Can you find the length AB? | (Fun Geometry Problem) |
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- Опубліковано 18 тра 2024
- Learn how to find the area of the length AB. Important Geometry, trigonometry, and algebra skills are also explained. Step-by-step tutorial by PreMath.com
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Once Theta is calculated to be 30 degrees.......
From bisector angle theory.
3/4 = CB/BD.
Thus CB = 3/7 x CD & BD = 4/7 x CD.
The ratio of areas follows due to common heights.
Area of triangle ABC = 3/7 x 3 root 3.
1/2 x 3 x AB x sin 30 = 3/7 x 3 root 3.
3/4 x AB = 3/7 x 3 root 3.
AB = 3/7 x 3 root 3 x 4/3.
12 root 3/7.
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Sin 120 is also √3/2. So theta becomes 60. So there is second answer too
Considering Theta as 60 the Answer is 12/7
@@laxmikantbondre338 While your answer is also correct, PreMath assumes angle CAD is acute.
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You are the best!
If triangle is obtuse scalene or acute scalene
Nice! φ = 30° → sin(φ) = 1/2 → cos(φ) = √3/2 → cos(2φ) = sin(φ) = 1/2 → sin(2φ) = cos(φ) = √3/2
∆ ACD → AC = 3; AD = 4; CD = BC + BD = a + b; BAC = DAB = θ; AB = k
(1/2)sin(2θ)12 = 6sin(2θ) = 3√3 → sin(2θ) = √3/2 = sin(2φ) → θ = φ
3/a = 4/b → b = 4a/3 → (1/2)sin(φ)3k = 3k/4 = (3/7)3√3 → k = 12√3/7
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Area triangie ACD=3*4*si2Q/2=3\/3. sin2Q=\/3/2. 2Q=60degrees. Q=30ddegrees. 3AB*sin30/2+4AB*sin30/2=3\/3.
АВ=2*3\/3/7sin30=12\/3/7!
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شكرا لكم على المجهودات
يمكن استعمال
12sinCAD =6(racine3)
...
CAD=60
CAB=30
7xsin (CAB)=6(racine3)
....
x=12/7 (racine3)
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One of the funniest questions ever!!! Congratulatons professor!!
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Thank you!
You are very welcome!
Thanks ❤️
A = 1/2 * a * b * sinC
3√3 = 1/2 * 4 * 3 * sin(2θ)
= 6 * sin(2θ)
(√3)/2 = sin(2θ)
2θ = sin⁻¹[(√3)/2]
2θ = 60°
θ = 30°
3√3 = [1/2 * 3 * AB * sin(30°)] + [1/2 * AB * 4 * sin(30°)]
= (1/2 * 3 * AB * 1/2) + (1/2 * AB * 4 * 1/2)
= 3/4(AB) + AB
= 7/4(AB)
(12√3)/7 = AB
So, the length of segment AB is (12√3)/7 units (exact), or about 2.97 units (approximation).
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Angle CAD can be larger than 90 degrees, so φ could be 60° as well; then the area would be 3√3 just the same but AB would be shorter:
3√3 = 3x(½sin(φ)) + 4x(½sin(φ)), sin(φ) = ½√3
--> 3√3 = x(¾√3) + x√3
--> x = 12/7 .
I got 2 answers, the other one is AB=12/7.
Same as me.
You are correct since he said at 1:01, 'This diagram may not be true to scale.'
There are two answers to this problem since the length of a CD can be sqrt 13 or sqrt 37, depending on the type of triangle.
It is sqrt 13 if it is an Acute scalene (and it looks like one), and it is sqrt 37 if it is an Obtuse scalene.
In other words, there is a triangle with sides 3, 4, and sqrt 13, area 3 sqrt 3, and one angle of 60 degrees.
AB in this case = 4 * sine 46.102/sine 103.898 = 2.969 or 2.97
But there is also another triangle with sides 3, 4, and sqrt 37 with area 3 sqrt 3 and one angle 120 degrees
AB in this case = 4 * sine 25.285/ sine 94.715 = 1.714 or 12/7
So, as said above, AB can either be sqrt 13 or sqrt 37.
He has to indicate whether it is obtuse scalene or acute scalene.
3√3=(1/2)*3*4*sin2θ..θ=30..posto α=ADC ..risulta..AB/sinα=4/sin(θ+α)...AB/sin(2θ+α)=3/sin(θ+α)..divido le equazioni..risulta ctgα=5/3√3..tgα=3√3/5..AB=4sinα/sin(α+θ)=2,969=12√3/7..mah,???
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DEIIAC
Extended AB meets DE at E.
AED triangle
AD=DE=4
Ang ACE =120
Cos 120=-1/2=(16+16-AE^2)/2*4*4
AE=4√3
Two triangles ACB and BDE are similar
AB/BE=AC/DE=AC/AD=3/4
AB=4√3*3/7=12√3/7
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U have proved angle CAB=30 degrees
Now I like to use angle bisector theorem in 🔺 ACD
CB/BD=3/4
Now area of ACB/area of ADB=3/4(as height is same and areas will be proportionate to bases)
Area of ACB=3√3*3/7=9√3/7
In other way
Area of ACB=1/2*3*AB*sin30
=3/4AB
Then 3/4AB=9√3/7
AB =9√3/7*4/3=12√3/7
Please comment. Unrest request.
