact that A+B+C=pi with the sum to product identity for sine and sine. Combine and simplify, you get 4sinAsinBsinC. Substitute back in your original expression and that's 2r^2.sinAsinBsinC
nice :) On a side note, i think if we equate the two formulas, we should end up getting a nice little property between angles. 2sinAsinBsinC = sin2A + sin2B + sin2C Itd be cool if this actually worked for any polygon besides triangle.
I think this is a pretty cool derivation for an expression for an area of a triangle inscribed in a circle. I just want to make one note about it; when you use the Law of Cosines in the triangle with vertices A, C, and the center of the circle, you assume that angle 2B is an angle in a triangle. This assumption translates to an assumption that 0 < 2B < 180 degrees, which is equivalent to 0 < B < 90 degrees. Therefore, the steps you have used in the derivation are rigorous only for angle B being acute, and since you argue the same things can be done with angle C to find an expression for c with r in it, the derivation also is only rigorous for the assumption as well that angle C is acute. I am pretty sure the final formula works for any triangle inscribed in the circle, but I think that there are a couple of additional cases to the problem to consider for doing the derivation by finding the sides of the triangle ABC you found, angle B being right or obtuse while angle C is still acute. I think the geometry and trigonometry of the derivation probably are slightly different to prove the final formula for angle B as right or obtuse. I hope you find this useful.
Construct a radius to each corner. The angle 'opposite' A (angle constructed by radii going to B and C ) is 2A, due to circle theorems. Area of each is 0.5 times the two sides times the Sine of the angle between them therefore, the area of the triangle is 0.5 r^2 (Sin 2A +Sin 2B + Sin 2C )
Once you have that the central angle is 2B, you could also extend the radius from C to a diameter and then use the fact that a triangle with one side a diameter is right. The angle next to b is 90 - B (because the central angle is 2B and the triangle is isosceles) and so the angle opposite b is also B, so since the hypotenuse is 2r you get b = 2r sin B.
Use inscribed angles to find angles from the circumcenter to each pair of points. Then write A=(r, 0), B=(r cos theta1, r sin theta1), and C = (r cos theta2, r sin theta2) and then just use the shoelace determinant to get it fast.
At the point where you've shown the area of a triangle as half the product of any two sides, times the sine of the included angle; and you've also divided the triangle into 3 (isosceles) subtriangles, joined at the center, O, of the circle; and you have all 3 apex angles of those; you can just apply that area formula to all 3 isosceles triangles, and get Area = ½r²[sin(2A) + sin(2B) + sin(2C)] Note that this works even when ∆ABC is obtuse, so that O is outside it. All that happens then, is that the obtuse angle contributes a negative area to ∆ABC, which is correct. So the really interesting thing here is, that when angles A + B + C = π, then from your result and this one, we get the identity: sin(2A) + sin(2B) + sin(2C) = 4 sin(A) sin(B) sin(C) Fred
I am ur biggest fan. The way u solve questions....it’s amazing thanks for coming on TH-cam. And for this video ....thanks. This would help me in many olympiads in my 10th grade
From AM-GM inequality (since sin alfa, sin beta, sin gamma are positive) we have that the product of those sines is less than or equal to (sin alfa + sin beta + sin gamma)^3 / 27 and because sine is concave on the interval (0, pi) we have from Jensen's inequality that sin alfa + sin beta + sin gamma
Skip the law of cosines. To calculate b, drop a perpendicular from the center (D) to chord b. Because ADC is isocelese, CAD = ADC. Right angles at crossing, imply angle on each side of perpendicular are equal and therefore b. Then opposite side is clearly 2 sin b. Same for c.
Here's how I'd do it. The triangle equals the circle minus three slices. The area of slice a is r²(2A - sin 2A)/2. Similarly for slices b and c. Subtract all three from the area of the circle, the r²(A+B+C) cancels the area of the circle, and you're left with r²(sin 2A + sin 2B + sin 2C)/2.
