Can you find the length X? | (Angle bisector) |

[@redfinance3403]

Although the diagram in this video does look like an angle of 60 degrees, it is bad to get accustomed to assuming things about the problem from the diagram. When you wrote sin^-1(sqrt(3)/2) that could have two solutions of 60 degrees and 120 degrees. From the diagram you saw it is 60 degrees, but in general this is a bad thing to do and you must consider the different solutions and eliminate possible solutions from other info in the problem. This actually has a significant impact on the problem, as with 60 degrees x is 12sqrt(3)/7 but with 120 degrees x is 12/7. For anyone doing competitive maths, please take note of this, it is really important.

[@prasoon7916]

you are right

[@redfinance3403]

@@LuisdeBritoCamacho I looked at your proof and you’re wrong. Solving for AB gives 2 solutions not one. This accounts for the possibility of theta being 60 degrees or 120 degrees.

[@redfinance3403]

@@LuisdeBritoCamachowhen AB = 3.606, angle = 60 degrees 😊

[@jimlocke9320]

At 4:05, there are a number of ways to find that the areas of ΔACD and ΔBCD are in the ratio 3:4. One way is to compare the formulas Area = (1/2) ab sin(c) and find that they are the same except for the sides of length 3 and 4. Alternatively, the angle bisector divides the side in the ratio of the other two sides, so, if AD is given length 3m, BD has length 4m. If these are the bases of ΔACD and ΔBCD, the height is common and areas are in the ratio 3:4. So, ΔACD has 3/7 the area of ΔABC, (1/2)(3)(x)sin(30°) = (3/7)(3√3), and x = (12√3)/7, as PreMath also found.

[@sergeyvinns931]

x=2,97? This is the heigt of the triangle?

[@jimlocke9320]

@@sergeyvinns931 No, x is not the height. Drop a perpendicular from C to AB. If AD and BD are treated as the bases of ΔACD and ΔBCD respectively, the length of the perpendicular is the common height, let's call it h. Area ΔACD = (1/2)(AD)(h) = (1/2)(3m)(h) = 3mh/2. Area ΔBCD = (1/2)(BD)(h) = (1/2)(4m)(h) = 4mh/2. Ratio of areas = 3:4. Knowing 2 sides and the area of ΔABC and AB is the third side, length AB can be found from Heron's formula. Then, h can be computed from Area ΔABC = (1/2)(AB)(h). However, we need x, not h. This is a harder way to compute x, but here goes. Compute m from AB = AD + BD = 3m + 4m = 7m. Then use the formula (CD)² = (AD)(BD). CD = x, so x² = 12m².

[@sergeyvinns931]

@@jimlocke9320 Я уже всё пересчитал, и нашёл ошибку в вычислении, которая была исправлена. Пифагор оказался прав, вся тригонометрия из его теоремы происходит, отсюда и подобие треугольникоа, здесь и произошёл сбой программы!

[@jimlocke9320]

@@sergeyvinns931 Thanks for the explanation!

[@adhiraj073]

My tution teacher was just telling me about 1/2absinc (while we were studying bio) for no apparent reason, I open yt later and this exact same formula. 'This world is a simulation'

[@misterenter-iz7rz]

Simply considering 1/2×3×4×sin C=3sqrt(3)), 】 sin C=sqrt(3)/2, C=60°, then 1/2×4×xsin30°=x=4/7 (3×sqrt(3))=12/7sqrt(3).😊

[@almosawymehdi3416]

AB = sqrt(13) with the cosinus law, then it is necessary to explain the bissector theorem : AD/DB = 3/4. Please one more like this for tomorrow.

[@jamestalbott4499]

Thank you!

[@ДмитрийИвашкевич-я8т]

sin120°=√3/2

[@Krestor1]

I tried to do it with heron's theorem and found out that line AB could either be sqrt(13) or sqrt(37) so please be a bit careful about that next time. Like everyone else has mentioned, sine sqrt(3)/2 could be 120 degrees too

[@valentinconito3777]

Well done ! But when sin(x) = sqrt(3)/2, then x = 60° or x=120°, then we got 2 distinct ways ;)

[@AndreasPfizenmaier-y7w]

Is it possible to solve with 3:AD=4:BD?

