First comment 😊❤️

@PreMath

2 жыл бұрын

Yes you are! Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards

@bahsarabi4234

2 жыл бұрын

❤️

@MathZoneKH

2 жыл бұрын

❤️😊

@durgalimbu9739

2 жыл бұрын

@@PreMath sir so long process ? cannot use short method ?

@hahatseck

2 жыл бұрын

I first use sine rule and have cosX=15/18 sinX=sqrt(11)/6 and Area=18^2/2*cos2Xsin2X+ 30^2/2*sinXcosX =18^2sinXcosX(2(cosX)^2-1)+ 30^2/2*sinXcosX =80sqrt(11)

18/Sin x = 30/Sin 2x. But Sin 2x = 2.Sin x . Cos x. So 18/Sin x = 30/2.Sin x . Cos x Multiplying both sides by Sin x we get:- 18 = 30/2Cos x. 2 Cos x = 30/18 Cos x =30/36 = 0.8333 Cos x(-1) = 33.56 degrees. So back to the triangle, the angle ABC is 180-33.56-67.12 = 79.32 degrees. Area = 1/2 x18 x30 x Sin 79.32 = 265.323 Ans.

@robertlynch7520

2 жыл бұрын

I got the same answer, the same way; see my write-up herein. My final part is a little different, computing 𝒄 and 𝒅 parts o the baseline, but ultimately, quite similar approaches.

@PreMath

2 жыл бұрын

Great tip! I'll make another vid with trigonometry and will be uploaded by tomorrow hopefully. Thanks Monty for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from Arizona, USA!

@armacham

2 жыл бұрын

That's basically the way I went except I didn't reduce it to decimals, I kept the radicals, so I was able to get an exact answer for sinx, cosx, sin2x, cos2x, the base, the height, and the area. The other identity to use is (sin x)^2 + (cos x)^2 = 1. With that, you can use the value of cos x to solve for the value of sin x. Knowing that cos x = 5/6, (cos x)^2 = 25/36, so (sin x)^2 = 11/36, so sin x = +- sqrt(11)/6. And you know sinx must be positive because angles in a triangle are between 0-180 degrees and sin x is always positive when x is between 0 and 180, so you can reject the negative solution. sinx = sqrt(11)/6. From there it's trivial to calculate sin2x and cos2x using the same formulas. With the values of sinx, cosx, sin2x, and cos2x, and the side lengths you're given at the start, it's easy to split the triangle into two right triangles and apply the formulas (sine = opposite/hyp, cos = adj/hyp) to calculate the length of the base and the height of the triangle and then apply the formula a = bh/2

@pankajkumarpandey6658

2 жыл бұрын

I have also said the same but you have solved completely. Excellent

@calspace

2 жыл бұрын

I have been out of school too long, so the final equation you use is not familiar to me. But I’ll follow the same up until that steps. Then I used the sin equations to find the base which was 32. I then found the semiperimeter, which is 40, and used Heron’s formula. Area = sqrt(p * (p-a) * (p-b) * (p-c))

As many have noted, using the sine law and the double angle formula for sine gives cos(x) = 5/6, and thus EC = 25, then using the observations you had you can easily see that AE = 7 and thus AC=32. The Pythagorean theorem used in triangle BCE gives BE, and you can finish with the standard Area formula. In the end you get an exact answer with radicals and you don't have to do any tricky algebra.

Profe - Sometimes it amazes me how many interesting problems you can pull out of a few triangles, squares and circles!

It’s obviously a 345 triangle. CX is 30 A 2x is 60, B must be 90. It’s a simple triangle base times height is 540. Half of that is your area = 270.

1) Use sine law to find value of angle x. 2) Use law of cosines to find the third side. 3) Use the area formula i.e (absinC)/2 -> This is just my way of doing it and there can be multiple ways to approach it. Btw, Kudos to them who solved it without using Trigonometry!

@chessdev5320

2 жыл бұрын

or simply after the 1st step, u can calculate angle B i.e (180°-3x) sinB=sin(180-3x)=sin3x Use triple angle formula as you already know sinx ;D And then just use the area formula:- =(18×30×sin3x)/2

@PreMath

2 жыл бұрын

Dear Kumar, in my next vid, I'll do this problem using trigonometry. No worries. I love trig as well. Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

@chessdev5320

2 жыл бұрын

@@PreMath well actually, your method is much better than me as mine method is just a list of formulas while your method is a good use of geometry 😁.

sine rule makes the process a lot simpler..

@nitinarora5719

2 жыл бұрын

1/2 ab sinC

@anonim6160

2 жыл бұрын

In which triangle

@anonim6160

2 жыл бұрын

nvm

@sumangupta1737

2 жыл бұрын

@@anonim6160 use sine rule and use the formula of sin3x to find sinx (i.e included angle of given sides..)

