Triangle BOC. Corner angles at B and C are 37.5 & 30 degrees respectively. Angle BOC = 180 - 37.5 - 30 = 112.5 degrees. 2 / sin 112.5 = OC / sin 37.5. OC = 2 sin 37.5 / sin 112.5 = 1.318. Triangle OFC. Sin 30 = OF / 1.318. 0.5 = OF / 1.318. OF = 0.659.= radius.
According to the sine theorem, the sides of a triangle can be found. sin(75)=sin(30+45)=sin(30)*cos(45)+ cos(30)*sin(45). r=2*S/(a+b+c). I don't like additional constructions.
Once known the 3 sides in some way (I used sines law) and knowing that the center of the circle is the point of intersection of all the bisectors, considering the bisector of angle in C we get COF and COD right triangle of 30,60,90 so CF = DC = r√ 3 (for two tangent theorem) BF = BE = 2 - r√ 3 AE = AD = √ 6 - 2 + r√ 3 AC = AD + DC = 1 + √ 3 √ 6 - 2 + r√ 3 + r√ 3 = 1 + √ 3 r = (1 - √ 2 + √ 3)/2
I didn't get close with this one. I started out wrong and it never got better. I will have to go through your video very slowly to seek understanding. Thank you.
BF=r/tan30º, CF=r/tan37.5º, BF+CF=2 => r=2 .tan30º.tan37.5º/(tan30º+tan37.5º) = 0.658918622... The "special triangles" are trigonometry in disguise, so why not go to the fullest?
I really DO appreciate the full geometric-and-special-triangles approach. There's elegance to it. However, as I can see others have concluded … why all the work? We're in a world of calculators and simple trigonometric relationships … using them directly delivers "the goods" without much complication. Dividing the top angle in 2 yields two 30-60-90 triangles up there, fully symmetric and similar. The "right leg" is √(3)R plain and simple. Likewise, dividing the bottom-right triangle in half yields another pair of similar triangles, the right leg of which has length R/(tan 37½°). And the given is that the whole right leg is 2 units long. So, [1.1] √(3)R + R/(tan 37½°) = 2 … factor out the common R [1.2] R( √3 ⊕ 1/(tan 37½°) ) = 2 … and divide by te () stuff to cancel [1.3] R = 2 / ( √3 ⊕ 1/(tan 37½°) ) … which my calculator says [1.4] R = 0.65892 Now, was that SO hard? Not really!!! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
What a strange problem! Just the process of rationalizing the fractional solution took as long as solving some entire problems on this channel. And the final answer is still ungainly. :)
Looooong, but Nice! "Luckily" 75=30+45... I missed that, so, I used Adj=Opp/Tan for triangles OFB & OFC with half angles 37.5 & 30 (because lines from outer point to circle are equal, creating similar equal triangles because of the radiuses (I prefer "radiuses"), and so, split the angle). r=OF. The sum of FB=r/Tan(37.5) & FC=r/Tan(30) is BC=2, so: r=2Tan(37.5)Tan(30)/(Tan(37.5)+Tan(30))~0.6590 as your solution (0.66).
Missing angle measure from sum of angles in triangle Missing side lengths from sine law By comparing two formulas for area one with sine and second with radius of inscribed circle
*In triangle OFC we have: FC = R. cotan (30°) = R. sqrt (3). In triangle OBF we have: FB = R. cotan (37.5°) So: 2 = BC = FC + FB = R (sqrt(3) + cotan (37.5°)), so R = 2 / (sqrt (3) + cotan (37.5°)). *Now let's calculate cotan (37.5°) cotan (75°) = cotan (90° - 15°) = tan (15°) = 2 - sqrt (3). (I presume this value is know, unless it is easy to obtain from tan (30°) and the formula giving tan (2x).) Now, we use the formula cotan (2x) = ((cotan (x))^2 - 1) / 2. cotan (x) If x is cotan (37.5°) then we have: 2 - sqrt (3) = (x^2 - 1) / x. This is a second degree equation: x^2 - 2 (2 - sqrt (3)) x + 1 = 0 delta ' = (2 + sqrt (3))^2 + 1 = 8 + 4 sqrt (3) = (sqrt (6) - sqrt (2))^2 So: x = 2 - sqrt (3) + sqrt (6) - sqrt (2) = cotan (37.5°), the other solution beeing negative . *We come back to R. R = 2 / (2 + sqrt 6) - sqrt (2)). We rationalize the denominator (easy but fastidious to write here with the computer). Finally we obtain: R = (1 - sqrt (2) + sqrt (3)) / 2.