Much quicker to do like this: It's known that abc=4R△ where a, b, and c are the side-lengths. -- (i) It's also known that a=2RsinA, b=2RsinB, and c=2RsinC Using these in (i), we get: (2RsinA)(2RsinB)(2RsinC)=4R△ Or, △=(2RsinA)(2RsinB)(2RsinC)/4R=2R²sinAsinBsinC
Hi blackpenredpen Can you help me with this problem? There is an isosceles triangle, in which the sine of the angle from the base is 3 time bigger that cosine of the angle from the top. What's the value of the sine of the angle from the base?
you could have done this withouut doing law of cosines by considering the triangle AOB. there it is an isosceles triangle. with an angle between them is 2B. now we draw a angle bisector to the angle2B. it bisects the opposite side perpendicularly and equally. so now take sin(B) which would be equal to b/2r. so, 2r ×sin(B)=b
The section 4:20 to 10:28 where you calculate b given the angle B and the radius r was needlessly complex. No need to invoke the law of the cosigns and complex algebra. Let Geometry do your work for you! Look at your diagrams! Let the center be labeled D. The triangle ADC is an isosceles triangle, have 2 sides of length r. Bisect the angle at D. This will split your isosceles triangle into 2 right triangles. The angle at the center will have measure B, because you have already noted that angle ADC has measure 2B because of central angle theorem. The hypotenuse of these right triangles is length r. The side opposite this angle has length r sin(B). Because right triangle and definition of sine. But the base is 2 of these segments. Thus the length of the base is 2r sin(B). This is much easier than fooling around with the law of the cosigns and algebra!
If you have the values if the three angles and r you can get the area of the three circular segments of the circle (AC, CB, BA) then take it out of the area of the whole circle in easy four steps, in another hand that's enjoyable good job 💜
You could have used S=a*b*c/4R. From sine rule we have a/sinA= b/sinB=c/sinC= 2R From there we have a=2RsinA,b=2RsinB ,c=2RsinC S=a*b*c/4R=2RsinA*2RsinB*2RsinC/4R=2R^2sinAsinBsinC
Now that you know the maximum triangle area possible in a full circle....... What about a semi-circle? Or maybe we can generalise.....if we have a sector of angle theta radians....what is the area of the largest triangle possible?
i found a triangle area formula if you know a side length and all 3 angles s = side length a1 = an angle adjacent to side s a2 = the other angle adjacent to side s o = the angle opposite side s formula : s^2 sin(a1) sin(a2) / 2 sin(o)
Thanks a lot bro you really helped me with my math homework and also helped me widen my interest in mathematics and also additional mathematics which is like a subject in IGCSE cambridge. Thanks
Haven't watched the video yet For the proof of the equilateral triangle being the biggest, I would say that for any chord in the circle, the point which will make the triangle with the largest area has to be the furthest away from the line so thet the height is the biggest From then, you can choose a random chord and start iterating, which will result in all of the heights in the triangle being equal due to symmetry, which means the triangle is equilateral
Sir when I solved I got expression like this Half of. R square (sin 2A +sin2B +sin2C) Is it correct I just use formula area of triangle = half xy sin theta where theta is angle between x and y
This was a pretty long approach honestly.. It's way way simpler if you use the "chord theorem" (that is, if an angle on a circumference alpha is opposite to a chord L then L = 2r*sin(alpha) ). Hence one has a = 2r*sin(A), b=2r*sin(B) and know since the height with respect to a is h = b*sin(C) we have area = 0.5*a*h = 0.5*2*r*sin(A)*2*r*sin(B)*sin(C) = 2*(r^2)*sin(A)*sin(B)*sin(C). It's almost a 1 line proof
If r*sin A•sin B•sin C = π Then P of tringle is 2πR and the area of tringle =area of circel when R=π/(sin A• •sin B•sin C).? Its could be really so cool
the double angle formula for cosine: cos(2b) = cos^2 (b) - sin^2 (b). then use the pythagorean identity to rewrite cos^2 (b) as 1 - sin^2 (b), so cos(2b) = 1 - 2sin^2 (b), which is the identity he used (the way you have it written isn't true, the left hand side should just be cos(2b). if you want more of these, google trig identities
Hello!! Awesome video once again. However, we can do this much simpler!! Draw in the radii. The angles at the centre will be twice the angles at the circumference. So, total area is: A= ((r^2)/2)(sin(2A) + sin(2B) + sin(2C)) Isn't this nicer!?
Sunny Dama it's certainly faster, but the result he got looks neater. When you have formulas and you work through you want to get a neater result. And rather than use trigonometric identities to convert your answer to his, perhaps he wanted to work with cosine rule to benefit viewers
Shreyas Sarangi , actually using the cosine rule was unnecessarily complicated. Much simpler would have been to drop a perpendicular to bisect the iscoceles triangle into two right-angled triangles giving sin B = b/2 / r immediately. On the other hand, if you combine the two methods, you get a neat proof of his formula for cos 2B.
