I solved without trigonometry tracing the height AH and seeing that AHB is a right triangle of 30,60,90 degree whose hypotenuse is 3x-5. It leads to the same quadratic equation without the cosine law
Just substitute 3X - 5 as “a”, then the 3 sides would be “a”, “a+1” and “a+2”. Then the calculations are simpler. At the end we don’t even need to change “a” back to X.
*Let's name t = 3.x -4, The side lenthes of the triangle are now BA = t - 1 , BC = t and AC = t + 1. The law of cosines in the triangle ABC gives: (t + 1)^2 = t^2 + (t - 1)^2 -2.t.(t - 1).cos(120°), or: t^2 +2.t + 1 = t^2 + t^2 -2.t + 1 +t^2 - t (as cos(120°) = -1/2). That gives: 2.t^2 - 5.t = 0 and then t = 5/2 (t cannot be equal to 0 as the lengthes are positive). So t = 3.x -4 = 5/2 (and x = 13/6). *The perimeter of the triangle is 3.t = 15/2. *The area of the triangle (1/2).BA.BC.sin(120°) = (1/2).t.(t -1).(sqrt(3)/2) = (1/2).(5/2).(3/2).(sqrt(3)/2) = (15/16).sqrt(3).
Sir 1) we may write the side 3x -4 =a Then 3x - 5=a -1 3x -3= a +1 2)Perimeter =3a 3)1/2*a*(a-1)sin120 4) Cos 120 =[(a-1)^2+a^2 -(a+1)^2]/[2a(a-1)] =(a^2-2a +1+a^2 -a^2-2a-1)/2a(a-1) =(a^2-4a)/2a(a-1) =(a -4)/(2a-2) > - 1/2 =(a-4)/(2a -2) > a=5/2 units 5)Perimeter =3a=3*5/2=15/2 units 6)Hence the legs of ang 120 degrees are a=5/2 and a-1=5/2 -1=3/2 Area =1/2*5/2* 3/2*sin 120 =1/2*5/2*3/2*√3/2 =1*5*3*√3/2*2*2*2 =15√3/16 sq units [ *** please note that this solution does not have any - ve value of length and no question of rejecting the - -ve arose. ]
Thanks chalenging. Draw a right line from A and mark it as F. We have a right triangle AFB in which B angle is 60 and A angle is 30. The side is FB is half of the cord AB. = 3x-5/2 focus on right triangle AFC and by phytagorus theorum AF^2+{( 3x-5/2/2)+(3x-4)}^2=(3x-3) ^2)===> X=2.16
Here there is another sol Drop a perpendicular AD on extended CB In 🔺 ADB is a special 🔺 of 30-60-90 So if BD =1 , AB =2 Hence AB =3x -5 = 2 > x =7/3 BC =3x -4 =3 AC = 3x -3 =4 Hence the sides of triangle ABC are 2,3,4 Perimeter = 2+3+5=9 units Area =1/2*2*3*sin120 =3sin 120 =3*√3/2=3√3/2sq units This have so many solutions as we take different values of BD
Thanks professor for sharing this interesting video! Merry Christmas to all who celebrate it. I hope to do not wrong to wish a happy winter solstice to everyone 😊 To my opinion this is not political, not dependent on what kind of belief one has, including atheists, even of no significance if earth is flat or a globe. It can be an occasion to send best wishes to anyone, like peace, health and happiness. For those who live in the north its the expectation of light and warmth will come back.
1/ Label AB= (3x-5)= a -> AC= (3x-5) +2=(a+2) and BC=(a+1) 2/ Drop the height AH to BC-> AH = a sqrt3/2 and BH= a/2 ( the triangle AHB is a 30/90/60 one) --> HC = a/2 + a+1= (3a+2)/2 By using the Pythagorean theorem sq (asqt3/2) + sq((3a+2)/2))= sq( a+2) --> 2sqa-a-3 = 0 -> a= 3/2 ( negative result rejected) --> 3x -5 = 3/2 x= 13/6 Perimeter= 7.5 units Area= 15sqr3/ 16😅😅😅
Let's do it again with another method: . .. ... .... ..... First of all we add point D such that ACD is a right triangle and B is located on CD. Then we have ∠ADB=90° and ∠ABD=180°−120°=60°. Therefore ABD is a 30°-60°-90° triangle and we can conclude: BD = AB/2 = (3x − 5)/2 AD = √3*BD = (√3/2)(3x − 5) Now we can apply the Pythagorean theorem to the right triangle ACD: AC = 3x − 3 = y + 2 AD = (√3/2)(3x − 5) = (√3/2)y CD = BC + BD = (3x − 4) + [(3x − 5)/2] = (y + 1) + y/2 = 3y/2 + 1 AC² = AD² + CD² (y + 2)² = [(√3/2)y]² + (3y/2 + 1)² y² + 4y + 4 = 3y²/4 + 9y²/4 + 3y + 1 0 = 2y² − y − 3 ... (see my first comment) 🙂
*Solução:* Seja y = 3x - 5. Daí, BC= y + 1 e AC = y + 2. Você pode usar a lei dos cossenos, porém , vamos construir uma perpendicular AD em relação ao lado BC. Assim, O ângulo ABD = 60° e, usando a definição de seno e cosseno no triângulo retângulo ∆ABD, temos: BD = y/2 e AD=(y√3)/2. Assim, DC = y/2 + y+1 = (3y+2)/2. Por Pitágoras no ∆ACD: (y+2)² = [(y√3)/2]² + [(3y+2)/2]² y²+4y+4 = 3y²/4+(9y²+12y+4)/4 y²+4y+4 = (12y²+12y+4)/4 y²+4y+4 = 3y²+3y+1 2y² - y -3 = 0, com y > 0. Resolvendo essas equação do 2° grau, obtemos y = 3/2, logo: AB=3/2, BC=5/2, AC=7/2 e AD=3√3/4. Temos: *_Perímetro=_* 3/2 + 5/2 + 7/2 = *15/2 U* *_Área=_* AD×BC/2 = = 5/2 × 3√3/8 = *15√3/16 U.Q*
First and foremost as a US citizen I follow the Constitution of the United States of America. If anyone is offended by that, I don't care! Be offended...🤣. @ 7:43 , never say never! Things get very strange very fast in this 4-dimensional space-time continuum.😊
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💯👍🎄 Merry Christmas 🎁 Be happy and all the best to you, Professor!
