Thanks my friend for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Thanks John dear for the visit! I put my heart and soul in this video. You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Thanks Amir dear for the visit! I put my heart and soul in this video. You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Thank you for a thorough and well explained proof of a formula that I never knew about. As for the the example, rearranging the formula is often not an easy task for students. I would encourage them to substitute the values first. After that there is only one unknown and the process of solving the equation becomes easier and more accessible. For the more able mathematicians in your audience, it’s also worth doing what you did. Thank you.
Draw a perpendicular line from point A to BC, and let the intersection point be E. Since triangles ABE and ACE are both right triangles and BC=21, we know that BE=5, EC=16, AE=12. We also know that DE=9, so using the triangle theorem for triangle AED, we can derive ED=15.
Stewart's Theorem will solve the problem nicely, as shown in the video. Another much easier solution can be found if you recognize a 13-20-21 triangle off the bat as two right triangles 5-12-13 and 16-12-20 (four times 3-4-5) back to back, sharing the side of length 12.
Wow, that's another pearl of wisdom, there seems to be a real sack full of geometric laws and theorems, and that's only the 2D stuff, makes you appreciate how amazing our world is, and how genius our creator giving us such an organism as the brain, si that we could try to understand it, we never will, of course, I'm convinced we were designed with some great plan in mind. Thanks for sharing, sometimes I can't believe how much better my mind works since following your lessons
Thanks Rahman dear for the visit! I put my heart and soul in this video. You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Thankfully we have mathematicians in the world. They are weird. Because of that they come up with the most esoteric relationships. Love it. My daughter is a math major at university. I rest my case. ☺
I have never heard about this theorem, the problem is easily solved by using Pythagoras theorem if we drop the hight from the angle A and consider two triangles with the common side (the hight).
I’ve come to this problem after two years and another method is drawing a parallel line to AB from D and the using Thales theorem and cosine law easily gives the answer
Choose X on BC with AX⊥BC & let AX = h. Area ABC = √{s(s-a)(s-b)(s-c)} = √{27 x 14 x 7 x 6} = √15876 = 126 = ½hx21 → h = 12 → BX = 5 → DX = 9 → AD = √(12² + 9²) = √125 = 15. No need for Stewart’s theorem (to be frank, premath, not only have I never heard of Stewart’s theorem, as so many here haven’t, but the result doesn’t seem to be general enough to deserve a special name).
Thanks my dear friend for the visit! I put my heart and soul in this video. You are awesome Ken 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
@@saurabhnegi4977 One can find cos(B) by the cosine theorem in triangle ABC. And by the same theorem from the triangle ABD to find AD.The formula is difficult to remember.
Hi can i ask a doubt why cos(a) = -cos(180-a) can i have a complete proof, anyway nice video i just saw this video on scrolling and it inspired me to check law of cosine and law of sine all by it's proofs. ❤️
*Estimated radius of the earth = 6,690 kms using Miles Davis’ Hopetoun Monument photo of the mountains in the background behind the Queensferry bridge and Stewart's theorem* We can use one of Miles Davis' photos, Stewart's theorem & Maple to compute the estimated radius of the earth. th-cam.com/video/imr2g89i3QY/w-d-xo.htmlm10s (Look at the frames around 12 mins 10 into seconds the video. These are mountains in Scotland. The video itself presents a qualitative argument that the Earth can't be flat using perspective. I took it further and estimated the radius of the Earth from the photo and using Google Earth and Peakfinder to obtain distances between the observation point, the bridge and the peaks). Stewart's Theorem - en.wikipedia.org/wiki/`Stewart%27s_theorem b^2*m+c^2*n = a(d^2+mn) Stewart's theorem can be used to estimate the radius of the earth from the photo as the earth is large relative to the distances between Hopetoun monument and mountains in the background, and that we can draw a line through Hopetoun monument through Queensferry to each of the mounting which is close to straight. The line through Hopetoun monument through Queensferry to Earl's Seat is almost straight. Also, each of the mountains used in the analysis appear to be "Almost" the same height as the Queensferry Bridge Central Tower. The variables for Stewart's theorem (see webpage, which are the same as it is in this video) are set up as follows: R = radius of the earth (this is what we are estimating) in kms. c = R +0.210, as Hopetoun monument is R + 0.210 kms from the earth's center. d = R +0.210, as the Queensferry Bridge Tower is R + 0.210 kms from the earth's center. b = R + height of the mountain (e.g. 570 kms for Meikle Bin). n = distance from Hopetoun monument to the Queensferry Bridge Tower. m = distance from the Queensferry Bridge Tower to the target mountain (e.g. 45.347 kms for Meilke Bin). a = n + m. *_Analysis 1:_* _Meikle Bin_ height 570 m (appears very slightly above the Tower Bridge on the photo). _Hopetoun Monument - Queensferry Bridge Tower_ distance 38.238 kms. _Queensferry Bridge Tower - Meikle Bin_ distance 45.347 kms. _Hopetoun Monument - Meikle Bin_ distance 83.7 kms (not used in the calculation). fsolve(38.238*(R+.570)^2+45.347*(R+.210)^2 = (45.347+38.238)*(45.347*38.238+(R+.210)^2), R); So, R = 5,263.955827 (kms). *_Analysis 2:_* _Earl's Seat_ height 578 m (appears very slightly above the Tower Bridge on the photo). _Hopetoun Monument - Queensferry Bridge Tower_ distance 38.238 kms. _Queensferry Bridge Tower - Earl's Seat_ distance 55.200 kms. _Hopetoun Monument - Earl's Seat_ distance 93.5 kms (not used in the calculation). fsolve(38.238*(R+.578)^2+55.200*(R+.210)^2 = (55.200+38.238)*(55.200*38.238+(R+.210)^2), R); So, R = 7,007.456001 (kms). *_Analysis 3:_* _Cairnoch Hill_ height 413 m (appears very slightly below the Tower Bridge on the photo). _Hopetoun Monument - Queensferry Bridge Tower_ distance 38.238 kms. _Queensferry Bridge Tower - Cairnoch Hill_ distance 42.557 kms. _Hopetoun Monument - Cairnoch Hill_ distance 80.9 kms (not used in the calculation). fsolve(38.238*(R+.413)^2+42.557*(R+.210)^2 = (42.557+38.238)*(42.557*38.238+(R+.210)^2), R); So, R = 8,468.636324 kms. *_Analysis 4:_* _Ben Lomond_ height 974 m (appears the same height as theTower Bridge on the photo). _Hopetoun Monument - Queensferry Bridge Tower_ distance 38.238 kms. _Queensferry Bridge Tower - Ben Lomond_ distance 78.684 kms. _Hopetoun Monument - Ben Lomond_ distance 116.5 kms (not used in the calculation). fsolve(38.238*(R+.974)^2+78.684*(R+.210)^2 = (78.684+38.238)*(78.684*38.238+(R+.210)^2), R); So, R = 6,020.278843 kms. *Let’s average these estimates* So, R (average) = (1/4)*(5263.955827+7007.456001+8468.636324+6020.278843); = 6,690.081748. So, the estimated radius of the earth (assuming it is a sphere) is *6,690 kms* (the average actual radius is 6,371 kms). Fairly close. It is over estimated by about 5%. Who need Eratosthenes and two sticks in the ground to measure the radius of the Earth when we have a Miles Davis photo and Stewart's theorem 😁.
Area ABC (formula Erone) = √p(p-a)(p-b)(p-c) , ove p è il semiperimetro di ABC e a , b, c sono le misure dei lati rispettivamente opposti a A,B,C Risulta area ABC =S =126, ma la stessa area è uguale anche al prodotto di BC per l’altezza AH=h , diviso 2 cioè S =21*h/2=126 da cui h=126*2/21=12 BH=(teor. Pitagora )= √(c2-h2)=√(169-144)=5 DH=BD-BH=14-5=9 AD=√(DH2+h2)=√(81+144)= √225= 15
Co oscendo i 3 lati conosco i 3 angoli del triangolo grande(teorama di carnot)... Quimdi dei triangoli piccoli conosco 2 lati e l'angolo compreso.... Perciò conosco il terzo lato x
Great proof! Thanks for sharing.
Thanks James for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Never heard of Stewart's theorem before. Very interesting! Thanks for the work you put in to your videos.
