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)
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.
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.
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.
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!
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
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))
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
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)
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)
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!
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
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😃
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
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.
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
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.
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
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.
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...
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.
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.
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... :)
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.
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 😀
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.
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!
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
@@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.
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
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.
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...)
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).
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
Also, as a question to Dr. Pre Math why in general are non-trigonometric solutions preferred? I think there is something fairly powerful about being able to generalize a solution trigonometrically, so that one might have ratios of 2.5× or 3.77871× between the angles, instead of this problem's very special 2×
Dear Robert, 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😃
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.
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.
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.
@@ЛидийКлещельский-ь3х 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.
@@ОлегПолканов-д1н @Олег Полканов Просто должно соблюдаться условие, что cos x = b/(2a), где а - сторона лежащая напротив угла х , b - напротив угла 2х.
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
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)
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.
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
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.
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.
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
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
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.
At 3:43 you have proved that triangle AEB is congurent to triangle BED by angle - angle - angle rule but i am totally confused that their is no rule of angle angle angle for congurency of triangles. Both teiangles are congurent not by AAA but AAS ok.😊
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.
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.
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.
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).
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
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.
Another solution. Let AE is bisector of angle A. Hence, angle AEB = 2x. Hence, ∆ BAC ~ ∆ BEA Therefore, BA/BE = BC/BA i. e. 18/BE = 30/18. Hence, BE = 54/5. By Pythagoras theorem to ∆ ABE, BE = 96/5 = CE. Also, BA/BC = AE/ AC i. e. 18/30 = (96/5)/CA Hence, CA = 32. Now apply Heron's formula or compute height from B on AC using Pythagoras theorem twice, get area of ∆ABC = 80√11.
we can use laws of sines 30/sin2x =18/sinx then we know that sin2x=2cosxsinx and we can have cosx and use laws of cosines to find the side length is 32 .Then use 1/2×30×32×sinx (we can use cosx to have sinx) the answer is the area
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.
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.]
how do you make an isosceles triangle? I believe there is a gap in your solution. Sticking in pythogas and the formula of tan(2x) with tanx you get the solution.
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.
At 3:39, We can use the Sine Rule, 18/sin(x)=30/sin(2x) sin(2x)=(5/3) sin(x)........................................(1) sin(2x)=2sin(x)cos(x) ...................................................standard relationship, (5/3)sin(x)=2sin(x)cos(x) cos(x)=5/6......................................................(2) cos^2 (x)+sin^2(x)=1.......................................................standard relationship, sin^2(x)=1-(5/6)^2=36/36-25/36=11/36 sin(x)= sqrt(11)/6...........................................(3) sin(2x)=(5/3)sin(x)=5sin(x)/18=5sqrt(11)/18......................(4) (sin(2x))^2+(cos(2x))^2=1, (25*11)/18^2+(cos(2x))^2=18^2/18^2 (cos(2x))^2=(18^2-25*11)/18^2=49/18^2 cos(2x)=7/18..................................................(5) Drop a perpendicular BE to intersect AC at E, Let BE=h sin(x)=sqrt(11)/6 from (3), h=30*sin(x)=5*sqrt(11)..................................(6) cos(x)=5/6=EC/30 EC=30*(5/6)=25.............................................(7) cos(2x)=7/18 from (5) AE=AB*cos(2x)=18*7/18=7.......................... (8) AC=AE+EC=7+25=32.....................................(9) Area 0f triangle ABC=(h*AC)/2=5*sqrt(11)*32)/2=80*sqrt(11) square units. and that is our answer. Let's try Heron's formula; a=18,b=30,c=32, s=40,s-a=22,s-b=10,s-c=8. Area=sqrt(s(s-a)(s-b)(s-c)) =sqrt(40*22*10*8) =sqrt(400*16*11) =20*4*sqrt(11) =80 sqrt(11) square units. This is a good place to stop.
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
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.
Hello Premath, There is a mathematical error in 3:50 to 3:54 . As you should know, U used AAA theory for congruent triangle which does not work in maths. U should use sine next time, simpler. Dani and Elli 😕😕
Let AL is bisector, then BAL ~ BCA, BL = 18^2/30=10,8, AL=LC=19,2, AC = AL * 30/18 = 32, Heron's formula: S = sqrt(40(40-18)(40-30)(40-32))=sqrt(2^3*5*2*11*2*5*2^3)=sqrt(2^8*5^2*11)=2^4*5*sqrt(11)=80*sqrt(11)
For my think , grade 7 & 8 ( Alberta , CAN ) not yet learn about sine , cos ....But follow all your steps ( apply grade 6, 7, 8 Math ) they can understand to solve this problem easier . Thank you .
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@@PreMath sir so long process ? cannot use short method ?
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)
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.
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.
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.
