At a quick glance, Using a similar method to the video, I used the exterior angle of ABC =6x+4x +26x = 36x. Then angle DBC = 180 - 40x. DBC is isosceles and angles BDC and DCB = 20x. Alpha = 26x -20x = 6x. Then ADC is isosceles and DBC is equilateral then 180 - 40x = 20 x. Then 60 * x = 180 and x = 3.
as AD=DB, draw a circle around the triangle with center D radius AD and since every triagngle can be drawn inside a circle, ABC lie on the circle. now by central angle theorem, 2angle C=angle D=52x in ∆adb, (4+4+52)x=180=> x=3
Nice. I always appreciate a good geometric solution, since that is my weak point. Here is my trigonometric solution: Let a = BC = AD = BD. Let c = AB. Since △ABD is isosceles, ∠DAB = ∠DBA = 4x and we can drop perpendicular from D that bisects AB at point E → AE = BE = c/2 cos(4x) = AE/AD = (c/2)/a → c/a = 2 cos(4x) Using law of sines in △ABC, we get: c/a = sin C / sin A = sin(26x) / sin(6x+4x) = sin(26x) / sin(10x) Now we set both expressions for c/a equal to each other and solve for x: sin(26x) / sin(10x) = 2 cos(4x) sin(14x+12x) = 2 sin(10x) cos(4x) sin 14x cos 12x + cos 14x sin 12x = 2 * 1/2 [sin(10x+4x) + sin(10x−4x)] sin 14x (1 − 2 sin²(6x)) + cos 14x (2 sin 6x cos 6x) = sin 14x + sin 6x sin 14x − 2 sin²(6x) sin 14x + 2 sin 6x cos 6x cos 14x = sin 14x + sin 6x −2 sin²(6x) sin 14x + 2 sin 6x cos 6x cos 14x = sin 6x 2 sin 6x (cos 6x cos 14x − sin 6x sin 14x) = sin 6x cos 6x cos 14x − sin 6x sin 14x = 1/2 cos(6x+14x) = 1/2 cos 20x = cos 60° 20x = 60° x = 3°
Would´nt it be much easier to add up the numbers of x´s + 4 x for the second angle in the yellow triangel coming to 40 x. The unmarked angel must be 60° as part o an equal sided triangel. 180° of the big triangel - 60° leaves 120°, which divided by 40 lead to X = 3° ?
Here's a more algebraic approach. All angle measurements are in degrees: Angle DAB = 4x; so angle CAB = 10x. Therefore angle ABC = 180 -- 36x and angle DBC = 180 -- 40x [1]. Now draw line CD. Angle BCD = angle BDC. If we let angle ACD = alpha, then angle DCB = angle BDC = 26x -- alpha. So angle DBC = 180 -- 2(26x -- alpha) = 180 -- 52x + 2(alpha) [2]. We now have two independent expressions for angle DBC. Equating expressions [1] and [2]: 180 -- 40x = 180 -- 52x + 2(alpha); subtract 180 from both sides: --40x = --52x + 2(alpha); add 52x to both sides: 12x = 2(alpha); so alpha = 6x. This means that triangle ADC is isosceles and line CD = line AD = line BD = line BC; so triangle BCD is equilateral, which I suspected from the beginning. So angle DCB = 60 degrees = 26x -- 6x = 20x; and x = 60/20 = 3 degrees. Veni, Vidi, Vici. 🤠
Hi i use diffrent method a bit so we dont know angle b we label it as 180-40x and we draw line from c to d and we know that it is an icoseles triangle so 180-(180-40x)/2 = 20x and when we remove from other side 26-20 =6x so another icoseles triangle 6x-6x and on the right it becomes a equilatoral so every angle becomes 60 when we do 20x=60° x=3
ADB is isosceles, so slot 4x in by A 10X + 26x + 4x + ax = 180 Find a. A line DC would make BCD isosceles with angles ax, bx, bx. I reckon that flipping the DA line into the DC position would give an equilateral, but I can't prove that without knowing the angles at B and C. If it did, the angle at C would be split into 6x and 20x. If 20x = 60deg, x = 3deg (sounds like some old singers) :). I will try it: Full angle at A would be 10x so 30deg C: 26x so 78deg B: 24x so 72deg. Well I'll be damned. x seems to be 3deg. It wasn't very mathematical - more second guessing and intuition. I imagine other answers may be possible though.