Connect OC ; OD; OF Let r is radius BF=a; CF=2-a tan(30)=OF/CF √3/3=r/2-a √3(2-a)=3r r==√3(2-a)/3 tan(75/2)=r/a 0.767=r/a r=0.767a 0.767a=√3(2-a)/3 a=0.86 So: r=(0.86)(0.767)=0.66 units. Thanks ❤❤❤
Вы не знаете простой , широко распространенной формулы в геометрии для вписанной окружности ? ДС=(АС+СВ-АВ)/2 = (\/3+1+2-\/6)/2 = (\/3 - \/6 + 3)/2 . Из прямоугольного треугольника ДСО - r/ДС = tg 30* (tg 30* = \|3/3) . r = ДС х tg 30* = (\/3 -\/6 + 3)/2 х \/3/3 = (3 - \/2х3х3 +3\/3)/2х3 = 3(1 -\/2 +\/3)/2х3 = (1-\/2+\/3)/2 . Теперь понимаю почему на Западе преподают простые вещи , доступные пониманию каждому среднему человеку , так сложно . Формула , которую привел , выводится элементарно простым школьником , обладающим средним умом , прошедшим школьную программу . Теперь и у нас правит олигархат , которым нужны необразованные , забитые люди - слепо их или их ставленников выбирать и работать на них за мизерную зарплату .
Too complicated for me, although I did follow the workings too the end Can't help thinking that I'd probably have more success helping NASA fix their current Voyager software glitch! 😊
Wow! I found an answer a completely different way, that also was somewhat shorter. First (key) is to recognize that a circle having tangents that intersect at some angle θ, bisects θ into two equal angles ½θ apiece. Thus ∠60° at the top is bisected into 30° + 30° by a line drawn thru the circle's midpoint. Likewise, using a radius from centerpoint to the tangent point of BC, we develop an upper triangle which is 30°-60°-90°. Further again, the line thru the center of the circle to base must also have a 75°∠ at the lower left. Here is the cool induction step: The line thru center to base must be the same length as the [2] given for BC because its an Isosceles triangle. Armed with that, then it is composed of the upper hypotenuse which must be 2𝒓 (30-60-90△, 1-√(3)-2 ratios), and the lower segment we'll call 𝒒 𝒒 = 𝒓 / sin 75° Because from center point to base is [𝒓] radius. The isosceles extension is 𝒒. So we then just add 'em together 𝒒 ⊕ 2𝒓 = 2 … substitute in 𝒓 / sin 75° ⊕ 2𝒓 = 2 … move stuff around 𝒓 = 2 / (1 / sin 75° ⊕ 2) 𝒓 = 0.65892 Which is ≈ to 0.66, matching our good math professor. Thanks again. ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅
I like your videos
thanks❤️
Triangle BOC.
Corner angles at B and C are 37.5 & 30 degrees respectively.
Angle BOC = 180 - 37.5 - 30 = 112.5 degrees.
2 / sin 112.5 = OC / sin 37.5.
OC = 2 sin 37.5 / sin 112.5 = 1.318.
Triangle OFC.
Sin 30 = OF / 1.318.
0.5 = OF / 1.318.
OF = 0.659.= radius.
Thanks ❤️
R/tg(30)+R/tg(37,5)=2 => R=0,6589
We note that
Thanks❤️
According to the sine theorem, the sides of a triangle can be found. sin(75)=sin(30+45)=sin(30)*cos(45)+ cos(30)*sin(45). r=2*S/(a+b+c). I don't like additional constructions.
What is S ?
