Learn how to find the area of triangle ABC by using the exterior angle theorem and the pythagorean theorem. Fast and easy explanation by PreMath.com
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@MathZoneKH2 жыл бұрын
First comment 😊❤️
@PreMath2 жыл бұрын
Yes you are! Thanks for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@bahsarabi42342 жыл бұрын
❤️
@MathZoneKH2 жыл бұрын
❤️😊
@durgalimbu97392 жыл бұрын
@@PreMath sir so long process ? cannot use short method ?
@hahatseck2 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)
@swinkscalibur85062 жыл бұрын
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.
@montynorth30092 жыл бұрын
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.
@robertlynch75202 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.
@PreMath2 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!
@armacham2 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
@pankajkumarpandey66582 жыл бұрын
I have also said the same but you have solved completely. Excellent
@calspace2 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))
@charlesbromberick42472 жыл бұрын
Profe - Sometimes it amazes me how many interesting problems you can pull out of a few triangles, squares and circles!
@chessdev53202 жыл бұрын
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!
@chessdev53202 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
@PreMath2 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😃
@chessdev53202 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 😁.
@michaelcrosby77152 жыл бұрын
3:41 There is no AAA Congruence theorem. The 2 triangles are congruent here, but you would need to use ASA or AAS congruence.
@mathsandsciencechannel2 жыл бұрын
Very great video sir. Making students gain confidence in math. Thank you sir
@Shinobi_OF_Tsushima2 жыл бұрын
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
@sumangupta17372 жыл бұрын
sine rule makes the process a lot simpler..
@nitinarora57192 жыл бұрын
1/2 ab sinC
@anonim61602 жыл бұрын
In which triangle
@anonim61602 жыл бұрын
nvm
@sumangupta17372 жыл бұрын
@@anonim6160 use sine rule and use the formula of sin3x to find sinx (i.e included angle of given sides..)
@sadtear7962 жыл бұрын
Thats right
@arpansit31552 жыл бұрын
Nice one sir keep it up
@UCE5YhzPOD2 жыл бұрын
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)
@timeonly14012 жыл бұрын
Beautiful!!
@johnfoggjr24582 жыл бұрын
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.
@kennethweigand46002 жыл бұрын
Loved this one! Thank you!!
@user-ly5bc4xd2s2 жыл бұрын
تمرين جميل. وشرح واضح مرتب . شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم وينصركم . تحياتنا لكم من غزة فلسطين .
@sie_khoentjoeng48862 жыл бұрын
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
@logos21142 жыл бұрын
hey we said it "u" teorem in turkey what you said it ? ( u(u-x)(u-y)(u-z) )^1/2
@sie_khoentjoeng48862 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'
@gurmukhsingh23582 жыл бұрын
It is called heron's formula here in india or simply hero's formula in local language
@logos21142 жыл бұрын
@@sie_khoentjoeng4886 😶 all people say it heron therom just l think it name is u teorem :// 😳
@versexx4132 жыл бұрын
By Heron's formulae
@acrade032 жыл бұрын
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)
@PreMath2 жыл бұрын
Thanks for sharing Yao You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@user-rt6si6pf5b2 жыл бұрын
I did same solution!
@khyatikumar38672 жыл бұрын
Awsm solution Thank you sir❤️
@colinratcliffe30742 жыл бұрын
me too!
@freemathacademy6632 Жыл бұрын
kzbin.info/www/bejne/ipqbeGqJiKufgKc
@kalaiselvi73742 жыл бұрын
Nice explanation Thank you.proof by using only common rules and theorems 👏👏
@VIVEKANANDA652 жыл бұрын
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
@devaroraa2 жыл бұрын
Interesting question 🔥
@prakashmadaksirashamrao59612 жыл бұрын
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.
@luigipirandello5919 Жыл бұрын
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 😀
@manishgandhi89342 жыл бұрын
Great video sir 👍I appreciate that.
@tablamurugesan2 жыл бұрын
Very nice explaination sir. Thanks.
@eleganttutorial58362 жыл бұрын
Nice Explanation sir
@ishnoorsingh18772 жыл бұрын
Perfect 🙌🙌
@sreedharaks31172 жыл бұрын
ಜೈ ಶ್ರೀ ರಾಮ್ 🙏 THAN Q "PRE Math "for presenting suuuuuper video of PUZZLE solution . very interesting.MAY GOD BLESS U ❤️!!!!
@dhrubajyotisarma62682 жыл бұрын
Geometry is very interesting as well as some hard ,nice video sir
@India-jq7pi2 жыл бұрын
Thank you sir
@mcorruptofficial65792 жыл бұрын
Hi dear, the problem can be solved by using bisection of BAC angle and so on Thanks for this example 👍
@theoyanto Жыл бұрын
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.
@johnbrennan33722 жыл бұрын
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.
@parameshbussu9244 Жыл бұрын
Nice Sir
@easy_s33512 жыл бұрын
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.
@ravikrpranavam2 жыл бұрын
Well explained
@adem-checkmate Жыл бұрын
Just awesome
@rabindranathrautaray76152 жыл бұрын
It is a very beautiful solution
@ddmm78932 жыл бұрын
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
@hnahler2 жыл бұрын
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).
@hakkisuperheld2 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...
@hung98372 жыл бұрын
Thank You Teacher
@MasterMathematicsSM2 жыл бұрын
Good . Explanation.. you are crossed 100 k.. congrats dear brother🎉🎉
@vcvartak71112 жыл бұрын
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)
@DanielNeedham25002 жыл бұрын
Exactly the way I've just done it
@dhrubajyotidaityari92402 жыл бұрын
I used this.