Thanks my friend for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
pythagorean theorem is the result of this theorem,when the triangle is isoscale.
I know , I have never heard about that theory!!!! I don’t know where it came from?
Well I had never even heard of Stewart's theorem until tonight and it's nice to see such an easy and elegant proof. It seems like a useful theorem.
U can use sin formula in this question to find the answer even faster
Very well explained step by step. Presentation faultless. A pleasure to read.
Thanks John dear for the visit! I put my heart and soul in this video.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
my favorite theorem.always liked it.
god power you.thanks.
whit respect.from Iran💐🙏
Thanks Amir dear for the visit! I put my heart and soul in this video.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
@@PreMath
you're so kind & professional.A real teacher.
In persian culture the teacher & the father are the most respectful persons.
Goodluck🙏💐
Thank you for a thorough and well explained proof of a formula that I never knew about. As for the the example, rearranging the formula is often not an easy task for students. I would encourage them to substitute the values first. After that there is only one unknown and the process of solving the equation becomes easier and more accessible. For the more able mathematicians in your audience, it’s also worth doing what you did. Thank you.
Thanks Hassan dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Love and prayers from the USA! 😃
Draw a perpendicular line from point A to BC, and let the intersection point be E. Since triangles ABE and ACE are both right triangles and BC=21, we know that BE=5, EC=16, AE=12. We also know that DE=9, so using the triangle theorem for triangle AED, we can derive ED=15.
DE = 9 ?,,
@@catherinekloos8531 14-5=9
Thanks dear for sharing this.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
you are still the best explainer one ever! love it.
Thanks dear friend for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Always good to see you dear
Didn't know about Mr. Stewart. Taught high school math years ago.
I used a,b,c and Law of Cosines to calculate cos C, then used LofC again with b,n and cos C, all of which are now known, to get d.
basically that's how u prove Stewart theorem using trigonometry.
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Stewart's Theorem will solve the problem nicely, as shown in the video. Another much easier solution can be found if you recognize a 13-20-21 triangle off the bat as two right triangles 5-12-13 and 16-12-20 (four times 3-4-5) back to back, sharing the side of length 12.
Great video. You really make math easier and simple for us. Thanks for the video♥️♥️♥️♥️
You are so welcome dear!
Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Wow, that's another pearl of wisdom, there seems to be a real sack full of geometric laws and theorems, and that's only the 2D stuff, makes you appreciate how amazing our world is, and how genius our creator giving us such an organism as the brain, si that we could try to understand it, we never will, of course, I'm convinced we were designed with some great plan in mind.
Thanks for sharing, sometimes I can't believe how much better my mind works since following your lessons
Nice one Prof.! But i have this question in mind: Is this Stewart Theorem also applicable to a right triangle?
We can find area of ABC (Heron) then we have area of ADC since it is 7:14 ... then we can find angle C that is acute . Then AD ...
Nice 👍
Thanks dear for the visit!
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
@@PreMathThank you. I do agree that this is a good way to present Stewar's formula.
Many many many many thanks for giving it. I have known it just now.
Thanks Rahman dear for the visit! I put my heart and soul in this video.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
I'm aspirants of CGL from India...when Gagan sir talk about this theorem...we came here and saw proof.. thanks a lot sir..for your hard work
Absolutely unknown theorem. Never told about at school. Interesting anyway. Thanks.
Thankfully we have mathematicians in the world. They are weird. Because of that they come up with the most esoteric relationships. Love it. My daughter is a math major at university. I rest my case. ☺
J Hill you are the best
Thanks dear for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Very informative video and nice explanation Sir.
Thanks Viyush dear friend for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Love and prayers from the USA! 😃
@@PreMath thanks sir
I first heard this theorem very nice theorem sir
We are all lifelong learners.
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I have never heard about this theorem, the problem is easily solved by using Pythagoras theorem if we drop the hight from the angle A and consider two triangles with the common side (the hight).