@@溫文爾雅-q3f s=0,5* a* b *sin(60)
S => area
Thanks ❤️🌹
inside circle radius = 2*Area / (a+b+c), now this I can understand
I think outside circle radius = (a+b+c) / 2
I taught and was taught that the legs made the right angle. What you called the Longest LEG should be called the HYPOTENUSE ( or longest side)
In the right triangle, the longest leg is called the HYPOTENUSE! Thanks ❤️
Once known the 3 sides in some way (I used sines law) and knowing that the center of the circle is the point of intersection of all the bisectors, considering the bisector of angle in C we get COF and COD right triangle of 30,60,90 so
CF = DC = r√ 3 (for two tangent theorem)
BF = BE = 2 - r√ 3
AE = AD = √ 6 - 2 + r√ 3
AC = AD + DC = 1 + √ 3
√ 6 - 2 + r√ 3 + r√ 3 = 1 + √ 3
r = (1 - √ 2 + √ 3)/2
Nice sharing my Allah bless you❤❤ stay connected🎉
Thank you, I will 🌹
Amazing explanation
I know that you will get millions of subscribers very soon keep it up
Thank you so much dear 😀❤️
Thank you for your explanation
I didn't get close with this one. I started out wrong and it never got better. I will have to go through your video very slowly to seek understanding. Thank you.
You can do it!
Thanks ❤️
BF=r/tan30º, CF=r/tan37.5º, BF+CF=2 => r=2 .tan30º.tan37.5º/(tan30º+tan37.5º) = 0.658918622...
The "special triangles" are trigonometry in disguise, so why not go to the fullest?
I want to thank you for the time it took to create this very interesting problem.
Many many thanks ❤️🌹
Very nice sharing Sar❤❤❤❤❤
Thanks ❤️
I really DO appreciate the full geometric-and-special-triangles approach. There's elegance to it.
However, as I can see others have concluded … why all the work? We're in a world of calculators and simple trigonometric relationships … using them directly delivers "the goods" without much complication.
Dividing the top angle in 2 yields two 30-60-90 triangles up there, fully symmetric and similar. The "right leg" is √(3)R plain and simple. Likewise, dividing the bottom-right triangle in half yields another pair of similar triangles, the right leg of which has length R/(tan 37½°).
And the given is that the whole right leg is 2 units long.
So,
[1.1] √(3)R + R/(tan 37½°) = 2 … factor out the common R
[1.2] R( √3 ⊕ 1/(tan 37½°) ) = 2 … and divide by te () stuff to cancel
[1.3] R = 2 / ( √3 ⊕ 1/(tan 37½°) ) … which my calculator says
[1.4] R = 0.65892
Now, was that SO hard? Not really!!!
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
HOW TO DEVELOP CRITICAL THINKING SKILLS IN MATHEMATICS
Practice and perseverance!
Just keep watching ...
Thanks ❤️
Dog barking @ 4:33 is as excited as I was with this problem. I mean I love all the math!...identities...rationalizing...ect. 🙂
Great observation! My neighbor's dogs. I guess they like math as well. 😀
Thanks ❤️
What a strange problem! Just the process of rationalizing the fractional solution took as long as solving some entire problems on this channel. And the final answer is still ungainly. :)
Since the circle is the incircle of the triangle, OBC and OCB are angle bisectors. So can't we just say that BF = rcot37.5 and CF = rcot30?
Looooong, but Nice!
"Luckily" 75=30+45... I missed that, so,
I used Adj=Opp/Tan for triangles OFB & OFC with half angles 37.5 & 30 (because lines from outer point to circle are equal, creating similar equal triangles because of the radiuses (I prefer "radiuses"), and so, split the angle). r=OF.
The sum of FB=r/Tan(37.5) & FC=r/Tan(30) is BC=2, so: r=2Tan(37.5)Tan(30)/(Tan(37.5)+Tan(30))~0.6590 as your solution (0.66).
Missing angle measure from sum of angles in triangle
Missing side lengths from sine law
By comparing two formulas for area one with sine and second with radius of inscribed circle
I focused on triangle ABC
*In triangle OFC we have: FC = R. cotan (30°) = R. sqrt (3).
In triangle OBF we have: FB = R. cotan (37.5°)
So: 2 = BC = FC + FB = R (sqrt(3) + cotan (37.5°)), so R = 2 / (sqrt (3) + cotan (37.5°)).
*Now let's calculate cotan (37.5°)
cotan (75°) = cotan (90° - 15°) = tan (15°) = 2 - sqrt (3). (I presume this value is know, unless it is easy to obtain from tan (30°) and the formula giving tan (2x).)
