Can you find area of the triangle? | (Without Trigonometry) |

[@kaziummeruman524]

LOVE THIS❤️‍🔥🥰

[@PreMath]

Excellent! Glad to hear that! Thanks ❤️

[@mariopopesco]

Another way without trigonometry : Draw bisector of angle BAC, AD. In triangle ACD , angle DAC is x, angle ADC is 2x. So triangle ADC is similar with triangle BAC. Triangle ABD has angles BAD and ABD equal with x, so is isosceles, so AD is equal with BD. Lets say AC is y and BD is a. So AD is a also. BD + CD is 12, so CD= 12- a From similar triangles we have : AD/AB = AC/BC = CD/AC, so a/10 = y/12 = 12-a/y > a = 10y/12, and in second equation we have y/12 = (12- 10y/12)/y -> y^2 = 144 - 10y, so y = 8 the only positive solution. We have all the sides, 8,10,12, so with Heron formula s = (8+10+12)/2 = 15 Aria = sqrt (15*7*5*3) = 15 sqrt7

[@PreMath]

Excellent! Thanks ❤️

[@gibbogle]

I did the same.

[@thinker821]

Sine rule: 10/sin(3x) = 12/sin(2x), expanding sin(2x) and sin(3x) and solving a quadratic in cos(x) gives cos(x) = either 3/4 or -1/3, the latter being impossible since 2x needs to be less than 180 degrees. Hence cos(x) = 3/4 and sin(x) = √7/4. Area = 1/2 c a sin B = 1/2 * 10 * 12 * √7/4 = 15√7.

[@valentinebestofthevoiceall9289]

Yes, I've chosen the same way. It's easier, i think.

[@rudychan8792]

Hey, this is my idea too^^ Sinus Rules! More Effective! Cut this video from 14' to 7' 😉

[@PreMath]

Thanks ❤️

[@DanielNeedham2500]

How do you deduce the remaining angle is 3x to use the Sine rule?

[@valentinebestofthevoiceall9289]

@@DanielNeedham2500 sin (180° - a) = sin a

[@JSSTyger]

You are one of the best geometry instructors on KZbin.

[@Abby-hi4sf]

So great to see how to solve it without trig! Beautiful question and solution

[@PreMath]

Glad you enjoyed it! Thanks ❤️

[@jimlocke9320]

At 1:00, there is an implication that side AB is the base. Actually, side BC or side AC can be treated as the base. However, PreMath's solution is dependent on using AB as the base, so that is the reason why he chose AB as the base. There are probably other solutions that use the other sides as the base. Also, it is possible that using line segments to divide ΔABC into 2 or more triangles would produce solutions where AB is not used as the base. For example, once PreMath has values for h and a, he could have used CP as a common base of triangles ΔACP and ΔBCP and computed and added their areas to equal the area of ΔABC. As for the trigonometric solutions provided in the comments, these are great, but the title states that trigonometry is not used.

[@PreMath]

Thanks ❤️

[@michaelkouzmin281]

Another solution using trigonometry: 1) Angle ACB = 180-3x; 2) sin(180-3x)=sin(3x); 3) 12/sin(2x)=10/sin(3x); 4) sin(3x)/sin(2x)=10/12=5/6; => cos(x)=3/4 => 5) sin(x) = =sqrt(1-(cos(x))^2)= sqrt(1-(5/6)^2) = sqrt(7)/4 6) AreaABC = 10*12/2*sin(x)= 60*sqrt(7)/4=15*sqrt(7) = approx 39.686 sq units.

[@PreMath]

Excellent! Thanks ❤️

[@aura_beast2678]

This is an application of inmo question 1998 . you can prove that a²= b(c+b) 144=10x + x² Just use quadratic formula to get x and appy herons formula Easier than your method

[@溫文爾雅-q3f]

Can you write down the specific calculation process?

[@phungcanhngo]

Thank you professor for amazing geometry.

[@gibbogle]

I drew a different line, bisecting CAB, meeting CB at D. Then triangles ABC and DAC are similar. Let w = AC, y = AD = BD, z = DC. z/w = w/12, i.e. z = (w^2)/12, and y/w = 10/12, i.e. y = (5/6)w But BC = BD + DC = y + z = (w^2)/12 + 5w/6 = 12, which gives w = 8 With s = (triangle perimeter)/2 = (a+b+c)/2 where a,b,c are the side lengths area = sqrt(s(s-a)(s-b)(s-c)) We have s = (12+10+8)/2 = 15 and area = sqrt(15*3*5*7) = sqrt(1575) = 15sqrt(7).

[@bigm383]

Quite convoluted but strangely satisfying!

[@PreMath]

Thanks ❤️

[@misterenter-iz7rz]

Using sin rule, trigonometric means yields a simpler solution, as 12/sin 2x=10/sin 3x, give cos x=3/4, so sin x=sqrt(7)/4, thus the area is 1/2×12×10×(sqrt(7)/4)=15sqrt(7).😊

[@PreMath]

Thanks ❤️

[@溫文爾雅-q3f]

能寫下具體的計算過程嗎?