[@msafasharhan]

HelloSir,you said regtangle ACDand cdb cncurent and they defferent areas how

[@adogonasidecar1262]

The most critical point here is that the angle bisector doesn't split the areas equally. It splits the angle and the base equally but not the areas. It's a very easy mistake to make

[@SalahGuel]

Thanks, but more algebra problems for us.

[@unknownidentity2846]

Let's find x: . .. ... .... ..... First of all we calculate the missing side length by using the formula of Heron for the area of the triangle: a = BC = 4 b = AC = 3 c = AB = ? s = (a + b + c)/2 = (4 + 3 + c)/2 = (7 + c)/2 A(ABC) = √[s*(s − a)*(s − b)*(s − c)] 3√3 = √{[(7 + c)/2]*[(7 + c)/2 − 4]*[(7 + c)/2 − 3]*[(7 + c)/2 − c]} 3√3 = √{[(7 + c)/2]*[(c − 1)/2]*[(c + 1)/2]*[(7 − c)/2]} 27 = [(7 + c)/2]*[(c − 1)/2]*[(c + 1)/2]*[(7 − c)/2] 27 = (7 + c)*(7 − c)*(c − 1)*(c + 1)/16 27*16 = (49 − c²)*(c² − 1) 36*12 = (49 − c²)*(c² − 1) ⇒ 36 = 49 − c² ∧ 12 = c² − 1 ⇒ c² = 13 ⇒ c = √13 ∨ 12 = 49 − c² ∧ 36 = c² − 1 ⇒ c² = 37 ⇒ c = √37 According to the angle bisector theorem we obtain: BD/AD = BC/AC = 4/3 ⇒ BD = (4/7)*AB ∧ AD = (3/7)*AB c = √13 ⇒ BD = (4/7)*√13 ∧ AD = (3/7)*√13 c = √37 ⇒ BD = (4/7)*√37 ∧ AD = (3/7)*√37 Now we are able to calculate the length of the angle bisector: CD² = x² = AC*BC − AD*BC First case: c = √13 x² = 3*4 − (3/7)*√13*(4/7)*√13 = 12 − 12*13/49 = 12*49/49 − 12*13/49 = 12*36/49 ⇒ x = (12/7)√3 Second case: c = √37 x² = 3*4 − (3/7)*√37*(4/7)*√37 = 12 − 12*37/49 = 12*49/49 − 12*37/49 = 12*12/49 ⇒ x = 12/7 Best regards from Germany

[@LuisdeBritoCamacho]

I was induced in error because I used Wolfram Alpha to calculate AB Length. It only returns with one solution : AB = 6,083. So, even WA got it wrong. This Problem has two Solutions : 1) Angle ACB = 60º ; X ~ 2,97 2) Angle ACB = 120º ; X ~ 1,7135 Thank you!!

[@redfinance3403]

@@LuisdeBritoCamacho no problem!

[@redfinance3403]

@@LuisdeBritoCamacho no problem!

[@LuisdeBritoCamacho]

@@redfinance3403, During the Construction of this Item it should have been stated that Angle ACB < 90º; or that is a convex angle. A simple statmente that would make all the difference. That's why I spend a lot of my time Wondering the better way to make a Good Mathematical Communication. Words are much too precious to be wasted. That's the little difference that makes the Big Difference.

[@redfinance3403]

@@LuisdeBritoCamacho fr

[@sergeyvinns931]

Angle ACB = 60 degrees, x=12\/3/7, запутался в дробях, доказывая с помощью теоремы Пифагора, голова совсем не варит от бессонницы.