Well … because I am not as smart, I found another solution that works, fairly abstractly. From each point A, B and C, draw a radius to center point O. Each length is 'radius' or 𝒓 if you prefer. Because of the central-vs-edge angle theorem, we can say that № 1.1: 2∠CAB = ∠COB; № 1.2: 2∠ACB = ∠AOB; № 1.3: 2∠BAC = ∠BOC; Alright, those central angles are twice the measure of the perimeter angles A, B an C, respectively. The next step however, is central to having this work… № 2.1: ∠AOB divided by 2 is 2∠ACB/2 which is ∠C Which is the same as the others № 2.2: ∠BOC divided by 2 is 2∠BAC/2 which is ∠A № 2.3: ∠AOC divided by 2 is 2∠ABC/2 which is ∠B however, each of the corresponding segments AB, BC and AC are each divided in half with those half-angles. Since the area of each sub-triangle is ½BH, then just need to find the B and H for each: № 3.1: AB = 2 × r cos C № 3.2: BC = 2 × r cos A № 3.3: AC = 2 × r cos B And the heights of each respective are № 4.1: O-to-AB = 2 × r sin C № 3.2: O-to-BC = 2 × r sin A № 3.3: O-to-AC = 2 × r sin B Then the triangles come together Area △ AOB = ½ BH = ½ × 2 × r cos C • r sin C = r² cos C sin C Area △ BOC = ½ BH = ½ × 2 × r cos A • r sin A = r² cos A sin A Area △ AOC = ½ BH = ½ × 2 × r cos B • r sin B = r² cos B sin B Which reduce, each, to Area △ ABC = r² ( cos C sin C + cos B sin B + cos A sin A ); And that is where I would be temped to stop. However, it is not as nice a form as the one RedPenBlackPen came up with. TESTING it with a whole lot of randomly chosen angles showed that the above is also the same as Area △ ABC = 2 r² sin A sin B sin C; So, there we are. Two different unequally memorable solutions. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
Couldn't you just use Law of Sinus a/sin(a)=b/sin(b)=c/sin(c)=2r where r is radius of the circle triangle is inscribed in You just made easier problem bit harder, but its ok, you can get either one of Laws from each other.
I have a question, if anyone can help: Find third side of a triangle which has sides a=3; b=5, area= 6. One answer is 4, but how can I find the other solution (it is 2*sqrt(13), but I do not know how, I just have it as an answer in my book).
I can find it geometrically. Construct concentric circles of radii 3 and 5 and center O. Construct a radius OA in the 3 circle. Construct a line perpendicular to OA, passing though A. Call the intersection of that line and the 5 circle B. OB is a radius of length 5. By the Pythagorean theorem, AB has length 4. First solution. Construct a line through B parallel to OA. Call the other intersection of that line with the 5 circle C. OC has length 5. Construct a line through O perpendicular to BC and call its intersection with BC D. OABD is a rectangle, so OD = AB = 4. Triangles CDO and BDO are right triangles with side 4 and hypotenuse 5, so CD and BD are both 3. BC = 6 ABC is a right triangle so AC^2 = AB^2 + BC^2. That is, AC^2 = 4^2 + 6^2 = 52 So AC = 2sqrt(13). Second solution.
Heron's formula is conventional. I didn't use it because I assumed it would result in a horrible quartic equation. But it turned to be two nested quadratics.
Heron's formula is probably the way to go here. The other one is using A= (0.5)absinC. You know A, a and b, so you'll get a value for sinC which gives you two values for C between 0° and 180°. You can then use the cosine rule to fine the side c. The first angle gives you 5 and the second one gives you the irrational answer.
I forgot there's no autocorrect on whiteboard...
How to find average area of above inscribed triangle in circle of radius r?
xD
blackpenredpen
Parker Triangle
mega crossover event
Here BPRP I thought I should let you know i found two other methods to solving this. You can check here:
cdchester.co.uk/wp-content/uploads/2018/06/Area-of-an-Inscribed-Triangle-4-Methods.pdf
*English* is important but *Math* is importanter.
The englisher you speak, the gooder you become.
Because Math is the language of universe and nature !!
İf you don't know English, You don't understant people speak english only ;)
But You don't know Mats, You don't understant anything !!!
Chris Sándor Kacsó Math is important but english is 1 times more important.
@@supercool1312 OMEGALUL
*Importanterer
You can do this alternative way to make things faster :
Step 1 : by using the law of sin.
