respect

@PreMath

14 күн бұрын

Thanks dear❤️

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.

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

Sin 120 is also √3/2. So theta becomes 60. So there is second answer too

@laxmikantbondre338

14 күн бұрын

Considering Theta as 60 the Answer is 12/7

@JamesDavy2009

14 күн бұрын

@@laxmikantbondre338 While your answer is also correct, PreMath assumes angle CAD is acute.

@PreMath

13 күн бұрын

Thanks for the feedback ❤️ You are the best!

@devondevon4366

12 күн бұрын

If triangle is obtuse scalene or acute scalene

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!

@PreMath

13 күн бұрын

Thanks for sharing ❤️

شكرا لكم على المجهودات يمكن استعمال 12sinCAD =6(racine3) ... CAD=60 CAB=30 7xsin (CAB)=6(racine3) .... x=12/7 (racine3)

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

I got 2 answers, the other one is AB=12/7.

@misterenter-iz7rz

14 күн бұрын

Same as me.

@devondevon4366

12 күн бұрын

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.

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

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

One of the funniest questions ever!!! Congratulatons professor!!

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

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 .

Thank you!

@PreMath

13 күн бұрын

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).

@PreMath

13 күн бұрын

Excellent! Thanks for sharing ❤️

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,???

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

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. ☺️☺️

@PreMath

13 күн бұрын

No worries. We are all lifelong learners. That's what makes our life exciting and meaningful! Take care ❤️

@thepi553

13 күн бұрын

@@PreMath ok sir, thanks☺️☺️

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).

@PreMath

13 күн бұрын

Thanks for the feedback ❤️

This one is easy !

@PreMath

13 күн бұрын

Excellent! Thanks for the feedback ❤️

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

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.😊😊😊

@PreMath

13 күн бұрын

Excellent! Thanks for the feedback ❤️

If Sin(2x) = sqrt(3)/2, then 2x=60° or 2x=120°, then we could have 2 possible solutions....

@PreMath

13 күн бұрын

Thanks for sharing ❤️

How to find the base of this triangle

@montynorth3009

14 күн бұрын

Cosine Law.

why cd is 5? how can be 5? what is the reason?

@PreMath

13 күн бұрын

No! CD is not 5. Cheers 😀

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

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

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.

@PreMath

13 күн бұрын

Thanks for the feedback ❤️