@sadtear796

2 жыл бұрын

Thats right

3:41 There is no AAA Congruence theorem. The 2 triangles are congruent here, but you would need to use ASA or AAS congruence.

I used the double angle formula and the fact that h=18sin(2x) and that sin(x) = h/30. Obviously cos(x) =sqrt(1 - (h^2)/900) and from there it is just rearranging to find h and then pythagoras' theorem to find the base. A nice quick and easy area problem - perfect for an A-level maths lesson starter

Very great video sir. Making students gain confidence in math. Thank you sir

I found a soultion with less steps: Draw AD - angle bissection of angle A -> triangles ABD and CBA are similar with coefficient 18/30 Now we can find BD = 18/30 * AB => DC = BC - BD => DA = DC (angles DAC=DCA) => AC = 30/18 * DA = 32 Now using Heron's formula p = 40 S = sqrt(40*10*8*22)=80*sqrt(11)

@timeonly1401

2 жыл бұрын

Beautiful!!

Thank you sir. Since the length of side is 18, 30 and 32, we also can calculate the area using: A = √s(s-a)(s-b)(s-c) which s=(a+b+c)/2 In this case, s=(18+30+32)/2=80/2=40 Then: A=√(40.(40-18).(40-30).(40-32)) A=√(40.22.10.8) = √70400 = 265.33

@logos2114

2 жыл бұрын

hey we said it "u" teorem in turkey what you said it ? ( u(u-x)(u-y)(u-z) )^1/2

@sie_khoentjoeng4886

2 жыл бұрын

@@logos2114 We know it is Heron theorems of my mind serves better 😃😃. Maybe 's' is anbreviate of semilength (half of sum of length) instead of 'u'

@gurmukhsingh2358

2 жыл бұрын

It is called heron's formula here in india or simply hero's formula in local language

@logos2114

2 жыл бұрын

@@sie_khoentjoeng4886 😶 all people say it heron therom just l think it name is u teorem :// 😳

@versexx413

2 жыл бұрын

By Heron's formulae

تمرين جميل. وشرح واضح مرتب . شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم وينصركم . تحياتنا لكم من غزة فلسطين .

Loved this one! Thank you!!

Very nice questions framed and give room to thought processes, wherein the math lovers try to solve in different methods. I did this using trigonometry and used 5 steps ... but for those learners, who haven't got into trigonometry can solve as explained in the video solution ... But this tribe grow to sustain the live for Maths

Nice one sir keep it up

Put AC on the x-axis with A at (0,0) C at (u,0); then vectorAB＝AC+CB we obtain 18sin2x＝30sinx. Therefore, cosx＝5/6; sin2x＝5sqrt(11)/18; cos2x＝7/18 and u＝32； total area＝determinat of〔AC AB〕 /2＝80sqrt(11)

@PreMath

2 жыл бұрын

Thanks for sharing Yao You are awesome 👍 Take care dear and stay blessed😃 Kind regards

@user-rt6si6pf5b

2 жыл бұрын

I did same solution!

@khyatikumar3867

2 жыл бұрын

Awsm solution Thank you sir❤️

@colinratcliffe3074

2 жыл бұрын

me too!

@freemathacademy6632

Жыл бұрын

kzread.info/dash/bejne/hp1sqZmMg6nSeqg.html

Another solution is to draw the angular bisector of the angle measuring 2x degrees and using angular bisector property and using similar triangles property and get BC =32.Though this method is a bit tortuous, it serves the purpose anyway.

Everyday I Watch your vídeos. Geômetry is a fun for me. A diversion like Cross words. My profession is ophthalmologist, but I love geometry as well. It is a hobby. Greetings from Brazil, South American , City capital mamed Brasília. Very nice brazilian people and beautiful country. Brazil is not only Amazonian. We Have very nice cities On-The south of the country. Came and visit this beautiful people with open arms for all. You must visit Iguaçu Falls near frontiers with Argentina and Paraguay. The bigger of the World.

@PreMath

Жыл бұрын

Wow, Great! Glad to hear that! Brazil is a beautiful country with beautiful people. So kind of you, my dear friend. You are very generous. Cheers! You are the best Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

Nice explanation Thank you.proof by using only common rules and theorems 👏👏

Excellent method. I did it using trigonometry. Area of triangleABC=1/2 by 18 by |ac| by sin 2x =1/2 by 30 by |ac| by sin x.This gives cos x =5 /6and sin x= (sq root11)\6. Area of the triangle ABC= 1/2 by 18 by30 by sin ( 180- 3x) = 1/2 by 18by30 by( 3sin x-4 sin cubed x).But sin x= (sq. root 11)/6 So area of triangle = 270 by (16 by sq root 11)/54 = 80 by sq root of 11.