For any triangle, the area will be 0.5*b*c*sin(A) where A is the angle between side b and c
Step 2 : by splitting the the isosceles triangle in two
the base and the angle will be split in exact half. Simply 0.5*b = r*sin(B)
Drestanto , I noticed that by combining the two methods we get a simple proof of the double angle formula for cos2A used in BPRP's method.
you'd have to prove that the triangle is indeed isosceles first then.
تمرين جميل جيد. رسم واضح. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
Also, Area= (1/2) r^2 [sin(2A)+sin(2B)+sin(2C)]
Which means If A+B+C=180, then *sin(2A)+sin(2B)+sin(2C) =4sin(A)sin(B)sin(C)*
Coool
I got that result as well by finding the area of each isosceles triangle and adding up the results
@@supertester23
RHS = 2sin(A)sin(B)(2sin(C)) (some manipulation of original RHS 4sin(A)sin(B)sin(C) where 2^2 = 4
= 2sin(C)(cos(A-B) - cos(A+B)) (product to sum, 2sin(A)sin(B) = cos(A-B) - cos(A+B))
= 2sin(C)cos((A-B)) - 2sin(C)cos((A+B)) (expansion)
= sin(C+(A-B)) + sin(C - (A-B)) - sin(C+A+B) - sin(C-A-B) (product to sum once again)
= sin(C + A - B) + sin(C - A + B) - sin(A + B + C) - sin(C - (A+B)) (expansion of input of sin())
= sin(pi - B - B) + sin(A + B + C - 2A) - sin(pi) - sin(C - (pi - C)) (using the fact A + B + C = pi)
= sin(pi - 2B) + sin(pi - 2A) - 0 - sin(C - pi + C) (sin(pi) = 0 and using the fact A + B + C = pi and expansion)
= sin(2B) + sin(2A) - sin(2C - pi) (sine is positive in the second quadrant)
= sin(2A) + sin(2B) + sin(pi - 2C) (sine is an odd function)
= sin(2A) + sin(2B) + sin(2C) (sin(pi-x) = sin(x))
= LHS
Alternatively, use the fact that BPRP's method and the method Heliocentric used to find the area must give the same area as they are the same triangle...
so (1/2)r^2[sin(2A) + Sin(2B) + sin(2C)] = 2r^2sin(A)sin(B)sin(C)
which means sin(2A)+sin(2B)+sin(2C) = 4sin(A)sin(B)sin(C)
of course, if A + B + C = 180 degrees since we are dealing with triangles.
@@supertester23 Since this formula using the double sines gives the area, we know it is equal to the area BPRP derived in the video. By cancelling r^2 and multiplying by 2 you arrange to get the equality OP derived, and we have this true only for A+B+C = 180 because this is only true when the angles are angles of a triangle, which sum to 180.
By using law of sines,
Once you get to Area = 1/2 bcsinA
b/sinB = c/sinC = 2r
b=2r sinB c=2rsinC
Area = 1/2 4r² sinB sinC sinA
= 2r²sinAsinBsinC
Wow, then, in this time we can use the law of sine, a/sin(A)=b/sin(B)=c/sin(C)=2r, now formula changes like S/abc=2r^2 sin(A)/a sin(B)/b sin(C)/c=2r^2 (1/2r)^3=1/4r, so we can know that S=abc/4r ! :)
Another Way:
Use Extended Law of Sines:
a/sinA=b/sinB=c/sin C=2R
And make appropriate substitutions for b, c.
Ah!!! That's very nice!
blackpenredpen Yeah, It comes in pretty handy in Olympiads.
You can use it to prove that abc/4R formula as well.
That is really nice. This makes me miss school level geometry. Great video my friend. Keep up the good work.
In a circle of radius 1 , for any inscribed triangle with sides length a , b , c We have : (a^4) + (b^4) + (c^4) - 2 * ( (a^2)*(b^2) + (a^2)*(c^2) + (b^2)*(c^2) ) + (a^2)*(b^2)*(c^2) = 0
Bro really rocking the supreme
To solve b, you could also break triangle OAC into 2 equal pieces and calculate the area of one of the pieces(little right triangle) 2 ways. You will get 1/4 bh = 1/2 rhsinB, which is b = 2rsinB, a bit easier than law of cosine.
Math is love, math is life😍
Maths is always so beautiful with you sir
I did the exact same thing you are going to do in the future, finding the largest inscribed triangle for any circumference, and it was a hell of an exercise! Good to see you started up just like I did a while back, and I'm very eager to see if you do anything better than I did, and I sincerely hope you destroy my proof so that I can learn from my mistakes! Keep making these awesome videos!