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
I solved without trigonometry tracing the height AH and seeing that AHB is a right triangle of 30,60,90 degree whose hypotenuse is 3x-5. It leads to the same quadratic equation without the cosine law
Just substitute 3X - 5 as “a”, then the 3 sides would be “a”, “a+1” and “a+2”. Then the calculations are simpler. At the end we don’t even need to change “a” back to X.
Wow Sirr ❤❤ Very Interesting Video❤❤❤ Thanks for sharing ❤❤
Thank you!
Perimeter = 7.5 , area =1.623 Sq.units(15× root3÷16)
Solution:
First, we've to calculate x, by using the Law of Cosines:
(3x - 3)² = (3x - 4)² + (3x - 5)² - 2 (3x - 4) (3x - 5) cos 120°
9x² - 18x + 9 = 9x² - 24x + 16 + 9x² - 30x + 25 + 9x² - 15x - 12x + 20
9x² - 18x + 9 = 27x² - 81x + 61
18x² - 63x + 52 = 0
x = (63 ± √225)/36
x = (63 ± 15)/36
x' = 78/36 = 39/18 = 13/6 (Accepted)
x" = 48/36 = 12/9 = 4/3 (Rejected)
Therefore x = 13/6
Now we have to calculate the sides length
3x - 5 = 3 (13/6) - 5 = 39/6 - 5 = (39 - 30)/6 = 9/6 = 3/2
3x - 4 = 3 (13/6) - 4 = 39/6 - 4 = (39 - 24)/6 = 15/6 = 5/2
3x - 3 = 3 (13/6) - 3 = 39/6 - 3 = (39 - 18)/6 = 21/6 = 7/2
Area = ½ × 3/2 × 5/2 × sin 120°
Area = ½ × 3/2 × 5/2 × √3/2
Area = (15√3)/16 Square Units ✅
Area ≈ 1.6237 Square Units ✅
Perimeter = 3/2 + 5/2 + 7/2
Perimeter = 15/2 Units ✅
Perimeter = 7.5 Units ✅
This has so many solutions. May see and comment on the two solutions offered by me
P=7,5
S=15√3/16≈1,624
I'm so glad to be one of your fallwers
Thanks dear ❤️🙏
You are the best! ❤️
Thanks for the feedback ❤️
*Let's name t = 3.x -4, The side lenthes of the triangle are now BA = t - 1 , BC = t and AC = t + 1.
The law of cosines in the triangle ABC gives: (t + 1)^2 = t^2 + (t - 1)^2 -2.t.(t - 1).cos(120°), or: t^2 +2.t + 1 = t^2 + t^2 -2.t + 1 +t^2 - t (as cos(120°) = -1/2).
That gives: 2.t^2 - 5.t = 0 and then t = 5/2 (t cannot be equal to 0 as the lengthes are positive). So t = 3.x -4 = 5/2 (and x = 13/6).
*The perimeter of the triangle is 3.t = 15/2.
*The area of the triangle (1/2).BA.BC.sin(120°) = (1/2).t.(t -1).(sqrt(3)/2) = (1/2).(5/2).(3/2).(sqrt(3)/2) = (15/16).sqrt(3).
Nice observation that simplifies the algebra!