Nice and awesome, this is great, many thanks, Sir! ∆ABC → AF = AD + DF = BE = BD + DE CAB = 10x; DAB = FAB = 4x; CAD = CAF = 6x; BCA = 26 x 4, 6, 10, 26 = integer → x = integer as well circle around ∆ ABC with center M and radius r = AM = BM = CM = EM = FM → BCA = 26x → BMA = 52x EBA = 4x → EMA = 8x FAB = 4x →FMB = 8x CAF = 6x → CMF = 12x φ = 30° → 52x < 6φ < 68x → 6φ/52 > 6φ/68 → 3,46 > 2,65 → x = 3 → CAB = 10x = φ → BCA = 26x = 13φ/5 → ABC = 12φ/5 btw: M = D
AD=BD, the angle DAB=4x so, the angle DBC=180-4x-4x-6x-26x=180-40x it was known as BD=BC, so angle BDC=BCD. prove that angle BDC=BCD=(180-DBC)/2=40x/2=20x then, the line CD must be equal to BC and BD. triangle BCD is equilateral triangle. 180-40x=20x x=3
An alternative method...perhaps. ∠DAB=4x° ⇒∠CAB=10x° ∠CAB +∠ACB +∠CBA =180° 10x° + 26x° + (4x°+ ∠CBD) = 180° 40x° + ∠CBD = 180° Triangle ABC is acute angle triangle. x=5 ⇒∠CBD negative. no answer. x=4 ⇒ ∠ACB = 26x4=104°. no answer x=1 ⇒ ∠ABC = 144°. no answer x=2 ⇒ ∠ABC = 108°. no answer. x=3 ⇒ ∠ABC = 72°. correct Answer. Makes ∠DAB=12°. ∠DAB=12°. ∠CAB=30°. ∠ACB=78°. ∠CBA=72° ∠DBC=60°. Also if x is an integer then we require 180/3=60×. We have 36x add 24x. Therefore x=3 is correct.
Yay! I solved the problem. I was playing around with all the interior angles of the inside triangles equaling 180 degrees, and then I saw triangle BCD was an equilateral triangle...(dohhhhhhhhhhh)....problem solved.
I was playing with the geometry when I found that CD == AD and CD == CB. With that and some other calculations I had made such that angle C = 20x + 6x, I saw that x must equal 3.
@@skwestFor isosceles triangle angle BCD=angle BDC=20x, then angle DCA=26x-20x=6x, so triangle ADC is isosceles triangle, so we have BD=DC=CB, CBD is equilateral triangle.
Nice Solution! But there is a lot easier way to do it: *1) All triangles are cyclic so imagine point A, B and C as points on the circumference* *2) Since, ΔADB is an isosceles triangle,* _∠DAB = 4x_. So, _∠ADB = 180-(4x+4x) = 180-8x_ *3) By Angle at the center theorem,* _∠ADB = 26x*2 = 52x_ *4) This gives us an equation:* _52x = 180-8x_ *5) Solve:* _52x = 180-8x , 52x+8x = 180 , 60x = 180 , x = 180/60 , _*_[x = 3]_*
Alternative: like by you at the begining AD=BD => DAB=4x Then: ADB=180°-8x => ADC+CDB=360°-(180°-8x) =180°+8x => CBD=360°-(26x+6x+180°+8x)=180°-40x. Because of BD=BC (isosceles triangle) follows (180°-(180°+40x))/2=20x => ACD=26x-20x=6x => AD=CD (=BD=BC) and there is an equliteral triangle => 60°=20x x=3
Nice solution sir🎉
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At a quick glance, Using a similar method to the video, I used the exterior angle of ABC =6x+4x +26x = 36x. Then angle DBC = 180 - 40x. DBC is isosceles and angles BDC and DCB = 20x. Alpha = 26x -20x = 6x. Then ADC is isosceles and DBC is equilateral then 180 - 40x = 20 x. Then 60 * x = 180 and x = 3.
You are the best teacher
as AD=DB, draw a circle around the triangle with center D radius AD and since every triagngle can be drawn inside a circle, ABC lie on the circle.
now by central angle theorem,
2angle C=angle D=52x
in ∆adb, (4+4+52)x=180=> x=3
Nice. I always appreciate a good geometric solution, since that is my weak point.
Here is my trigonometric solution:
Let a = BC = AD = BD. Let c = AB.