When you are a math god, but can't spell similarly lol
great video! amazing as always :)
He also thinks 3 is smaller than 2
AndDiracisHisProphet sometimes
maybe depends on the metric?
I would really like to be a math god than not being able to spell
LOL
steve* uses blue
me* unsubscribes
User unknown :(
blackpenredpen come on, you know I can't do that ))
: )
sorry for the late reply, i just saw ur reply
Genius at work....love your channel.
Balaji Shankar thank you!
Arc AC was supposed to be the 2B since B is inscribed about the circle. By the central angle theorem if O is the center then
damn, I got so close I got
r^2 * [sin(2A) + sin(2B) + sin(2C)]
That said, wouldn't the formula i got be simpler for calculus?
Ruby Duby, you're not far away. Use the fact that A+B+C=pi with the sum to product identity for sine and sine. Combine and simplify, you get 4sinAsinBsinC. Substitute back in your original expression and that's 2r^2.sinAsinBsinC
nice :)
On a side note, i think if we equate the two formulas, we should end up getting a nice little property between angles.
2sinAsinBsinC = sin2A + sin2B + sin2C
Itd be cool if this actually worked for any polygon besides triangle.
Ruby Duby it's actually your furmula multiplied by 1/2
It's actually the same thing and it is correct.
I think this is a pretty cool derivation for an expression for an area of a triangle inscribed in a circle. I just want to make one note about it; when you use the Law of Cosines in the triangle with vertices A, C, and the center of the circle, you assume that angle 2B is an angle in a triangle. This assumption translates to an assumption that 0 < 2B < 180 degrees, which is equivalent to 0 < B < 90 degrees. Therefore, the steps you have used in the derivation are rigorous only for angle B being acute, and since you argue the same things can be done with angle C to find an expression for c with r in it, the derivation also is only rigorous for the assumption as well that angle C is acute. I am pretty sure the final formula works for any triangle inscribed in the circle, but I think that there are a couple of additional cases to the problem to consider for doing the derivation by finding the sides of the triangle ABC you found, angle B being right or obtuse while angle C is still acute. I think the geometry and trigonometry of the derivation probably are slightly different to prove the final formula for angle B as right or obtuse. I hope you find this useful.
Construct a radius to each corner. The angle 'opposite' A (angle constructed by radii going to B and C ) is 2A, due to circle theorems. Area of each is 0.5 times the two sides times the Sine of the angle between them therefore, the area of the triangle is 0.5 r^2 (Sin 2A +Sin 2B + Sin 2C )
Man , You are awesome. Please upload more geometric video like euclidian, Non-euclidian and other geometry.
I love Lagrange Multipliers.
Not Your Average Nothing me too. Coming tmr!
Once you have that the central angle is 2B, you could also extend the radius from C to a diameter and then use the fact that a triangle with one side a diameter is right. The angle next to b is 90 - B (because the central angle is 2B and the triangle is isosceles) and so the angle opposite b is also B, so since the hypotenuse is 2r you get b = 2r sin B.
Very nice presentation.
Furthermore, because it is equilateral triangle, angles A,B & C are same. So it can be SinA^3.
Use inscribed angles to find angles from the circumcenter to each pair of points. Then write A=(r, 0), B=(r cos theta1, r sin theta1), and C = (r cos theta2, r sin theta2) and then just use the shoelace determinant to get it fast.
At the point where you've shown the area of a triangle as half the product of any two sides, times the sine of the included angle;
and you've also divided the triangle into 3 (isosceles) subtriangles, joined at the center, O, of the circle;
and you have all 3 apex angles of those;
you can just apply that area formula to all 3 isosceles triangles, and get
Area = ½r²[sin(2A) + sin(2B) + sin(2C)]
Note that this works even when ∆ABC is obtuse, so that O is outside it. All that happens then, is that the obtuse angle contributes a negative area to ∆ABC, which is correct.
So the really interesting thing here is, that when angles A + B + C = π, then from your result and this one, we get the identity:
sin(2A) + sin(2B) + sin(2C) = 4 sin(A) sin(B) sin(C)
Fred
What is that thing in your hand
What an interesting and pleasant result.
I am ur biggest fan. The way u solve questions....it’s amazing thanks for coming on TH-cam. And for this video ....thanks. This would help me in many olympiads in my 10th grade
7:15 That's vector subtraction
MIND BLOWN
what's the largest area of a triangle inscribed in a circle with radius r?