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😃
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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
I have also said the same but you have solved completely. Excellent
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))
3:41 There is no AAA Congruence theorem. The 2 triangles are congruent here, but you would need to use ASA or AAS congruence.
Profe - Sometimes it amazes me how many interesting problems you can pull out of a few triangles, squares and circles!
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
hey we said it "u" teorem in turkey what you said it ? ( u(u-x)(u-y)(u-z) )^1/2
@@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'
It is called heron's formula here in india or simply hero's formula in local language
@@sie_khoentjoeng4886 😶 all people say it heron therom just l think it name is u teorem :// 😳
By Heron's formulae
تمرين جميل. وشرح واضح مرتب . شكرا جزيلا لكم استاذنا الفاضل والله يحفظكم ويرعاكم ويحميكم وينصركم . تحياتنا لكم من غزة فلسطين .
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)
Beautiful!!
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)
Thanks for sharing Yao
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
I did same solution!
Awsm solution
Thank you sir❤️
me too!
th-cam.com/video/Ti8G7SSFlJE/w-d-xo.html
sine rule makes the process a lot simpler..
1/2 ab sinC
In which triangle
nvm
@@anonim6160 use sine rule and use the formula of sin3x to find sinx (i.e included angle of given sides..)
Thats right
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!
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
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😃
@@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 😁.
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
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.
Very great video sir. Making students gain confidence in math. Thank you sir
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)
Exactly the way I've just done it
I used this.
Yes but so long !!!
How do you use that to find the value of x using algebra
@@DanielNeedham2500 algebra method is not easy.because
X= arc cos 5/6.
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
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.
How is the sine rule less rigorous?
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.
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
rak tama khoya la3ziz
momkin l9awha b sinx w cosx (we use sinx=h/30 , sin2x=h/18 and sin²x+cos²x=1.....)
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.
Brilliant, I got the correct strategy, just needed guidance clipping it all together, great detailed explanation as always. Thanks again 👍🏻
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...
3:43 the angle-angle-angle theorem indicates similar triangles only, not congruent triangles.
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.
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.
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.
Angle, angle, angle congruency for triangles? It should have been HL. AAA is used for triable similarity.
Hi dear, the problem can be solved by using bisection of BAC angle and so on
Thanks for this example 👍
What math sofware do you use to draw the figer?
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... :)
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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.
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ಜೈ ಶ್ರೀ ರಾಮ್ 🙏 THAN Q "PRE Math "for presenting suuuuuper video of PUZZLE solution . very interesting.MAY GOD BLESS U ❤️!!!!
this can be done with those theorems as well as using trig. but which method is quicker, which method is widely accepted?
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.
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.
It was a very challenging problem, but you provided its outstanding solution. Kudos to you dear.
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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
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
18/sinx = 30/sin2x, we get cosx=5/6.
I also did the same method.
@@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.
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
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.
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...)
Good . Explanation.. you are crossed 100 k.. congrats dear brother🎉🎉
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).
Indeed, I've done the same approach.
I suspect PreMath prefers always solving problems the LONG way because it may help to increase youtube monetization and ad dollars...
Nice explanation Thank you.proof by using only common rules and theorems 👏👏
What is the guarantee that BD of length 18 will meet AC at D
You can also calculate it from: P=1/2a*b*sin(
Where did you find angle-angle-angle theorem about congruency? Say Hypotenuse-leg or angle-angle-side.
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
Very nice explaination sir. Thanks.
Also, as a question to Dr. Pre Math why in general are non-trigonometric solutions preferred? I think there is something fairly powerful about being able to generalize a solution trigonometrically, so that one might have ratios of 2.5× or 3.77871× between the angles, instead of this problem's very special 2×
Dear Robert, 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😃
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.
Nice problem, could you tell me what app do you use to make your videos? I mean, what "board" do you use?
Thanks BDH for the feedback. You are awesome 👍 Take care dear and stay blessed😃
We use Camtasia Techsmith utility. Thanks for asking
Nice question and very nice solution. Thank you, professor. Have a nice day.
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.
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.
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.
@@ЛидийКлещельский-ь3х 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.
@@ОлегПолканов-д1н @Олег Полканов
Просто должно соблюдаться условие, что
cos x = b/(2a),
где а - сторона лежащая напротив угла х ,
b - напротив угла 2х.
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
Very good solution
After first line to make them isotriangle rest one straight forward process
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal.... '' ''
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)
Nice solution：D
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.
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
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.
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.
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
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
SIR ,
you have did a mistake at 3:41 as there is so called theorem like AAA for congreunce , they will be congruent by ASA or RHS
Nice one sir keep it up
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.
Really nice, this time the puzzle was more difficult what make it more "edcatif" i dont speak english very nice😅
At 3:43 you have proved that triangle AEB is congurent to triangle BED by angle - angle - angle rule but i am totally confused that their is no rule of angle angle angle for congurency of triangles. Both teiangles are congurent not by AAA but AAS ok.😊
Angle angle angle can not prove congruency?