Since △ABD is isosceles, ∠DAB = ∠DBA = 4x
and we can drop perpendicular from D that bisects AB at point E → AE = BE = c/2
cos(4x) = AE/AD = (c/2)/a → c/a = 2 cos(4x)
Using law of sines in △ABC, we get:
c/a = sin C / sin A = sin(26x) / sin(6x+4x) = sin(26x) / sin(10x)
Now we set both expressions for c/a equal to each other and solve for x:
sin(26x) / sin(10x) = 2 cos(4x)
sin(14x+12x) = 2 sin(10x) cos(4x)
sin 14x cos 12x + cos 14x sin 12x = 2 * 1/2 [sin(10x+4x) + sin(10x−4x)]
sin 14x (1 − 2 sin²(6x)) + cos 14x (2 sin 6x cos 6x) = sin 14x + sin 6x
sin 14x − 2 sin²(6x) sin 14x + 2 sin 6x cos 6x cos 14x = sin 14x + sin 6x
−2 sin²(6x) sin 14x + 2 sin 6x cos 6x cos 14x = sin 6x
2 sin 6x (cos 6x cos 14x − sin 6x sin 14x) = sin 6x
cos 6x cos 14x − sin 6x sin 14x = 1/2
cos(6x+14x) = 1/2
cos 20x = cos 60°
20x = 60°
x = 3°
Easy method 👌 sir
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Would´nt it be much easier to add up the numbers of x´s + 4 x for the second angle in the yellow triangel coming to 40 x. The unmarked angel must be 60° as part o an equal sided triangel.
180° of the big triangel - 60° leaves 120°, which divided by 40 lead to X = 3° ?
Here's a more algebraic approach. All angle measurements are in degrees:
Angle DAB = 4x; so angle CAB = 10x. Therefore angle ABC = 180 -- 36x and angle DBC = 180 -- 40x [1].
Now draw line CD. Angle BCD = angle BDC. If we let angle ACD = alpha, then angle DCB = angle BDC = 26x -- alpha. So angle DBC = 180 -- 2(26x -- alpha) = 180 -- 52x + 2(alpha) [2].
We now have two independent expressions for angle DBC. Equating expressions [1] and [2]:
180 -- 40x = 180 -- 52x + 2(alpha); subtract 180 from both sides:
--40x = --52x + 2(alpha); add 52x to both sides:
12x = 2(alpha); so alpha = 6x. This means that triangle ADC is isosceles and line CD = line AD = line BD = line BC; so triangle BCD is equilateral, which I suspected from the beginning.
So angle DCB = 60 degrees = 26x -- 6x = 20x; and x = 60/20 = 3 degrees.
Veni, Vidi, Vici. 🤠
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Hi i use diffrent method a bit so we dont know angle b we label it as 180-40x and we draw line from c to d and we know that it is an icoseles triangle so 180-(180-40x)/2 = 20x and when we remove from other side 26-20 =6x so another icoseles triangle 6x-6x and on the right it becomes a equilatoral so every angle becomes 60 when we do 20x=60° x=3
ADB is isosceles, so slot 4x in by A
10X + 26x + 4x + ax = 180
Find a.
A line DC would make BCD isosceles with angles ax, bx, bx.
I reckon that flipping the DA line into the DC position would give an equilateral, but I can't prove that without knowing the angles at B and C.
If it did, the angle at C would be split into 6x and 20x.
If 20x = 60deg, x = 3deg (sounds like some old singers) :).
I will try it: Full angle at A would be 10x so 30deg
C: 26x so 78deg
B: 24x so 72deg.
Well I'll be damned. x seems to be 3deg.
It wasn't very mathematical - more second guessing and intuition. I imagine other answers may be possible though.
Once we find that
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Nice and awesome, this is great, many thanks, Sir!
∆ABC → AF = AD + DF = BE = BD + DE
CAB = 10x; DAB = FAB = 4x; CAD = CAF = 6x; BCA = 26 x
4, 6, 10, 26 = integer → x = integer as well
circle around ∆ ABC with center M and radius r = AM = BM = CM = EM = FM →
BCA = 26x → BMA = 52x
EBA = 4x → EMA = 8x
FAB = 4x →FMB = 8x
CAF = 6x → CMF = 12x
φ = 30° → 52x < 6φ < 68x → 6φ/52 > 6φ/68 → 3,46 > 2,65 → x = 3 →
CAB = 10x = φ → BCA = 26x = 13φ/5 → ABC = 12φ/5
btw: M = D
AD=BD, the angle DAB=4x
so, the angle DBC=180-4x-4x-6x-26x=180-40x
it was known as BD=BC, so angle BDC=BCD.
prove that
angle BDC=BCD=(180-DBC)/2=40x/2=20x
then, the line CD must be equal to BC and BD.
triangle BCD is equilateral triangle.