Hi dear, the problem can be solved by using bisection of BAC angle and so on Thanks for this example 👍

Interesting question 🔥

ಜೈ ಶ್ರೀ ರಾಮ್ 🙏 THAN Q "PRE Math "for presenting suuuuuper video of PUZZLE solution . very interesting.MAY GOD BLESS U ❤️!!!!

May be using sine rule it would have been easier. 18/sin(x)=30/sin(2x) .we can get sin(180-3x) area =1/2 *18*30 *sin(180-3x)

@DanielNeedham2500

2 жыл бұрын

Exactly the way I've just done it

@dhrubajyotidaityari9240

2 жыл бұрын

I used this.

@ddmm7893

2 жыл бұрын

Yes but so long !!!

@DanielNeedham2500

2 жыл бұрын

How do you use that to find the value of x using algebra

@dhrubajyotidaityari9240

2 жыл бұрын

@@DanielNeedham2500 algebra method is not easy.because X= arc cos 5/6.

Great video sir 👍I appreciate that.

Geometry is very interesting as well as some hard ,nice video sir

Alternatively you can draw a line from point A to BC so that it divides angle A into two equal angles of x degrees. Call the intersection with BC point D. In triangle ABC angle B is 180-3x degrees. In triangle ABD angle D then is 180-(180-3x)-x=2x degrees. As you can see triangle ABD and triangle ABC are similar triangles (two same angles and one same side). Using the law of sines in triangle ABC we can state that 18/sin x=30/sin 2x and so 18 sin 2x=30 sin x and sin x=18/30 sin 2x. Using that same law in triangle ABD we can state that 18/sin 2x=BD/sin x and so 18 sin x=BD sin 2x. Substituting for sin x gives 18(18/30 sin 2x)=BD sin 2x and so BD=324/30=10.8. That means CD=30-10.8=19.2. If you now consider triangle ACD you have angle A and angle C which are both x degrees and so triangle ACD is an isosceles triangle. That means AD=CD=19.2. Since triangles ABD and ABC are similar triangles we can now state that AB/BC=AD/AC so 18/30=19.2/AC which gives AC=32. Using Heron's Formula we can then calculate the area is √(s(s-a)(s-b)(s-c)) with s being half the perimeter. So s=1/2*(18+30+32)=40 and we get area=√(40(40-30)(40-32)(40-18)=√70400=80√11.

18/sinx = 30/sin2x, we get cosx=5/6.

@ankaiahgummadidala1371

2 жыл бұрын

I also did the same method.

@krabkrabkrab

2 жыл бұрын

@@ankaiahgummadidala1371 me too. I did it entirely mentally. It directly gives cosx=5/6, so EC=25. Then cos2x=2cos^2x-1=7/18, so AE=7 and the base, AC=32. h^2=30^2-EC^2=275. Half base times height is 16*sqrt(275). This method is way quicker.

@lazzatbastar3497

2 жыл бұрын

But not full answer. I mean cosx=5/6 does not give us AC length yet. Only knowing length of 2 sides and sin or cos of angle between two sides can let us calculate area of triangle. In exact S= 1/2* 18 * AC *sin x So we get cos x= 5/6 from sine theorem for triangle : sinx/18 = sin 2x/30. In exact sin x/sin 2x = 18/30, sin x /(2*sinx*cosx)= 18/30, 2*cos x=30/18, from where cos x= 5/6. Then we can calculate sin x from (1-(cos x)^2)^(1/2. So sin x=(11/36)^(1/2) Then using this information calculate AC which is sin(180- 3x)/AC = sin x/18 from sine theorem sinx/18=sin 2x/30 = sin (180-3x)= AC, where sin x = (11/36)^(1/2) Where sin(180-3x)= sin 3x which equals sin x*cos 2x+sin 2x*cos x. Where sin 2x= 2*(cos x)^2-1 = 1-2*(sinx)^2. We know already values for cos x, sin x At the end replace value for sin 3x and sin x in sin 3x/AC = sin x/18 and get AC=32. So area of triangle is S=1/2* 18*32 *sin x. Where sin x is (11/36)^(1/2) to get S= 1/2*18*32*(11/36)^(1/2)= 48*(11)^1/2

Brilliant, I got the correct strategy, just needed guidance clipping it all together, great detailed explanation as always. Thanks again 👍🏻

@yeetboi4877

Жыл бұрын

The strat in the vid sucked i solved with system of sines.

A área do triângulo ABC é a metade do polígono formado pelos vértices A,B,C e "D". Uma vez achados os lados do polígono ABCD que são AB=18 e AC=32, basta concluir que a área do triângulo é A=(18*32)/2=288. E outro modo de achar o ângulo x é pela lei dos senos : (18/sen x) = (30/sen 2x), em uma única equação. Encontrando "x" se determina o ângulo ^B, e em seguida o lado AC. Dessa forma seria muito mais simples e limpo.