That's coming up tomorrow.
It has to be equilateral tho.
A = ¾ r² √3
From AM-GM inequality (since sin alfa, sin beta, sin gamma are positive) we have that the product of those sines is less than or equal to (sin alfa + sin beta + sin gamma)^3 / 27 and because sine is concave on the interval (0, pi) we have from Jensen's inequality that sin alfa + sin beta + sin gamma
Netherman 41 Nice proof :)
Can you send the link?
i like math,physic and chemistry but i like physics more but bro your channel is superb for math ,,👍👍👍👌
The same result can be obtained by simply using Area=(1/2)*a*b*sinC and the extended sine law
Skip the law of cosines. To calculate b, drop a perpendicular from the center (D) to chord b. Because ADC is isocelese, CAD = ADC. Right angles at crossing, imply angle on each side of perpendicular are equal and therefore b. Then opposite side is clearly 2 sin b. Same for c.
Here's how I'd do it. The triangle equals the circle minus three slices. The area of slice a is r²(2A - sin 2A)/2. Similarly for slices b and c. Subtract all three from the area of the circle, the r²(A+B+C) cancels the area of the circle, and you're left with r²(sin 2A + sin 2B + sin 2C)/2.
Much quicker to do like this:
It's known that abc=4R△ where a, b, and c are the side-lengths. -- (i)
It's also known that a=2RsinA, b=2RsinB, and c=2RsinC
Using these in (i), we get:
(2RsinA)(2RsinB)(2RsinC)=4R△
Or, △=(2RsinA)(2RsinB)(2RsinC)/4R=2R²sinAsinBsinC
Little idea to make the formula need less input: just replace the angle C by 180° - A - B, like
Area = 2r² * sin A * sin B * sin (180° - A - B)
Beautiful! I don't like planimetrics very much but this is outstanding!
how can we find area of the inscribed triangle in the trapezoid??
given vertices are: (8,6) (4,0) and (0,3)
Did you edit the video because there is an excessive amount of motionblurr when you turn your head.
Hi blackpenredpen
Can you help me with this problem?
There is an isosceles triangle, in which the sine of the angle from the base is 3 time bigger that cosine of the angle from the top.
What's the value of the sine of the angle from the base?
This man loves his video's comments.
Very, very satisfying. Thanks!!
Fav math teacher 😃
you could have done this withouut doing law of cosines by considering the triangle AOB. there it is an isosceles triangle. with an angle between them is 2B. now we draw a angle bisector to the angle2B. it bisects the opposite side perpendicularly and equally. so now take sin(B) which would be equal to b/2r. so, 2r ×sin(B)=b
The section 4:20 to 10:28 where you calculate b given the angle B and the radius r was needlessly complex. No need to invoke the law of the cosigns and complex algebra. Let Geometry do your work for you! Look at your diagrams!
Let the center be labeled D. The triangle ADC is an isosceles triangle, have 2 sides of length r. Bisect the angle at D. This will split your isosceles triangle into 2 right triangles. The angle at the center will have measure B, because you have already noted that angle ADC has measure 2B because of central angle theorem. The hypotenuse of these right triangles is length r. The side opposite this angle has length r sin(B). Because right triangle and definition of sine. But the base is 2 of these segments. Thus the length of the base is 2r sin(B).
This is much easier than fooling around with the law of the cosigns and algebra!
If you have the values if the three angles and r you can get the area of the three circular segments of the circle (AC, CB, BA) then take it out of the area of the whole circle in easy four steps, in another hand that's enjoyable good job 💜
You could have used S=a*b*c/4R.
From sine rule we have a/sinA= b/sinB=c/sinC= 2R
From there we have a=2RsinA,b=2RsinB ,c=2RsinC
S=a*b*c/4R=2RsinA*2RsinB*2RsinC/4R=2R^2sinAsinBsinC
Blackpenredpen uses bluepen(*´∇｀*)
And Ur explaination is just like ur smile.
Now that you know the maximum triangle area possible in a full circle.......
What about a semi-circle? Or maybe we can generalise.....if we have a sector of angle theta radians....what is the area of the largest triangle possible?
I salutes your brain . I am from india
You are the Wizard of mathematics.
i found a triangle area formula if you know a side length and all 3 angles
s = side length
a1 = an angle adjacent to side s
a2 = the other angle adjacent to side s
o = the angle opposite side s
formula : s^2 sin(a1) sin(a2) / 2 sin(o)
Side length ‘a’ be like
''Why I'm still here?! Just to suffer''
When a video math is great, it's no problem the language!
actually do u have a curriculum of what u teach in ur vids?
