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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.

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!!

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

Sorry @PreMath. The question itself is wrong. It should be "Given the Circle Area = π. If arcs AB, BC and CA are 1/4th, 5/12th and 1/3rd of the circle circumference respectively, find out the area of the Triangle ABC.

Hi,this has truly helped me 100% .Thanks so much

Thank you!

Thnku from india❤

This video is a testament to the power of human connection. Heartwarming!

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.

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

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'

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

sin120°=√3/2

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).😊

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.

CD is bisector so AC/BC=AD/BD 3/4=AD/BD AD=3a ; BD=4a X^2=AC.CB+AD.BD X^2=12+12a^2 Let LACB=2x 1/2(3)(4)sin(2x)=3√3 So x=30° Cos(60)=(3°2+4^2-AB^2)/2(3)(4) AB=√13 7a=√13 So a=√13/7 So x=√12+12a^2 x=2√186/7 units=3.9units.❤❤❤

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.

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

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

Nice

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 😅

Thanks, but more algebra problems for us.

I can relate! Woohoo! 🙂

r=8, diameter=16 the side of the square = s area of the square = s² s² + 9s² = 16² 10s² = 256 s² = 25.6

Very good

Thank you so much Sir🙏

Don't put any advanced math, physics, engineering degrees on job applications. Sometimes you just need a job to pay the bills and they won't hire you as a truck driver if you are being honest..

Thank you professor for amazing geometry.

Good morning sir

Kinda solved in my head in 10 seconds: Let E be the center point of the square making the diagonals 4, the sides 2sqrt2, and each 1/4 of the square, one of which is our triangle, 2 square units..

I am going to be very quick 01) Draw a Vertical Line from E. Define Point F between Point A and Point B. 02) Triangle ADE is congruent with Triangle AEF 03) 2 / AD = EF / 2 04) AD * EF = 4 05) AD = AB 06) Green Area = 4 / 2 = 2 ANSWER : Area of Green Triangle equal to 2 Square Units.

Thank you!

Here's a different kind of solution. The locus of the right angle of the yellow triangle is a semicircle inside the square whose diameter is the left side of the square. If we assume that the area of the green triangle is determined uniquely by the available data, then we can safely place the right angle at the center of the square, making the green triangle an isosceles right triangle with legs 2 and area (1/2)(2)(2), which is 2. The solutions given in the video confirm the above assumption.

2^3^4

This is too hard

If F is the projection of E onto AB, let y be the length of EF. The area G of the green triangle is 1/2*x*y. Triangle AEF is similar to triangle DAE. Thus 2/x=y/2 => y=4/x so G=1/2*x*(4/x)=2.

DE is not given so the solution must be the same for all valid values of DE and we can solve special cases. monthnorth3009 chose DE = 2 and found the area of the green triangle to be 2. A simpler special case is DE = 0. ΔADE is now a line segment equal to the side of the square, so AD = AB = 2 and green triangle has area 2. However, we no longer really have a triangle. To make our case more legitimate, let DE be extremely small but >0. As we make DE smaller, we can make AD as close as we want to AE but never make them equal as long as DE > 0. Eventually, we can make DE small enough to treat the difference between AD and AE as negligible. In mathematics, AD = AE is called the limit as DE approaches 0.

I like this problem very much! Here’s my solution : let x be the side length of the square. draw a perpendicular to AD from E, let the intersection at AD be labelled as G. Let GE = h. draw another perpendicular to AB from E and label its intersection at AB as H. HE = sqrt(2^2 - h^2) = sqrt(4 - h^2) by Pythagoras. The area of triangle AED is either (AD)(GE)/2 or (AE)(ED)/2 so (AD)(GE)/2 = (AE)(ED)/2, so ED = xh/2. By Pythagoras, ED^2 + AE^2 = AD^2. So (xh/2)^2 + 2^2 = x^2. Solving for h^2 gives h^2 = 4 - 16/x^2. Now, HE = sqrt(4 - h^2) = sqrt(16/x^2) = 4/x. Finally, area of green triangle equal to (AB)(HE)/2 = 1/2 * x * 4/x = 2 units squared! When you can’t solve with similar triangles, look for Pythagoras. When you can’t solve by Pythagoras, solve by trigonometry. When you can’t solve by trigonometry, solve by coordinate geometry 😂

Both methods are wonderful.

Holly shit! Beautiful question!!!

DE can also be 2 units which fully satisfies the 90 degree angle requirement, and then each yellow corner angle will be 45 degrees. Yellow and green will be congruent with areas 1/2 x 2 x 2. 2 units.

Usual trick, we suppose the diagonal is 4, side is 4/sqrt(2), area is 1/4×(4/sqrt(2))^2=2.😂😂😂

1/Drop the height BH to AE (extended) We have: angle AEE = angle BAH ( having sides perpendicular) so the two triangles ADE and HAB are congruent so, AE=BH=2 --> Area of the green triangle= 1/2 x2x2= 2 sq units😅

Pretty killer

Trig identities are more fun! 🙂

Super Duper👍

1.extend AE→AH丄BH △ADE,△BAH congruent(AAS) →AE=BH=2 2.△AEB=AE·BH/2=2😊

Nice! ∎ABCD → AB = AF + BF = BC = CD = AM + DM = r + r; AE = 2; sin⁡(EFA) = 1; EF = h; ADE = FAE = δ → sin⁡(δ) = 2/2r = 1/r = h/2 → hr = 2 → h = 2/r → rh = 2 = area ∆ AEB

Так это тот же болт. только вид с боку! h/2 = 2/x=sina, тоже тригонометрия.

Интересно. какой второй метод. Первый решается устно!