My final answer is 80(sqrt(11)). I think the trick is remembering the double angle formula for sine. The other side length is 32.

Nice Explanation sir

Perfect 🙌🙌

Good . Explanation.. you are crossed 100 k.. congrats dear brother🎉🎉

Спасибо. НО , можно чуть иначе. (1) A=0.5*18*30*sin(180*-3*x)=270*sin(3*x). Известно , что (2) sin(3*x)=3*sin(x)-4*[sin(x)]^3. По теореме синусов для треугольника ABC , получаем 18/sin(x)=30/sin(2*x). Отсюда cos (x)=5/6 , а (3)sin(x)=sqrt(11)/6. Подставляем (3) в (2) , потом в (1) - получаем Ваш ответ. С уважением, Лидий.

@user-rk5eh2sh9v

2 жыл бұрын

По мне так бесконечность решений, ибо со сторонами 18 и 30 и зависимостью углов что угол а в два раза больше угла б - бесконечное множество треугольников. Допустим угол х - 30, тогда площадь 270. А у треугольника с углами 36, 72 и 72 который внезапно равнобедренный можно опустить высоту из угла С на сторону АВ, которая будет равна корень из (900-81) а площадь этот корень помножить на 1/2*18. И очевидно что это не 270. Вот уже 2 треугольника удовлетворяющих условию и с разными площадями.

@user-pd7js7cy9m

2 жыл бұрын

Увы! Нельзя «допустить» , что один угол 30* , а другой 60*. Ибо : 18/sin30* НЕ РАВЕН 30/sin60*. А мы все верим в ТЕОРЕМУ СИНУСОВ. С уважением, Лидий.

@user-rk5eh2sh9v

2 жыл бұрын

@@user-pd7js7cy9m То есть выходит, что существует только одна комбинация и углов где один больше другого в два раза и при которых стороны 30 и 18. Я просто визуально прикинул что если тянуть на за точку В этого треугольника будет меняться высота, длина основания и углы, но меняться они будут не произвольно, а относительно длины сторон. То есть да, значения углы могут быть какие угодно, но так что бы один угол был больше другого в два раза только один вариант. Спасибо.

@sv6183

Жыл бұрын

@@user-rk5eh2sh9v @Олег Полканов Просто должно соблюдаться условие, что cos x = b/(2a), где а - сторона лежащая напротив угла х , b - напротив угла 2х.

Very nice explaination sir. Thanks.

Why don't we opt for the sine rule... It's quite easier... Although if you are willing to go for a rigorous solution then you use the method shown in the video

@ankaiahgummadidala1371

2 жыл бұрын

Following sine rule is a better method than what is followed in this video. You will arrive at the solutio with less time and effort.

@MarieAnne.

2 жыл бұрын

How is the sine rule less rigorous?

Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal.... '' ''

Very good solution After first line to make them isotriangle rest one straight forward process

Thank you for this solution But I used another way which is the sin law And I got cos x Considering the third angle(180-3x)we can get the area 1/2*18*30*sin(180-3x)=sin3x =sin (2x+x) Using the sum formula and double angle we get the same result I am glad you wrote your opinion

@cosmosz8125

2 жыл бұрын

rak tama khoya la3ziz

@cosmosz8125

2 жыл бұрын

momkin l9awha b sinx w cosx (we use sinx=h/30 , sin2x=h/18 and sin²x+cos²x=1.....)

@user-pt7wn8pm1j

Жыл бұрын

Solving by sin(3x) is not a good idea in this case. reason 1, sin(x) = sqrt[1- (cosx)^2] = sqrt(11)/6, a "sqrt" in included. reason 2, sin(3x) = 3*(sinx)-4*[(sinx)^3]. sin3x need to deal with "sqrt" and "(sqrt)^3". This solution need complex calculation.

y+2x+x=180 => y=180-3x [sin(2x)*18*AC/2]=sin(x)*30*AC/2 9*2sin(x)cos(x)=15 , sin(2x)=2sin(x)cos(x) cos(x)=5/6 => x=33.557 S=(18*30sin(180-3*33.557)/2)=265.33

Really nice, this time the puzzle was more difficult what make it more "edcatif" i dont speak english very nice😅

You can also calculate it from: P=1/2a*b*sin(

The problem actually has 2 solutions and you only consider one. Demonstration 1) sin (x) / 18 = sin (2x) / 30 => cos (x) = 5/6 2) AB² = AC² + BC²-2.AC.BC.cos (x) 18² = AC² + 30²-60.AC.5 / 6 AC²-50.AC-576 = 0 Delta = 625-576 = 49: 1st case: AC = 25 + 7 = 32 It is the solution that you consider and that gives Area = 80V11 and 2nd case: AC = 25-7 = 18 You do not consider this solution which gives according to the formula of Heron Area = 45V11