[@溫文爾雅-q3f]

Can you write down the specific calculation process?

[@misterenter-iz7rz]

@user-cb8lx4ot4y s×in 3x/sin 2x=10/12=5/6, sin 3x=sin x(3-4sin^2 x),sin 2x=2sin xcos x,....

[@misterenter-iz7rz]

You successfully solve it without any means of trigonometry. 😮

[@PreMath]

Yes! Thanks ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome! Thanks ❤️

[@prossvay8744]

Area of the triangle=1/2(12)(10)sin(41.41)=39.69 square units.❤❤❤ Thanks sir.

[@PreMath]

Great job! You are very welcome! Thanks ❤️

[@溫文爾雅-q3f]

Can you write down the specific calculation process?

[@giuseppemalaguti435]

Teorema dei seni 10/sin3x=12/sin2x....risulta cosx=3/4...A=12*10*sinx/2=15√7

[@PreMath]

Excellent! Thanks ❤️

[@溫文爾雅-q3f]

Can you write down the specific calculation process?

[@davidstaphnill9636]

Could you explain please why you replaced the 2 in the numerator with 1/4😊

[@shrikrishnagokhale3557]

I was wrong in taking a=18instead of 8.Corrected myself.Thanks.

[@shrikrishnagokhale3557]

Using Heron's formula my answer is 40√2sq.unit

[@alster724]

After seeing the quadratic equation, I did basic factoring instead of quadratic formula

[@joeschmo622]

Magic!

[@PreMath]

Glad to hear that! Thanks ❤️

[@professorrogeriocesar]

Problema maravilhoso. Muito obrigado. Fiz de outra forma, traçando a bissetriz relativa ao ângulo A. Muito obrigado e parabéns!!

[@PreMath]

You are very welcome! Thanks ❤️

[@ОльгаСоломашенко-ь6ы]

Можно без дополнительных построений. Применяя теорему синусов и синус тройного угла, находит тригонометрическую функцию x. Удобнее cos( x)=0.75. Зная cos( x) находим sin( x), высоту, проведённую к стороне AB.

[@PreMath]

Thanks ❤️

[@溫文爾雅-q3f]

Can you write down the specific calculation process?

[@ОльгаСоломашенко-ь6ы]

@@溫文爾雅-q3f BC/sin(2x)=AB/sin(180-3x), 12/sin(2x)=10/sin(3x), we use the reduction formula. sin(3x)=3sin(x)-4sin^3(x). Substitute 12/(2sin(x)*cos(x))=10/(3sin(x)-4sin^3(x)). We reduce everything that is possible and get 3/cos(x)=5/(3-4sin^2(x)). We replace the sine with the cosine on the right side of the equality using the basic trigonometric identity sin^1(x)+cos^2(x)=1. 3/cos(x)=5/(3-4+4cos^2(x)). 3/cos(x)=5/(4cos^2(x)-1). After the transformations, we obtain the quadratic equation 12cos^2(x)-5cos(x) -3=0. This equation has 2 roots, one of which is negative, which cannot be. cjs(x)=0.75. We find sin(x)=√1-0,75:2=√7/4 and the area according to the formula.

[@溫文爾雅-q3f]

​@@ОльгаСоломашенко-ь6ы Great, this is what I needed, thank you!

[@ОльгаСоломашенко-ь6ы]

@@溫文爾雅-q3f According to the sine theorem CB/sin(2x)=AB/sin(180-3x), 12/sin(2x)=10/sin(3x), we use the reduction formula. sin(3x)=3sin(x)-4sin^3(3x). 12/(2sin(x)*cos(x))=10/(3sin(x)-4sin^3(x). We're cutting everything we can. 3/cos(x)=5/(3-4sin^2(x)). We use the basic trigonometric identity sin^2(x)+cos^2(x)=1. 3/cox(x)=5/(3-4sin^2(x)). After the transformations, we get a quadratic equation with respect to cos(x), which has two roots, one of which is negative, which is impossible. So cos(x)=0.75. sin(x)=√1-(0,75)^2=√7/4. S=0.5*12*10 * sin(x)=15√7.

[@marcgriselhubert3915]

The area of the triangle is (1/2).BA.BC.sin(x) = (1/2).10.12.sin(x) = 60.sin(x). Now let(s calculate sin(x) We have: sin(x)/AC = sin(2.x)/12 = sin(180° -3.x)/10 in triangle ABC, so 10.sin(2.x) = 12.sin(3.x), or 5.sin(2.x) = 6.sin(3.x) Then: 5.2.sin(x).cos(x) = 6.(3.sin(x) - 4.(sin(x))^3) We divide by sin(x) which is non equal to 0: 10.cos(x) =18 - 24.(sin(x))^2 We simplify by 2 and replace (cos(x))^2 by 1 - (sin(x))^2, we obtain: 5.cos(x) = 9 -12.(1 -(cos(x)^2) or: 12.(cos(x)^2 -5.cos(x) -3 =0 Delta = (-5)^2 -4.12.(-3) = 25 +144 = 169 = (13)^2, so cos(x) = (5 -13)/24 = -1/3 which is rejected as negative, or cos(x) = (5 +13)/24 = 18/24 = 3/4 Then (cos(x)) = 9/16 and (sin(x))^2 = 1 - (9/16) = 7/16 and sin(x) = sqrt(7)/4 (as sin(x)>0, 0

[@PreMath]

Thanks ❤️

[@marcgriselhubert3915]

At the last line, please naturaly read : .... Area = 60.sin(x) = 15.sqrt(7).