[@rabotaakk-nw9nm]

🤣🤣🤣

[@LuisdeBritoCamacho]

STEP-BY-STEP RESOLUTION PROPOSAL : 01) Using Heron's Reverse Formula to calculate AB. 02) AB = 6,083 03) Using Triangle Bisector Theorem 04) 3 / 4 = AD / BD 05) 3*BD = 4*AD 06) AD + BD = 6,087 06) AD = 2,607 07) BD = 3,476 08) 2,607 + 3,476 = 6,083 09) Coordinates of Point C = (2,466 ; 1,708) 10) Coordinates of Point D = (2,607 ; 0) 11) Distance Formula to Calculate X 12) X = sqrt [(2,466 - 2,607)^2 + (1,708 - 0)^2] 13) X = sqrt (0,019881 + 2,917264) 20) X = sqrt (2,937145) 21) X = 1,713486 22) ANSWER : X Length approx. equal to 1,7135 Linear Units.

[@redfinance3403]

@@LuisdeBritoCamacho AB has 2 solutions my guy

[@LuisdeBritoCamacho]

@@redfinance3403 wich are?

[@redfinance3403]

​@@LuisdeBritoCamacho 3.606 and 6.083

[@LuisdeBritoCamacho]

@@redfinance3403, You are right. Nevertheless this is an ambiguous Problem, with two Solutions; although PreMath only presented one.

[@redfinance3403]

@@LuisdeBritoCamacho yes I agree completely. A lot of his problems are good, but some are lacking information to provide a definitive answer.

[@phungpham1725]

1/ Label the amgle ACB as 2 Alpha and AD=m, BD=n We have: Area of triangle ABC= 1/2 x 3x 4 sin (2alpha)=3sqrt3-> sin (2alpha)= 3 sqrt3/2 --> 2 alpha= 60 degrees-> alpha = 30 degrees. 2/ Consider the two trisngles ACD and BCD They have the same height ( from the vertex C to AB) so their the ratio of their areas= ratio of the two bases= m/n CD is the bisector of angle ACB so, m/n = 3/4 Let S and S’ be the area of the triangle ACD and BCD we have: S/S’ = 3/4-> S/3=S’/4=(S+S’)/(3+4)=3sqrt3/7 --> Area of ACD=9sqrt3/7 --> 1/2 . 3. x. sin 30 degrees= 9 sqrt3/7 --> x = 12sqrt3/7 😅

[@anthonycheng1765]

How about 2*alpha = 120 degrees?

[@sergeyvinns931]

@@anthonycheng1765 2 alpha = 60 degrees!!!!!!!!!!! Accountant!

[@phungpham1725]

@@anthonycheng1765 Sorry about my hasty post! There is another result! Angle ACB can be 120 degrees as well. So, sin alpha= sin 60 dregrees= sqrt3/2. x = 12/7 units

[@murdock5537]

Nice! φ = 30°; ∆ ABC → AB = AD + BD = a + b; AC = 3; BC = 4; CD = x = ? BCD = DCA = δ; (1/2)sin⁡(2δ)12 = 3√3 → sin⁡(2δ) = √3/2 = sin⁡(2φ) → δ = φ → cos⁡(2φ) = sin⁡(φ) = 1/2 3/a = 4/b → b = 4a/3 → a + b = 7a/3 → 49a^2/9 = 9 + 16 - 2(3)(4)cos⁡(2φ) → a = 3√13/7 → b = 4√13/7 → ab = 36(12)/49 → x^2 = 12 - ab = 12√3/7 or: φ = 30°; 3/a = 4/b → b = 4a/3 → a + b = 7a/3 → 49a^2/9 = 25 - 12 → a = 3√13/7 → b = 4√13/7 → (1/2)sin⁡(2δ)12 = 3√3 → sin⁡(δ) = sin⁡(φ) = 1/2 area ∆ ADC = (3/7)(3√3) = 9√3/7 → (1/2)sin⁡(φ)3x = 3x/4 = 9√3/7 → x = (9/7)(4/3)√3 = 12√3/7