After finding the length of AC - which is 32 - you can use the cosine rule and rearrange to find the angle 2x 2x=Cos-¹(18²+32²-30²/2(18)(32)) which gives you 67.1... (1dp) then use ½abSinC to find the area ½×18×32×Sin(67.1...) = 265.32...

denoting AB by c and AC by b one gets a = BC = c+ 2d ( say) So height AD *AD = c*c - d*d = b*b - (c+d)^2 simplifying d= b*b/(2c) - c so AD *AD = c*c - d*d =(2c - b*b/(2c))(b*b)/(2c)) so desired area =(b*b+ 2c*c)/(2c))AD/2 For c = 18 b = 30, b*b/(2c)= 25 Hereby AD = 5√(11) and a = 7+7 +18 Therefore deaired area = 80√(11) Another method b/ sin(2x) = c/ sin(x) or b/(2c) = cos(x) and so forth

It is a very beautiful solution

Draw a line from C to the extension of the side BA so the angle between AC and this new line is X, because angle A is 2X, the other angle mising from this new triangle has to be X too, the side opposite to the first x is equal in length to AC because is an isosceles triangle, apply similarity and 32 as the length of AC. Then use heron's formula for the area.

Thank you sir

Can't believe you have chosen this very long and complicated method. You can do it in two lines using sine rule to get angle X, then angle B = 180 - 3X , the use area = 1/2 (18) (30) sin B.

You may also use heron's formula √s(s-a)(s-b)(s-c) where s = half of parameter and a,b,c are sides of triangle

Here is another way 1.extend CA to CD such that AD=AB=18 2.triangle ABD is similar to triangle BCD 3.AB:BD=BC:CD , then CD=50 AC=32 4.AB=18,BC=30,AC=32 ,use Heron’s formula get the answer , 80sqrt(11)

@Andy-ly1ww

2 жыл бұрын

Nice solution：D

Also such a way: (1) Bisect angle A. Let D is intersection point with BC such that BC=BD+DC . Then ADC is isosceles triangle AD=DC then (2) ABD ~ CBA --> AB/CB=BD/BA=AD/CA --> 18/30=BD/18=AD/AC where AD=BC-DB=30-BD --> BD=18^2/30 AD=30-18^2/30=(30^2-18^2)/30 and AC=AD*30/18=(30^2-18^2)/18=12*48/18=32 (3) Half perimeter p=(AB+BC+CA)=(18+32+30)/2=40 . Applying Heron relation S=Sqrt(40*(40-18)*(40-32)*(40-30))=Sqrt(4*10*2*11*8*10)=80*Sqrt(11) is ANSWER

Well explained

At 3:39 the theorem that you used to prove that the two triangles is not a valid congruence theorem because AAA theorem is valid for similar triangles and not for congruent triangles. The triangles can be said to be congruent by RHS, ASA or SAS congruence theorems.

Just awesome

this can be done with those theorems as well as using trig. but which method is quicker, which method is widely accepted?

Thank You Teacher

Law of sines and double angle for sine - value of cosine of x Pythagorean identity - value of sine of x From sum of angles in triangle on the Euclidean plane is 180 so we need the value of sin(3x) sin(3x) can be calculated using double angle identity for both sine and cosine and then sin(3x) can be calculated using sin of sum

Nice Sir

Angle, angle, angle congruency for triangles? It should have been HL. AAA is used for triable similarity.

Nice question and very nice solution. Thank you, professor. Have a nice day.

I used Law of Sines to calculate value of x, which is 33.557 degrees. Then, 30sin33.557 = 16.583 which is the altitude. The base is obtained by adding 18cos67.115 and 30cos33.557, which equals 32. Then 1/2basexheight = 265.328 units.

Different approach and much faster: drop a vertical of length h from B to the base b to get two triangles. Use the sine expressions for x and 2x based on h and the given length (18 and 30) and take sin(2x)=2sin(x)cos(x). It follows that cos(x) = 5/6. Then divide b into the parts left and right of the vertical. Using cos(x) and Pythagoras they are 25 and 7, so b=32. Further using Pythagoras, h = 5*sqrt(11). A=0.5*b*h=80*sqrt(11).

@hakkisuperheld

2 жыл бұрын

Indeed, I've done the same approach.

@okeuwechue9238

Жыл бұрын

I suspect PreMath prefers always solving problems the LONG way because it may help to increase youtube monetization and ad dollars...

Thanks very much🙏🙏🙏🙏

Good solution

Apply sine rule first we get angle A is equal to 180 minus 3x, then with the help of sine rule we find cosx =5/6, sinx = root 11 upon 6, after that area of triangle is equal to 1/2*18*30*sinA which is equivalent to 80 root 11.