Marvelous!
Thanks a lot bro you really helped me with my math homework and also helped me widen my interest in mathematics and
also additional mathematics which is like a subject in IGCSE cambridge.
Thanks
where is the next video
If triangle is on the plane in Euclidean space we can get rid of one angle
Area = 2r^2sinAsinBsin(A+B)
Wait you had 2r^2 and then in under the root sign it became 4r^2? Shouldn't it have been (2r)^2 then?
Haven't watched the video yet
For the proof of the equilateral triangle being the biggest, I would say that for any chord in the circle, the point which will make the triangle with the largest area has to be the furthest away from the line so thet the height is the biggest
From then, you can choose a random chord and start iterating, which will result in all of the heights in the triangle being equal due to symmetry, which means the triangle is equilateral
Prove the central angle please I love your videos 🎊🎊💪🏼
XmatterX it's already in the description :)
More geometry and Calc III PLEASE!!!!
Well with only r you can't tell what's the area. You must know any 3 of (the side lengths or the angles).
Sir when I solved I got expression like this
Half of. R square (sin 2A +sin2B +sin2C)
Is it correct I just use formula area of triangle = half xy sin theta where theta is angle between x and y
Does this work if the center of the circle is not inside of the inscribed triangle?
Great question and I should have covered that as well.
And the answer is yes.
Came be done using lami theorem and area=ABC/4r
This was a pretty long approach honestly.. It's way way simpler if you use the "chord theorem" (that is, if an angle on a circumference alpha is opposite to a chord L then L = 2r*sin(alpha) ).
Hence one has a = 2r*sin(A), b=2r*sin(B) and know since the height with respect to a is h = b*sin(C) we have area = 0.5*a*h = 0.5*2*r*sin(A)*2*r*sin(B)*sin(C) = 2*(r^2)*sin(A)*sin(B)*sin(C).
It's almost a 1 line proof
Nicola Rares Franco good
thanks and by the way, I really like the style of your videos!
If r*sin A•sin B•sin C = π
Then P of tringle is 2πR and the area of tringle =area of circel when R=π/(sin A• •sin B•sin C).? Its could be really so cool
Math rules and you rock!
Brilliant
law of cosine was overkill a bit :p you can use twice the projection with sin(B) on b. :)
I love this video
5:44 proof, please
AndDiracisHisProphet it's in the description already! :)
thanks^^
Great video. Can someone please explain to me why 1-cos(2b)=1-2sin^2(B)? Where can I find more manipulations like that?
the double angle formula for cosine: cos(2b) = cos^2 (b) - sin^2 (b). then use the pythagorean identity to rewrite cos^2 (b) as 1 - sin^2 (b), so cos(2b) = 1 - 2sin^2 (b), which is the identity he used (the way you have it written isn't true, the left hand side should just be cos(2b). if you want more of these, google trig identities
short form for similarly used to be "///ly" don't know if this is widely used though? saves you worrying about spelling !!!!
Hello!!
Awesome video once again.
However, we can do this much simpler!!
Draw in the radii. The angles at the centre will be twice the angles at the circumference. So, total area is:
A= ((r^2)/2)(sin(2A) + sin(2B) + sin(2C))
Isn't this nicer!?
Sunny Dama it's certainly faster, but the result he got looks neater. When you have formulas and you work through you want to get a neater result. And rather than use trigonometric identities to convert your answer to his, perhaps he wanted to work with cosine rule to benefit viewers
Shreyas Sarangi , actually using the cosine rule was unnecessarily complicated. Much simpler would have been to drop a perpendicular to bisect the iscoceles triangle into two right-angled triangles giving sin B = b/2 / r immediately. On the other hand, if you combine the two methods, you get a neat proof of his formula for cos 2B.
So, find circumcenter first, and you don't need the circle. Cool. :)
msolec2000 Point of concurrence of sides' perpendicular bisectors? Something like that?
You could simplify the formula, using Sin law.
Nice, this is the sinuses theorem, isn't it?, I remember when a was young, thanks mate. Now the Heron formula, ;-)
Nice video
Well … because I am not as smart, I found another solution that works, fairly abstractly.
From each point A, B and C, draw a radius to center point O. Each length is 'radius' or 𝒓 if you prefer.