2.969
A different approach
Area or A = 3 sqrt 3 a=3 b =4 , and c =p
A^2 = 27
Let's find the unknown side using Heron's formula Area = sqrt [s (s-a)(s-b)(s-c)]
A^2 = s(s-a)(s-b)(s-c)
s = ( a+ b+ c )/2
Let the third side = p
Hence s= ( 3+ 4 + p)/2
s = 3.5 + 0.5 p
27 = 3.5 +0.5p (3.5 + 0.5p -3)(3.5 + 0.5p-4)( 3.5 + 0.5p -p)
27 = 3.5 + 0.5p ( 0.5 + 0.5p)( -0.5 + 0.5p ) (3.5-0.5p)
27 = (3.5 + 0.5p)(3.5 - 0.5p)( 0.5 + 0.5p)(-0.5 + 0.5p) Notice two difference of squares
27 =(12.25-0.25p^2)( -0.25 +0.25 p^2)
27 = -3.0625 +0.0625p^2 -0.0625p^4 + 3.0625p^2
27 = - 3.0625 + 3.125 p^2 - 0.0625 p^4
0 = -27 - 3.0625 + 3.125p^2 - 0.0625p p^4
0= 0.0625 p^4 - 3.125 p^2 + 3.0625 + 27 multiply both sides by -1
0= 0.0625 p^4 - 3.125 p^2 + 30.0625
let n =p^2 then
0= 0.0625 n^2 - 3.125 n + 30.0625
Using the quadratic formula calculator n = 13 and n= 37
Hence p^2 = 13 and p^2 = 37
p = sqrt 13 or p= sqrt 37 ie the length of CD = sqrt 13 or sqrt 37
Since the triangle appears to be an acute scalene, p = sqrt 13.
(But it could have also been an Obtuse scalene triangle in this p= sqrt 37 since he said the diagram is not to scale)
Since there are three sides, the Law of Cosine can be used, and
the angles are 60, 73.898 and 46.102 degrees
Hence A = 60 degrees
and theta = 30 degrees
Hence, to find length AB, use 30 degrees, 4, and 46.102 degrees and the law of sines
180 - 30 + 46.102 = 103.898 degree
AB = 4. ( sin 46.102)
------------------
sin 103.898
AB = 4 ( 0.74230643)
AB = 2.969 Answer
Once we found sin(2θ) =√3/2, we know that cos(2θ) = ±1/2. Then by cosine rule CD² = 25±12, so either CD=√13, or CD=√37. Given all three sides of the triangle, we can easily find the length of bisector, either through the well-known formula, or using equalities: BC:BD=AC:AD along with AB² = AC×AD - BC×BD (BTW, that's how the formula for bisector is obtained).
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I had been solving this question from 1 hour but i could solve then i started the video and saw that solution is too easy🤣🤣. I laughed on myself for a while and learnt the new approach for solving this type of question.
Thanks sir for giving such a question.
☺️☺️
No worries. We are all lifelong learners. That's what makes our life exciting and meaningful!
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@@PreMath ok sir, thanks☺️☺️
This one is easy !
Excellent!
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6sin 2t=3sqrt(3), sin 2t=sqrt(3)/2, 2t=60 or 120, t=30 or 60, 1/2×3×AB×sin 30=(3/7)×3sqrt(3), AB=(4/3)×(9/7)sqrt(3)=(12/7)sqrt(3) or 1/2×3×AB×sin 60=(9/7)sqrt(3), AB=(4/3sqrt(3))(9/7)sqrt(3)=12/7.😊😊😊
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If Sin(2x) = sqrt(3)/2, then 2x=60° or 2x=120°, then we could have 2 possible solutions....
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Let's face this challenge:
.
..
...
....
.....
The area of the yellow triangle can be calculated in the following way:
A(ACD) = (1/2)*AC*AD*sin(∠CAD)
3√3 = (1/2)*3*4*sin(∠CAD)
√3/2 = sin(∠CAD)
⇒ ∠CAD = 60°
Now we are able to calculate the length of AB:
A(ACD)
= A(ABC) + A(ABD)
= (1/2)*AB*AC*sin(∠CAD/2) + (1/2)*AB*AD*sin(∠CAD/2)
= (1/2)*AB*(AC + AD)*sin(∠CAD/2)
3√3
= (1/2)*AB*(3 + 4)*sin(60°/2)
= (1/2)*AB*7*sin(30°)
= (1/2)*AB*7*(1/2)
⇒ AB = 12√3/7
Best regards from Germany
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How to find the base of this triangle
Cosine Law.
why cd is 5? how can be 5? what is the reason?
No! CD is not 5.
Cheers 😀
Something is not right formulating this Problema.
This cannot be a Right Triangle (3 ; 4 ; 5).
CD cannot be 5. If CD = 5, so the Area should be A = (a * b) / 2 ; A = (3 * 4) / 2 ; A = 12 / 2 and A = 6 Square Units and not, A = 3*sqrt(3) Square Units or A ~ 5,196 Square Units.
Thanks.
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