Sir it can easily be solved by sine rule. From that cos x=5/6,sin x=√11/6. Area of triangle=0.5*18*30*sin(180-3x)=0.5*18*30*sin 3x=80*√11

More geometry questions please.

I did this using trig exclusively. sin x/18 =sin2x/30 implies sinx/18=2sinxcosx/30 thus cos x = 5/6 and sinx = 11^(1/2)/6 then the area is A = .5(18)(30)sin(180-3x) = 270(sin (180-3x)) the expression sin (180-3x) can be shown to equal sinx(3-4(sinx)^2) = 11^(1/2)/6(3-4*11/36)= 11^(1/2)/6(3-44/36)=11^(1/2)/6(64/36)=11^(1/2)/6(16/9) so A =270*11^(1/2)(16/54)= 80*11^(1/2).

good one...same time Sin formula can be used here and easier

What is the guarantee that BD of length 18 will meet AC at D

30/sin(2x) = 18/sin(x) gives cos(x)=5/6 by using sin(2x) = 2sin(x)cos(x). Then the final side follows from rule of cosines: 18^2 = 30^2 + c^2 - 2*30*c*(5/6), so c^2 - 50c +576=0 from which c=32 follows. Then apply standard Heron formula for the area. No need for any drawings or helping triangles.

Very good

It was a very challenging problem, but you provided its outstanding solution. Kudos to you dear. Love and Prayers from India!! ❤️

@PreMath

2 жыл бұрын

Thank you so much Leonardo for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊 Enjoy every moment of your life 🌻 Greetings from the USA!

Trigonometry also provides a relatively straight forward solution. № 1.1: 𝒉 = 𝒂 sin 2𝒙 … where (𝒂 = 18;) № 1.2: 𝒉 = 𝒃 sin 𝒙 … where (𝒃 = 30;) Expanding, rearranging № 2.1: 18 sin 2𝒙 = 30 sin 𝒙 … rearranging № 2.2: 18 ÷ 30 = (sin 𝒙) / (sin 2𝒙) Remembering that (sin 2θ = 2⋅cos θ⋅sin θ), then № 2.3: 18 ÷ 30 = ( sin 𝒙 ) / ( 2 cos 𝒙 ⋅ sin 𝒙 ) … cancelling № 2.4: 18 ÷ 30 = 1 / ( 2 cos 𝒙 ) … inverting № 2.5: 30 ÷ 18 = 2 cos 𝒙 … inverting, and moving the '2' around № 2.6: 30 ÷ 36 = cos 𝒙 … and solving № 2.7: arccos( 30 ÷ 36 ) = 𝒙 … numerically № 2.8: 𝒙 = 33.56° Well! Now we're armed with a nice big shotgun shell: № 1.3: 𝒉 = 30 sin (𝒙 → 33.56°) № 1.4: 𝒉 = 16.584; Just got to figure the length of the base line to determine △ area: № 3.1: 𝒄 = 𝒂 cos 2𝒙 № 3.2: 𝒅 = 𝒃 cos 𝒙 № 3.3: 𝒄 = 7.0 № 3.4: 𝒅 = 25.0 № 4.1: base = 𝒄 + 𝒅 № 4.2: base = 7 + 25 № 4.3: base = 32 Since we have the height (𝒉 → 16.584) № 5.1: area △ABC = ½(base ⋅ height) 𝒖² № 5.2: area △ABC = ½(32 ⋅ 16.584) 𝒖² № 5.3: area △ABC = 265.33 𝒖² And that is that! ⋅-⋅-⋅ Just saying, ⋅-⋅-⋅ ⋅-=≡ GoatGuy ✓ ≡=-⋅

@rafaelmayoral5571

2 жыл бұрын

Cos-1 x = 30/36 x = 33º 33` h=16.58, a =7 b = 25 A = 265.28

Good solution, but solution with trigonometry is shorter. I first used law of sines: 18:sinx=30:sin2x. Then replaced sin2x with 2sinx cosx, thus we can find cosx=5/6 and sinx=√11/6 . The square of the triangle is equal to 0.5*18*30*sin(180-3x)=270*sin3x anb then use formula for sin3x.

This problem can be solved much more easily with the Euclid catheter set. the squares over the cathetus result in the square over the hypothenuse. If you divide the square above the hypotenuse with one of the smaller squares of the cathete, you get the second length that you need to calculate the height using the Pythagorean theorem. the rest is then simple, base x height divided by 2.

@okeuwechue9238

Жыл бұрын

I suspect that PreMath is always solving these problems the *long* way because of the youtube algo (- longer videos and more "user time" spent watching the vids may increase monetization and advertising dollars...)