Because of the central-vs-edge angle theorem, we can say that
№ 1.1: 2∠CAB = ∠COB;
№ 1.2: 2∠ACB = ∠AOB;
№ 1.3: 2∠BAC = ∠BOC;
Alright, those central angles are twice the measure of the perimeter angles A, B an C, respectively. The next step however, is central to having this work…
№ 2.1: ∠AOB divided by 2 is 2∠ACB/2 which is ∠C
Which is the same as the others
№ 2.2: ∠BOC divided by 2 is 2∠BAC/2 which is ∠A
№ 2.3: ∠AOC divided by 2 is 2∠ABC/2 which is ∠B
however, each of the corresponding segments AB, BC and AC are each divided in half with those half-angles.
Since the area of each sub-triangle is ½BH, then just need to find the B and H for each:
№ 3.1: AB = 2 × r cos C
№ 3.2: BC = 2 × r cos A
№ 3.3: AC = 2 × r cos B
And the heights of each respective are
№ 4.1: O-to-AB = 2 × r sin C
№ 3.2: O-to-BC = 2 × r sin A
№ 3.3: O-to-AC = 2 × r sin B
Then the triangles come together
Area △ AOB = ½ BH = ½ × 2 × r cos C • r sin C = r² cos C sin C
Area △ BOC = ½ BH = ½ × 2 × r cos A • r sin A = r² cos A sin A
Area △ AOC = ½ BH = ½ × 2 × r cos B • r sin B = r² cos B sin B
Which reduce, each, to
Area △ ABC = r² ( cos C sin C + cos B sin B + cos A sin A );
And that is where I would be temped to stop. However, it is not as nice a form as the one RedPenBlackPen came up with. TESTING it with a whole lot of randomly chosen angles showed that the above is also the same as
Area △ ABC = 2 r² sin A sin B sin C;
So, there we are.
Two different unequally memorable solutions.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Prove Morley's theorem.
Thank You
You're great!
Thanks
Couldn't you just use Law of Sinus
a/sin(a)=b/sin(b)=c/sin(c)=2r where r is radius of the circle triangle is inscribed in
You just made easier problem bit harder, but its ok, you can get either one of Laws from each other.
👌 very nice
Love your Videos! My first Intuition would have been to use polar cordinates. Maybe you'll find the time to cover that in an other video :)
Thanks sir
very nice!
1:03 that triangle definitely doesn't look better xD
Hey nice free hand circle
soooo good!
I have a question, if anyone can help: Find third side of a triangle which has sides a=3; b=5, area= 6. One answer is 4, but how can I find the other solution (it is 2*sqrt(13), but I do not know how, I just have it as an answer in my book).
Crazy Drummer I am at the gym so I don't have pen and paper to work this out. Have u tired Heron's formula?
I can find it geometrically.
Construct concentric circles of radii 3 and 5 and center O.
Construct a radius OA in the 3 circle.
Construct a line perpendicular to OA, passing though A. Call the intersection of that line and the 5 circle B.
OB is a radius of length 5. By the Pythagorean theorem, AB has length 4. First solution.
Construct a line through B parallel to OA.
Call the other intersection of that line with the 5 circle C.
OC has length 5.
Construct a line through O perpendicular to BC and call its intersection with BC D.
OABD is a rectangle, so OD = AB = 4.
Triangles CDO and BDO are right triangles with side 4 and hypotenuse 5, so CD and BD are both 3.
BC = 6
ABC is a right triangle so AC^2 = AB^2 + BC^2. That is, AC^2 = 4^2 + 6^2 = 52
So AC = 2sqrt(13). Second solution.
Thank you both for the ideas (solutions), but I am certain that there is a way of doing it using convencional formulas.
Heron's formula is conventional. I didn't use it because I assumed it would result in a horrible quartic equation. But it turned to be two nested quadratics.
Heron's formula is probably the way to go here. The other one is using A= (0.5)absinC. You know A, a and b, so you'll get a value for sinC which gives you two values for C between 0° and 180°. You can then use the cosine rule to fine the side c. The first angle gives you 5 and the second one gives you the irrational answer.
Area is also equal to:
A=abc/4r
"Always Be Careful 4 Rapist"
The most confusing part is... why did he use the law of cosines to prove the law of sines? He could have used it right away, couldn't he?
You have assumed that the interior of the triangle contains the centre of the circle. Proof may have to be adapted if this is not the case.
In the central angle theorem video he has proven it's also true when the center is outside the inscribed angle.