[@CloudBushyMath]

Interesting

[@PreMath]

Glad to hear that! Thanks ❤️

[@wackojacko3962]

One needs to have a measured sense of reason to solve this problem. 🙂

[@PreMath]

Thanks ❤️

[@sergiocasavecchia5819]

Who grants that point T really exists ? And if angle CAB is greater than 90° ? You should demonsrrate first that 2x < 90° (without using trigonomerry ...)

[@himadrikhanra7463]

30 square unit ? Angles are in arithmetic progression

[@pralhadraochavan5179]

Good night sir

[@PreMath]

Same to you❤️

[@sorourhashemi3249]

How do you know the angle CTA is 2x?

[@johnbrennan3372]

By construction he made it equal to 2x

[@unknownidentity2846]

Let's find the area: . .. ... .... ..... Let's introduce a point D on BC such that ∠BAD=x. Then we have two triangles ABD and ACD with: ∠BAD = x ∠ABD = x ∠ADB = 180° − ∠BAD − ∠ABD = 180° − 2x ∠CAD = x ∠ADC = 180° − ∠ADB = 180° − (180° − 2x) = 2x ∠ACD = 180° − ∠CAD − ∠ADC = 180° − x − 2x = 180° − 3x Therefore ABD is an isosceles triangle (AD=BD) and the triangles ABC and ACD are similar: AB/AD = AC/CD = BC/AC AC/CD = BC/AC ⇒ AC² = CD*BC AB/AD = AC/CD AB²/AD² = AC²/CD² AB²/AD² = CD*BC/CD² AB²/AD² = BC/CD AB²/AD² = BC/(BC − BD) AB²/AD² = BC/(BC − AD) 10²/AD² = 12/(12 − AD) 100/AD² = 12/(12 − AD) 25/AD² = 3/(12 − AD) 300 − 25*AD = 3*AD² 3*AD² + 25*AD − 300 = 0 AD = [−25 ± √(25² + 4*3*300)] / (2*3) = [−25 ± √(625 + 3600)] / 6 = (−25 ± √4225) / 6 = (−25 ± 65) / 6 Obviously only one of the two solutions is useful: AD = (−25 + 65)/6 = 40/6 = 20/3 Since ABD is an isosceles triangle, it can be divided into two congruent right triangles. So we can conclude: cos(x) = (AB/2)/AD = (10/2)/(20/3) = 5*3/20 = 3/4 sin(x) = √(1 − cos²(x)) = √(1 − 9/16) = √(7/16) = √7/4 Finally we are able to calculate the area of the triangle ABC: A(ABC) = (1/2)*AB*BC*sin(x) = (1/2)*10*12*√7/4 = 15√7 Best regards from Germany

[@unknownidentity2846]

I also made a comment with an alternative solution to the problem you published yesterday. I'm looking forward to your opinion.

[@PreMath]

Excellent! Thanks ❤️

[@StephenRayWesley]

(10)^2=100 (12)^2=144 2x(45°)=90°x. 3x(15°)=45°x 3(15°)=45°x {90°x+45°x+45°x}=180°x^3 b{100+144}=244 ,{180°x^3-244}=0√76°x^3 4^√19x^2 4^√19^1x^3 √4^√1^√1x^3√ 2^2x^3 √2^1^2x^3 √1^√1^2x^3 23 (x+2x-3)

[@PreMath]

Thanks ❤️

[@rudychan8792]

Looks Complicated, Messy, Too Long. Sinus Rules, Please: sin(3x) ÷ 10 = sin2x ÷ 12 sin3x / sin2x = 10/12 ... do the rest, you will get 12p"-5p -3 = 0 ; p = cos x = 3/4 Turn into sin x = √7 ÷ 4 Area = (1/2)•10•12•(√7/4) = 15√7 = 39,7 sq units Much-much Easier\Practical. --- Anyway, i'll give you thumbsup, anyway^ "Longtime no see" Dec'23 today Mar'24 😉 👍

[@PreMath]

Good to see you again 😀 Thanks ❤️

[@adept7474]

СК || АВ. АК - bisector ∠ВАС. СН ⟂ АВ. АСКВ - Isosceles trapezoid. СК = АС = ВК = х, АН = y. х² - y² = 12² - (10-y)², х = 10 - 2y. (10 - 2y)² - y² = 12² - (10 - y)². y² - 15y + 14 = 0. y = 1. СН = √(12² - 9²) = √63. S(АВС) = (10√63)/2 = 15√7.

[@SkinnerRobot]

I really like how you let AC = 10 - 2y , and the rest is so easy. Bravo! Thumbs up.

[@adept7474]

@@SkinnerRobot Thanks

[@PreMath]

Thanks ❤️