@ddmm78932 жыл бұрын
Yes but so long !!!
@DanielNeedham25002 жыл бұрын
How do you use that to find the value of x using algebra
@dhrubajyotidaityari92402 жыл бұрын
@@DanielNeedham2500 algebra method is not easy.because X= arc cos 5/6.
@jesusantoniocarhuashuerta46622 жыл бұрын
Good solution
@3outas_math823 Жыл бұрын
Very good
@sidimohamedbenelmalih71332 жыл бұрын
Really nice, this time the puzzle was more difficult what make it more "edcatif" i dont speak english very nice😅
@luismiguelguevarahinojosa66622 жыл бұрын
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.
@qudretastanov29802 жыл бұрын
Thanks very much🙏🙏🙏🙏
@user-wm1vk1cv2t2 жыл бұрын
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-ly1ww2 жыл бұрын
Nice solution:D
@mauriciobrito11602 жыл бұрын
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.
@JSSTyger2 жыл бұрын
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.
@7mtm7862 жыл бұрын
Very good solution After first line to make them isotriangle rest one straight forward process
@janiewiemkto23682 жыл бұрын
You can also calculate it from: P=1/2a*b*sin(
@crazyrapidshorts81122 жыл бұрын
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
@mshanmukhavalli45672 жыл бұрын
Nice sum
@3nelad2 жыл бұрын
18/sinx = 30/sin2x, we get cosx=5/6.
@ankaiahgummadidala13712 жыл бұрын
I also did the same method.
@krabkrabkrab2 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.
@lazzatbastar34972 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
@user-tt4ep6fr4d2 жыл бұрын
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
@cosmosz81252 жыл бұрын
rak tama khoya la3ziz
@cosmosz81252 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.
@mrsir8968 Жыл бұрын
thank you
@DxRzYT2 жыл бұрын
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...
@GillAgainsIsland12 Жыл бұрын
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.
@fouadhammout6512 жыл бұрын
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal.... '' ''
@luigipirandello59192 жыл бұрын
Nice question and very nice solution. Thank you, professor. Have a nice day.
@c8h1822 жыл бұрын
More geometry questions please.
@ramaprasadghosh7172 жыл бұрын
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
@bocaj.4552 жыл бұрын
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.
@holyshit9222 жыл бұрын
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
@phuhuynhtoanphothong468 Жыл бұрын
Verrygood
@fedorhotabich Жыл бұрын
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
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...)
@zsus1 Жыл бұрын
good one...same time Sin formula can be used here and easier
@user-pd7js7cy9m2 жыл бұрын
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-rk5eh2sh9v2 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-pd7js7cy9m2 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-rk5eh2sh9v2 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.
@_basu_63202 жыл бұрын
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
@ankaiahgummadidala13712 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?
@maneeshaliyanapatabendy14812 жыл бұрын
this can be done with those theorems as well as using trig. but which method is quicker, which method is widely accepted?
@AbouTaim-Lille2 жыл бұрын
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)
@hennobrandsma47552 жыл бұрын
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.
@johnfoggjr24582 жыл бұрын
A very complicated way of arriving at answer.This must the new math today.
@johnfoggjr24582 жыл бұрын
I guessed 270 just by looking at it.
@NinjaJacky2 жыл бұрын
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.
@billbutler8560 Жыл бұрын
nice
@nazimusik2 жыл бұрын
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
@robertlynch75202 жыл бұрын
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 ✓ ≡=-⋅
@rafaelmayoral55712 жыл бұрын
Cos-1 x = 30/36 x = 33º 33` h=16.58, a =7 b = 25 A = 265.28
@taekwondomeerut7562 жыл бұрын
Good
@govindashit65242 жыл бұрын
Very difficult problem, but you solved it nicely. Thank you sir.
@TheEulerID2 жыл бұрын
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.
@arunprasad1022 Жыл бұрын
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.
@ufukkoyuncu34082 жыл бұрын
Trigonometrik çozümlede kolayca bulunur ama sizin yaptiğiniz sekilde sentetik cozüm her zaman en güzeli cünkü daha yaraticı
@raymondruiz58392 жыл бұрын
Angle, angle, angle congruency for triangles? It should have been HL. AAA is used for triable similarity.
@mkryan23972 жыл бұрын
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.]
@lusalalusala29662 жыл бұрын
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.
@johngreen3543 Жыл бұрын
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).
@julioricardoaguilarsilva15762 жыл бұрын
Por semejanza sale un poco más rápido. El ejercicio desarrollado usa mucha 95% de álgebra pero es un buen aporte.
@Hertog_von_Berkshire2 жыл бұрын
I thought my method was laborious but I hadn't watched the video yet!!
@shrovitz9692 жыл бұрын
We could have Lao used the law of sines but this solution is much more elegant.
@cuongtu60882 жыл бұрын
good
@sparky46952 жыл бұрын
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... :)
@PreMath2 жыл бұрын
Thanks dear for the feedback. You are awesome 👍 Keep smiling😊
@konstantinjoukovski7062 Жыл бұрын
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.
@williamwingo47402 жыл бұрын
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.
@DhirajSingh-mn6zu2 жыл бұрын
What is the guarantee that BD of length 18 will meet AC at D
@andrewng6222 жыл бұрын
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.
@giovannimonaco16132 жыл бұрын
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.
@mohammadshahabeezurrahman66462 жыл бұрын
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