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 ✓ ≡=-⋅
Cos-1 x = 30/36 x = 33º 33` h=16.58, a =7 b = 25 A = 265.28
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
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
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.
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.
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.
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).
Onde está a tradução ?
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
sin(c)=sin(3x)
good one...same time Sin formula can be used here and easier
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
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.
Loved this one!
Thank you!!
It is a very beautiful solution
Another solution.
Let AE is bisector of angle A. Hence, angle AEB = 2x.
Hence, ∆ BAC ~ ∆ BEA
Therefore, BA/BE = BC/BA
i. e. 18/BE = 30/18.
Hence, BE = 54/5.
By Pythagoras theorem to ∆ ABE, BE = 96/5 = CE.
Also, BA/BC = AE/ AC
i. e. 18/30 = (96/5)/CA
Hence, CA = 32.
Now apply Heron's formula or compute height from B on AC using Pythagoras theorem twice, get area of ∆ABC = 80√11.
Geometry is very interesting as well as some hard ,nice video sir
You are such a super Teacher , i bet your students looked forward to your lectures !
Thanks Man !
we can use laws of sines
30/sin2x =18/sinx
then we know that
sin2x=2cosxsinx and we can have cosx and use laws of cosines to find the side length is 32 .Then use
1/2×30×32×sinx (we can use cosx to have sinx) the answer is the area
Trigonometrik çozümlede kolayca bulunur ama sizin yaptiğiniz sekilde sentetik cozüm her zaman en güzeli cünkü daha yaraticı
Very difficult problem, but you solved it nicely.
Thank you sir.
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.
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.]
It can also be done by trigonometry
how do you make an isosceles triangle? I believe there is a gap in your solution. Sticking in pythogas and the formula of tan(2x) with tanx you get the solution.
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.
Draw a bisectrix if angle 2x. You obtain two similar triangles. You can then find the third side equal to 32. Then Heron formula.
At 3:39,
We can use the Sine Rule,
18/sin(x)=30/sin(2x)
sin(2x)=(5/3) sin(x)........................................(1)
sin(2x)=2sin(x)cos(x) ...................................................standard relationship,
(5/3)sin(x)=2sin(x)cos(x)
cos(x)=5/6......................................................(2)
cos^2 (x)+sin^2(x)=1.......................................................standard relationship,
sin^2(x)=1-(5/6)^2=36/36-25/36=11/36
sin(x)= sqrt(11)/6...........................................(3)
sin(2x)=(5/3)sin(x)=5sin(x)/18=5sqrt(11)/18......................(4)
(sin(2x))^2+(cos(2x))^2=1,
(25*11)/18^2+(cos(2x))^2=18^2/18^2
(cos(2x))^2=(18^2-25*11)/18^2=49/18^2
cos(2x)=7/18..................................................(5)
Drop a perpendicular BE to intersect AC at E,
Let BE=h
sin(x)=sqrt(11)/6 from (3),
h=30*sin(x)=5*sqrt(11)..................................(6)
cos(x)=5/6=EC/30
EC=30*(5/6)=25.............................................(7)
cos(2x)=7/18 from (5)
AE=AB*cos(2x)=18*7/18=7.......................... (8)
AC=AE+EC=7+25=32.....................................(9)
Area 0f triangle ABC=(h*AC)/2=5*sqrt(11)*32)/2=80*sqrt(11) square units.
and that is our answer.
Let's try Heron's formula;
a=18,b=30,c=32, s=40,s-a=22,s-b=10,s-c=8.
Area=sqrt(s(s-a)(s-b)(s-c))
=sqrt(40*22*10*8)
=sqrt(400*16*11)
=20*4*sqrt(11)
=80 sqrt(11) square units.
This is a good place to stop.
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
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.
Hello Premath,
There is a mathematical error in 3:50 to 3:54 . As you should know, U used AAA theory for congruent triangle which does not work in maths. U should use sine next time, simpler. Dani and Elli 😕😕
You can also use trig:
18/sin(x)=30/sin(2x)
x=33.6
area = 1/2ab*sin(C)
Let AL is bisector, then BAL ~ BCA, BL = 18^2/30=10,8, AL=LC=19,2, AC = AL * 30/18 = 32, Heron's formula: S = sqrt(40(40-18)(40-30)(40-32))=sqrt(2^3*5*2*11*2*5*2^3)=sqrt(2^8*5^2*11)=2^4*5*sqrt(11)=80*sqrt(11)
Interesting question 🔥
For my think , grade 7 & 8 ( Alberta , CAN ) not yet learn about sine , cos ....But follow all your steps ( apply grade 6, 7, 8 Math ) they can understand to solve this problem easier . Thank you .