I did it trigonometrically: Drop BE perpendicular from B to AC at E; then BE = your "h" = 30 sin x = 18 sin 2x; so 30 sin x = 18 sin (2x). By the double-angle formula, sin 2x = 2 sin x cos x; so 30 sin x = 18 (2 sin x cos x) = 36 sin x cos x; thus 30 sin x = 36 sin x cos x. Divide both sides by 36 sin x and we have cos x = (30 sin x) / (36 sin x) = 30/36 = 5/6. Now the right-hand line segment EC = 30 cos x; so 30 cos x = 30 (5/6) = 150/6 = 25. By Pythagoras, h^2 = 30^2 - 25^2 = 900 - 625 = 275; so h = sqrt(275) = sqrt[(25)(11)] = 5 sqrt(11). Now for AE which you call "a": invoking Pythagoras again, a^2 = 18^2 - h^2 = 324 - 275 = 49; so a = sqrt(49) = 7. So the base of the triangle is 25 + 7 = 32, and the height (altitude) is 5 sqrt(11), just as your solution indicates; and finally, area = (1/2)(32)(5 sqrt(11) = (16)(5)(sqrt(11) = 80 sqrt(11). Since this is the exact solution, I'll skip the decimal approximation. Thank you, ladies and gentlemen; I'll be here all week.

18/sinx = 30/sin2x = c/sinc . This relation can be used to determine the angle x and thus all the angles. ( sin2x = 2sinx cosx leads directly to cos x as sinx≠0 cancels out from both side) And then after knowing the 3rd angle we can use the last formula to determine the third side. Finally the area of the triangle can be calculated easily using the formula : A² = s(s-a) (s-b)(s-c) . Where a, b, c are the 3 sides, 2s the perimeter. Simple

@seroujghazarian6343

Жыл бұрын

sin(c)=sin(3x)

Dá para resolver por Trigonometria! Temos que sen2x = h/18 => senx = h/30 ----> (1) Onde a identidade: sen2x = sen(x + x) = senx.cosx + senx.cosx = 2senx.cosx ----> (2) Substituindo (1) em (2): 2(h/30).cosx = (h/15)cosx ----> (3) Sabemos que sen2x = h/18 onde substituindo na identidade sen2x = 2senx.cosx ,obtemos o valor de cosx = ? Logo: h/18 = 2(h/30).cosx => cosx = 5/6 ----> (4) Da relação fundamental da Trigonometria, podemos determinar o valor da altura "h". (sen²x + cos²x = 1) => sen²x + (5/6)² = 1 => sen²x = 11/36 ---> Comparando com (1),temos o "h" determinado! Da seguinte forma: h/30 = Raiz quadrada de 11 dividida por 6, o que nos dá: h = 5 . que multiplica a Raiz quadrada de 11. ////////////////////////////////// Temos que, a área do triângulo ABC é igual a: (18 + 2u)h/2 = ? ; Onde u = Raiz quadrada de (18 + h).(18 - h) --> (5) Desenvolvendo, temos: Área do triangulo ABC = (18 + 2u)5.(Raiz quadrada de 11)/2 O que nos dá: 5. (Raiz quadrada de 11).(9 + u) ---> (6) Onde substituindo u por (5) em (6), teremos a área do triângulo = 80 que multiplica a raiz quadrada de 11.

Namaste sirji.

Actually the area of this 18 by 30 right triangle is 18 times 30 divided by 2.

Nice sum

Trigonometrik çozümlede kolayca bulunur ama sizin yaptiğiniz sekilde sentetik cozüm her zaman en güzeli cünkü daha yaraticı

thank you

Nice problem, could you tell me what app do you use to make your videos? I mean, what "board" do you use?

@PreMath

2 жыл бұрын

Thanks BDH for the feedback. You are awesome 👍 Take care dear and stay blessed😃 We use Camtasia Techsmith utility. Thanks for asking

I believe law of sines can be used to avoid so much working, and making so many triangles and all this bunch of separate solutions

Thnx for the vid. However, I'm still not sure why you chose to solve this the LOOOONG way by first constructing the isosceles triangle. A far simpler technique is to simply split the triangle into two right-angled triangles then apply the double angle formula and Pythag to find the side lengths and hence the total area.