I’ve come to this problem after two years and another method is drawing a parallel line to AB from D and the using Thales theorem and cosine law easily gives the answer
Sir I am writing competition exams in India very useful sir thanks sir❤️
Choose X on BC with AX⊥BC & let AX = h. Area ABC = √{s(s-a)(s-b)(s-c)} = √{27 x 14 x 7 x 6} = √15876 = 126 = ½hx21 → h = 12 → BX = 5 → DX = 9 → AD = √(12² + 9²) = √125 = 15. No need for Stewart’s theorem (to be frank, premath, not only have I never heard of Stewart’s theorem, as so many here haven’t, but the result doesn’t seem to be general enough to deserve a special name).
Thank you sir, well explained👌🙏
Learnt something new from you.... ❤️❤️❤️
Thanks Hasan dear for the visit!
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
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@@PreMath ❤️❤️
Thank you very much, very helpful
Very elegant and clear.
Thanks my dear friend for the visit! I put my heart and soul in this video.
You are awesome Ken 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
Nice one - very good explanation
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very well done, thanks for sharing bro
Thanks dear friend for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Always good to see you dear
Можно найти cos(B) по теореме косинусов в треугольнике ABC. И по той же теореме из треугольника ABD найти AD.Формула сложная для запоминания.
Прокомментируйте, пожалуйста, на английском, чтобы вас могли понять
@@saurabhnegi4977 One can find cos(B) by the cosine theorem in triangle ABC. And by the same theorem from the triangle ABD to find AD.The formula is difficult to remember.
Very useful sir.
Glad to hear that
Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
What is the difference between Stewart's Theorem and Apollonius's Theorem?
I solved this problem using law of cosines twice
Fist time I used law of cosines for cosine of angle , second time for length of d
Very well! You are my idol!
Can u find the answer by using coordinate geometry or by using some other methods
Teach us derivation of ladder theorem
Why to memorize such a lengthy formula, if one can do the calculation with the COS-rule at points B or C?
Solved using cosine rule 🤟🤟🤟 , but great proof................ and what about proof of other theorems i.e, ladder theorem
Thanks Aryan dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
I'm working on other theorem as well. Stay tuned.
@@PreMath Always 😇😇😇 , but bring some olympiad level problem plzzzz
Thank you
Hi can i ask a doubt
why cos(a) = -cos(180-a)
can i have a complete proof,
anyway nice video i just saw this video on scrolling and it inspired me to check law of cosine and law of sine all by it's proofs. ❤️
Well d=-15 (negative 15) is also a solution. Thank you Sir. 👍
That,s not real. A length is positive in this case. Else AD is out of the triangle.
@@eladiond No, that's it's real is also negative area!!! 😁😁👍👍
🙏🙏🙏.. amazing 👍
*Estimated radius of the earth = 6,690 kms using Miles Davis’ Hopetoun Monument photo of the mountains in the background behind the Queensferry bridge and Stewart's theorem*
We can use one of Miles Davis' photos, Stewart's theorem & Maple to compute the estimated radius of the earth.
th-cam.com/video/imr2g89i3QY/w-d-xo.htmlm10s
(Look at the frames around 12 mins 10 into seconds the video. These are mountains in Scotland. The video itself presents a qualitative argument that the Earth can't be flat using perspective. I took it further and estimated the radius of the Earth from the photo and using Google Earth and Peakfinder to obtain distances between the observation point, the bridge and the peaks).
Stewart's Theorem - en.wikipedia.org/wiki/`Stewart%27s_theorem
b^2*m+c^2*n = a(d^2+mn)
Stewart's theorem can be used to estimate the radius of the earth from the photo as the earth is large relative to the distances between Hopetoun monument and mountains in the background, and that we can draw a line through Hopetoun monument through Queensferry to each of the mounting which is close to straight. The line through Hopetoun monument through Queensferry to Earl's Seat is almost straight. Also, each of the mountains used in the analysis appear to be "Almost" the same height as the Queensferry Bridge Central Tower.
The variables for Stewart's theorem (see webpage, which are the same as it is in this video) are set up as follows:
R = radius of the earth (this is what we are estimating) in kms.
c = R +0.210, as Hopetoun monument is R + 0.210 kms from the earth's center.
d = R +0.210, as the Queensferry Bridge Tower is R + 0.210 kms from the earth's center.
b = R + height of the mountain (e.g. 570 kms for Meikle Bin).
n = distance from Hopetoun monument to the Queensferry Bridge Tower.
m = distance from the Queensferry Bridge Tower to the target mountain (e.g. 45.347 kms for Meilke Bin).
a = n + m.