180-40x=20x
x=3
Thanks for sharing! Cheers!
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An alternative method...perhaps.
∠DAB=4x° ⇒∠CAB=10x°
∠CAB +∠ACB +∠CBA =180°
10x° + 26x° + (4x°+ ∠CBD) = 180°
40x° + ∠CBD = 180°
Triangle ABC is acute angle triangle.
x=5 ⇒∠CBD negative. no answer.
x=4 ⇒ ∠ACB = 26x4=104°. no answer
x=1 ⇒ ∠ABC = 144°. no answer
x=2 ⇒ ∠ABC = 108°. no answer.
x=3 ⇒ ∠ABC = 72°. correct Answer.
Makes ∠DAB=12°. ∠DAB=12°.
∠CAB=30°. ∠ACB=78°. ∠CBA=72°
∠DBC=60°.
Also if x is an integer then we require 180/3=60×. We have 36x add 24x. Therefore x=3 is correct.
Nice solve, where are you from?
Thank you , the Proof was great.
You are very welcome!
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Find x? Just circle the letter x. Done
❓❔
👍
Very easy proof it was
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At a quick glance the exterior angle theorem saves some steps.
Yay! I solved the problem. I was playing around with all the interior angles of the inside triangles equaling 180 degrees, and then I saw triangle BCD was an equilateral triangle...(dohhhhhhhhhhh)....problem solved.
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
I was playing with the geometry when I found that CD == AD and CD == CB. With that and some other calculations I had made such that angle C = 20x + 6x, I saw that x must equal 3.
.Too many x😮, angle B=180-40x, so angle D =20x, thus the triangle CBD is an equilateral triangle, then 20x=60, x=3.😊
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@@skwestFor isosceles triangle angle BCD=angle BDC=20x, then angle DCA=26x-20x=6x, so triangle ADC is isosceles triangle, so we have BD=DC=CB, CBD is equilateral triangle.
This is correct.
Nice Solution!
But there is a lot easier way to do it:
*1) All triangles are cyclic so imagine point A, B and C as points on the circumference*
*2) Since, ΔADB is an isosceles triangle,* _∠DAB = 4x_. So, _∠ADB = 180-(4x+4x) = 180-8x_
*3) By Angle at the center theorem,* _∠ADB = 26x*2 = 52x_
*4) This gives us an equation:* _52x = 180-8x_
*5) Solve:* _52x = 180-8x , 52x+8x = 180 , 60x = 180 , x = 180/60 , _*_[x = 3]_*
A priori, D n'est pas forcément le centre du cercle !
x=30,x=9,x=3...solo quest'ultima è corretta
Excellent!
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The problem is absurd. Put the value x=3 and check, you get wrong things.
We should find x? Let's do it:
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....
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AD = BD:
∠BAD = ∠ABD = 4x
∠BAC = ∠BAD + ∠DAC = 4x + 6x = 10x
∠ABC = 180° − ∠BAC − ∠ACB = 180° − 10x − 26x = 180° − 36x
∠CBD = ∠ABC − ∠ABD = (180° − 36x) − 4x = 180° − 40x
BC = BD:
∠BCD = ∠BDC = (180° − ∠CBD)/2 = [180° − (180° − 40x)]/2 = 20x
∠ACD = ∠ACB − ∠BCD = 26x − 20x = 6x
∠CAD = ∠ACD = 6x:
CD = AD (= BC = BD)
Since CD = BC = BD, BCD is an equilateral triangle:
∠BCD = 60° = 20x
x = 3°
This result is checked for the other angles in this triangle:
∠BDC = 60° = 20x
x = 3° ✓
∠CBD = 60° = 180° − 40x
40x = 120°
x = 3° ✓
Best regards from Germany
Excellent!
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Alternative: like by you at the begining AD=BD => DAB=4x
Then: ADB=180°-8x
=> ADC+CDB=360°-(180°-8x)
=180°+8x
=> CBD=360°-(26x+6x+180°+8x)=180°-40x. Because of BD=BC (isosceles triangle) follows (180°-(180°+40x))/2=20x
=> ACD=26x-20x=6x => AD=CD (=BD=BC) and there is an equliteral triangle
=> 60°=20x x=3