Maybe this is not the simpliest solution, but one ... Let's assume that the (B) altitude line of the triangle is perpendicular to the A to C line and its values can be calculated with the sine function knowing the side lengths (18 & 30) and angles ( x & 2x). So the high (H) of the triangle is the distance of AC line and B point. And from the left side sin 2x= H/18 and from the right side sinx=H/30. H= 18*sin2x and H= 30*sinx >> 18*sin2x=30*sinx and we know that sin 2x = 2sinx*cosx Well 36*sinx*cosx=30*sinx : 6 where sinx doesn't equal 0! 6*cosx=5 cosx=5/6 =0.8333 and x=33.5573' So the High of triangle is 30*sin33.5573=16.583 or 18*sin67.1146=16.583 And the distance of A to C point is 18*cos67.1146 + 30*cos33.5573 = 7.000+25 = 32 So the area of the triangle is 32*16.583/2 = 265.328

one correction at 3:45 proving triangles congruent by AAA axiom is not true and only applies in some special cases because it is for similar triangle not congruent , and congreunt triangle are similar triangle but similar triangle are not necessarily congruent. there should be at least one side equal for a triangle to be congruent

I noted that h = 30 * sin(x). Also, h = 18 * sin (2x). But sin(2x)=2 * sin(x) * cos (x) therefore h = 36 * sin(x) * cos(x) = 30 * sin(x). Cancel out the sin(x) and remove common factors and you get cos(x) = 5/6. Therefore the length of EC is 5 x 30 / 6 = 25. By Pythagoras's theorem, l = √(900-625) = √275. Now apply Pythagoras's theorem to the triangle AEB and you get AE^2 = 324 - 275 = 49 therefore the length of AE is 7. Thus the length of the base AC is 7+25 = 32. Multiply the base by half the height and we get the area = 32 * √275 / 2 which simplifies slightly to 80 * √11. Of course this requires the use of a trigonometric identity, but I think it's simpler.

Another way to solve this problem is to find h as follows. h=30sin(x)=18sin(2x)=36sin(x)cos(x). Because sin(x) cannot be 0, we have cos(x)=5/6, therefore sin(x)=sqrt(11)/6, and then h=30sin(x)=5sqrt(11). now use Pythagoras to find the two segments from A to the foot of h and from there to C.

At 3:47 The Congruency might be by Side-Angle-Angle Axiom and Not Angle-Angle-Angle because AAA Theorem is for Similar Triangles.... And thus if all the angles of the tri. AEB and BED respectively congruent then it can be this way that the scale of the triangles are different. So, I guess it's SAA as both have BA=BD, BAE=BDE and BEA=BED... :)

@PreMath

2 жыл бұрын

Thanks dear for the feedback. You are awesome 👍 Keep smiling😊

3:43 the angle-angle-angle theorem indicates similar triangles only, not congruent triangles.

@MarieAnne.

2 жыл бұрын

You're right. He seems to have skipped a step. Angle-angle-angle indicates similar triangles. But since we have similar triangles that share a corresponding side, they are congruent.

We could have Lao used the law of sines but this solution is much more elegant.

Very difficult problem, but you solved it nicely. Thank you sir.

Io l’ho fatto diversamente, ho preso subito in considerazione l’altezza del triangolo, che divide tale triangolo in 2 triangoli rettangoli, così posso mettere l’altezza in relazione con il seno dei suoi 2 angoli opposti ad essa (2x e x) ottenendo h=30senx e h=18sen2x seguendo i teoremi fondamentali della trigonometria. Poi con la proprietà transitiva si ottiene 30senx=18sen2x e poi con la regola di duplicazione del seno si ottiene 30senx=18(2senxcosx). Così senx scompare (si semplifica) comunque però posso ottenere il coseno e rimane cosx=30/36 che semplificato ulteriormente fa 5/6. Dal coseno ottengo seno considerando il teorema di Pitagora applicato in trigonometria si ottiene che la somma del quadrato del seno e del quadrato del coseno fa 1, con la formula inversa senx sarà uguale a radice di 11 il tutto diviso 6. Con questi dati ottengo l’altezza con la formula precedente (h=30senx) ma anche la base, che non è altro che la somma dei cateti adiacenti ai due angoli considerati che sono rispettivamente c1=18cos2x (il coseno di 2x per la regola di duplicazione e la differenza tra il quadrato del coseno e poi quello del seno) e c2=30cosx. La somma dei due cateti viene la base e poi il mezzo prodotto tra la base e l’altezza viene 80 per radice di 11.

From ratio theory of the triangle ABC, we can write, (Sinx/18) = (sin2x/30) Or, 2sinx.cosx = 30.sinx/18 [as, sin2A=2sinA.cosA] Or, cosx = 30/36 = 5/6 ......(i) Again, cosx = b/30 [b= length bet'n perpendicular foot point & triangle's point "C"] So, b/30 = 5/6 Or, b = 25 And if height be "h" then, h^2 = 30^2 -b^2 = (30+25)(30-25) [b=25] = 55×5 = 11×5×5 So, h = 5.rt11 So, a = rt {18^2-(5.rt11)^2} = rt(324-275) = rt.49 Or, a = 7 [here, a = AC-b] So, AC = 7+25 = 32 Therefore, Area = 0.5×h× AC = 0.5×5.rt11×32 = 80.rt11 = 265.33 sqr unit [Ans.]