*_Analysis 1:_*
_Meikle Bin_ height 570 m (appears very slightly above the Tower Bridge on the photo).
_Hopetoun Monument - Queensferry Bridge Tower_
distance 38.238 kms.
_Queensferry Bridge Tower - Meikle Bin_
distance 45.347 kms.
_Hopetoun Monument - Meikle Bin_
distance 83.7 kms (not used in the calculation).
fsolve(38.238*(R+.570)^2+45.347*(R+.210)^2 = (45.347+38.238)*(45.347*38.238+(R+.210)^2), R); So, R = 5,263.955827 (kms).
*_Analysis 2:_*
_Earl's Seat_ height 578 m (appears very slightly above the Tower Bridge on the photo).
_Hopetoun Monument - Queensferry Bridge Tower_
distance 38.238 kms.
_Queensferry Bridge Tower - Earl's Seat_
distance 55.200 kms.
_Hopetoun Monument - Earl's Seat_
distance 93.5 kms (not used in the calculation).
fsolve(38.238*(R+.578)^2+55.200*(R+.210)^2 = (55.200+38.238)*(55.200*38.238+(R+.210)^2), R);
So, R = 7,007.456001 (kms).
*_Analysis 3:_*
_Cairnoch Hill_ height 413 m (appears very slightly below the Tower Bridge on the photo).
_Hopetoun Monument - Queensferry Bridge Tower_
distance 38.238 kms.
_Queensferry Bridge Tower - Cairnoch Hill_
distance 42.557 kms.
_Hopetoun Monument - Cairnoch Hill_
distance 80.9 kms (not used in the calculation).
fsolve(38.238*(R+.413)^2+42.557*(R+.210)^2 = (42.557+38.238)*(42.557*38.238+(R+.210)^2), R);
So, R = 8,468.636324 kms.
*_Analysis 4:_*
_Ben Lomond_ height 974 m (appears the same height as theTower Bridge on the photo).
_Hopetoun Monument - Queensferry Bridge Tower_
distance 38.238 kms.
_Queensferry Bridge Tower - Ben Lomond_ distance 78.684 kms.
_Hopetoun Monument - Ben Lomond_
distance 116.5 kms (not used in the calculation).
fsolve(38.238*(R+.974)^2+78.684*(R+.210)^2 = (78.684+38.238)*(78.684*38.238+(R+.210)^2), R);
So, R = 6,020.278843 kms.
*Let’s average these estimates*
So, R (average) = (1/4)*(5263.955827+7007.456001+8468.636324+6020.278843); = 6,690.081748.
So, the estimated radius of the earth (assuming it is a sphere) is *6,690 kms* (the average actual radius is 6,371 kms).
Fairly close. It is over estimated by about 5%.
Who need Eratosthenes and two sticks in the ground to measure the radius of the Earth when we have a Miles Davis photo and Stewart's theorem 😁.
Nice I haven’t heard it
Keep watching Rs dear
Ans:15
thanks a lot
AD^2=12^2+9^2 =144+81 AD=V225=15
I used Pythagoras' thm'
Area ABC (formula Erone) = √p(p-a)(p-b)(p-c) , ove p è il semiperimetro di ABC e a , b, c sono le misure dei lati rispettivamente opposti a A,B,C
Risulta area ABC =S =126,
ma la stessa area è uguale anche al prodotto di BC per l’altezza AH=h , diviso 2 cioè S =21*h/2=126 da cui h=126*2/21=12
BH=(teor. Pitagora )= √(c2-h2)=√(169-144)=5
DH=BD-BH=14-5=9
AD=√(DH2+h2)=√(81+144)= √225= 15
Amazing
tks
Thanks Asus dear for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Co oscendo i 3 lati conosco i 3 angoli del triangolo grande(teorama di carnot)... Quimdi dei triangoli piccoli conosco 2 lati e l'angolo compreso.... Perciò conosco il terzo lato x
Please all theorems in one video. No need proof
cool