Now, we use the formula cotan (2x) = ((cotan (x))^2 - 1) / 2. cotan (x)
If x is cotan (37.5°) then we have: 2 - sqrt (3) = (x^2 - 1) / x. This is a second degree equation: x^2 - 2 (2 - sqrt (3)) x + 1 = 0
delta ' = (2 + sqrt (3))^2 + 1 = 8 + 4 sqrt (3) = (sqrt (6) - sqrt (2))^2
So: x = 2 - sqrt (3) + sqrt (6) - sqrt (2) = cotan (37.5°), the other solution beeing negative .
*We come back to R. R = 2 / (2 + sqrt 6) - sqrt (2)). We rationalize the denominator (easy but fastidious to write here with the computer).
Finally we obtain: R = (1 - sqrt (2) + sqrt (3)) / 2.
Thanks❤️
Connect OC ; OD; OF
Let r is radius
BF=a; CF=2-a
tan(30)=OF/CF
√3/3=r/2-a
√3(2-a)=3r
r==√3(2-a)/3
tan(75/2)=r/a
0.767=r/a
r=0.767a
0.767a=√3(2-a)/3
a=0.86
So: r=(0.86)(0.767)=0.66 units. Thanks ❤❤❤
Thanks ❤️
why don't you use the law of sin and 30.60.90 special triangle to find the radius? your way is very complicated.
Вы не знаете простой , широко распространенной формулы в геометрии для вписанной окружности ? ДС=(АС+СВ-АВ)/2 = (\/3+1+2-\/6)/2 = (\/3 - \/6 + 3)/2 . Из прямоугольного треугольника ДСО - r/ДС = tg 30* (tg 30* = \|3/3) . r = ДС х tg 30* = (\/3 -\/6 + 3)/2 х \/3/3 = (3 - \/2х3х3 +3\/3)/2х3 = 3(1 -\/2 +\/3)/2х3 = (1-\/2+\/3)/2 .
Теперь понимаю почему на Западе преподают простые вещи , доступные пониманию каждому среднему человеку , так сложно . Формула , которую привел , выводится элементарно простым школьником , обладающим средним умом , прошедшим школьную программу . Теперь и у нас правит олигархат , которым нужны необразованные , забитые люди - слепо их или их ставленников выбирать и работать на них за мизерную зарплату .
Sanırım 10:56 dan sonrası çok da gerekli değil! Çünkü o noktada da ondalık hesap yapılabilirdi.
Too complicated for me, although I did follow the workings too the end
Can't help thinking that I'd probably have more success helping NASA fix their current Voyager software glitch!
😊
😀
Thanks ❤️
Never stop trying. It wards off dementia.
as clear as the Mississippi river
tan37.5°=R/BF; tan30°=R/CF; BF+CF=2=R(1/tan37.5°+1/tan30°)
R=2/(1/tan37.5°+1/tan30°)=0.6589...
Thanks ❤️
Wow! I found an answer a completely different way, that also was somewhat shorter.
First (key) is to recognize that a circle having tangents that intersect at some angle θ, bisects θ into two equal angles ½θ apiece.
Thus ∠60° at the top is bisected into 30° + 30° by a line drawn thru the circle's midpoint. Likewise, using a radius from centerpoint to the tangent point of BC, we develop an upper triangle which is 30°-60°-90°. Further again, the line thru the center of the circle to base must also have a 75°∠ at the lower left.
Here is the cool induction step:
The line thru center to base must be the same length as the [2] given for BC because its an Isosceles triangle. Armed with that, then it is composed of the upper hypotenuse which must be 2𝒓 (30-60-90△, 1-√(3)-2 ratios), and the lower segment we'll call 𝒒
𝒒 = 𝒓 / sin 75°
Because from center point to base is [𝒓] radius. The isosceles extension is 𝒒. So we then just add 'em together
𝒒 ⊕ 2𝒓 = 2 … substitute in
𝒓 / sin 75° ⊕ 2𝒓 = 2 … move stuff around
𝒓 = 2 / (1 / sin 75° ⊕ 2)
𝒓 = 0.65892
Which is ≈ to 0.66, matching our good math professor.
Thanks again.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Col teorema dei seni calcolo i 2 lati mancanti..poi r=S/p,p semiperimetro...r=(3+√3)/(3+√3+√6)=0,6589...sempre siano i calcoli corretti
Thanks ❤️
Il denominatore va diviso per 2 (semiperimetro)
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