Perimeter = (3x - 3) + (3x - 4) + (3x - 5) = 9x - 12
Area = (height)(base)/2 = (Sin60°)(3x - 5)(3x - 4)/2 = (√3)(3x - 5)(3x - 4)/4
[(Sin60°)(3x - 5)]^2 + [(Cos60°)(3x - 5)]^2 = (3x - 5)^2
To find x
(3x - 3)^2 = (3x - 4)^2 + (3x - 5)^2 - 2(3x - 4)(3x - 5)(Cos120°)
18x^2 - 63x + 52 = 0
x = [63 +/- √(63)(63) - 4(18)(52)] / 36 = [21 +/-√(21)(21) - 4(2)(52)] / 12 = 13/6 or 4/3 (the base 3x - 4 cannot be zero)
Perimeter = 9(13/6) - 12 = 39/2 - 12 = 7.5
Area = (√3)(13/2 - 5)(13/2 - 4)/4 = (√3)(3/2)(5/2)/4 = (15√3)/16
Bom dia Mestre
Irei usar a Lei dos cossenos e produtos notáveis p resolver essa questão
Grato
Sir
1) we may write the side
3x -4 =a
Then 3x - 5=a -1
3x -3= a +1
2)Perimeter =3a
3)1/2*a*(a-1)sin120
4)
Cos 120 =[(a-1)^2+a^2 -(a+1)^2]/[2a(a-1)]
=(a^2-2a +1+a^2 -a^2-2a-1)/2a(a-1)
=(a^2-4a)/2a(a-1)
=(a -4)/(2a-2)
> - 1/2 =(a-4)/(2a -2)
> a=5/2 units
5)Perimeter =3a=3*5/2=15/2 units
6)Hence the legs of ang 120 degrees
are a=5/2 and a-1=5/2 -1=3/2
Area =1/2*5/2* 3/2*sin 120
=1/2*5/2*3/2*√3/2
=1*5*3*√3/2*2*2*2
=15√3/16 sq units
[ ***
please note that this solution does not have any - ve value of length and no question of rejecting the - -ve arose. ]
Yes, substituting a, (a-1) & (a+1) for the side lengths simplifies the Cosine Rule calculations and eliminates the need to solve for x.
@georgebliss964 There is another solution in comments.
Here I said the problem may have a lot of solutions.
Please see it and comment
Thanks chalenging. Draw a right line from A and mark it as F. We have a right triangle AFB in which B angle is 60 and A angle is 30. The side is FB is half of the cord AB. = 3x-5/2 focus on right triangle AFC and by phytagorus theorum AF^2+{( 3x-5/2/2)+(3x-4)}^2=(3x-3)
^2)===> X=2.16
Here there is another sol
Drop a perpendicular AD on extended CB
In 🔺 ADB is a special 🔺 of 30-60-90
So if BD =1 , AB =2
Hence AB =3x -5 = 2
> x =7/3
BC =3x -4 =3
AC = 3x -3 =4
Hence the sides of triangle ABC are 2,3,4
Perimeter = 2+3+5=9 units
Area =1/2*2*3*sin120
=3sin 120
=3*√3/2=3√3/2sq units
This have so many solutions as we take different values of BD
Thanks professor for sharing this interesting video!
Merry Christmas to all who celebrate it.
I hope to do not wrong to wish a happy winter solstice to everyone 😊
To my opinion this is not political, not dependent on what kind of belief one has, including atheists, even of no significance if earth is flat or a globe.
It can be an occasion to send best wishes to anyone, like peace, health and happiness. For those who live in the north its the expectation of light and warmth will come back.
So kind of you dear🙏❤️
Merry Christmas! Wishing you a happy holiday season as well! 😊🙏
Stay blessed 😀
Merriest Christmas Team Premath
Very nice, many thanks, Sir! Merry Christmas!
φ = 30° → cos(4φ) = -cos(6φ - 4φ) = -cos(2φ) = -sin(φ) = -1/2
sin(4φ) = sin(6φ - 4φ) = sin(2φ) = cos(φ) = √3/2
∆ ABC → ABC = 4φ; AB = 3x - 5; BC = 3x - 4; AC = 3x - 3; area & perimeter ∆ ABC = ?
3x - 3 ∶= a → 3x - 4 = a - 1 → 3x - 5 = a - 2 →
a^2 = (a - 1)^2 + (a - 2)^2 - 2(a - 1)(a - 2)cos(4φ) = (a - 1)^2 + (a - 2)^2 + (a - 1)(a - 2) →
a^2 - 9a + 7 = 0 → a1,a2 = (1/4)(9 ± 5) → a1 = 1 → a - 2 < 0 ≠ solution →
a2 = 7/2 = 3x - 3 → x = 13/6 → 3x - 3 = 7/2 → 3x - 4 = 5/2 → 3x - 5 = 3/2 →
perimeter ∆ ABC = 15/2 → area ∆ ABC = (1/2)sin(4φ)(a - 1)(a - 2) = 15√3/16
1/ Label AB= (3x-5)= a
-> AC= (3x-5) +2=(a+2) and BC=(a+1)
2/ Drop the height AH to BC-> AH = a sqrt3/2 and BH= a/2 ( the triangle AHB is a 30/90/60 one)
--> HC = a/2 + a+1= (3a+2)/2
By using the Pythagorean theorem
sq (asqt3/2) + sq((3a+2)/2))= sq( a+2)
--> 2sqa-a-3 = 0
-> a= 3/2 ( negative result rejected)
--> 3x -5 = 3/2
x= 13/6
Perimeter= 7.5 units
Area= 15sqr3/ 16😅😅😅
4+5+6=15permeter 8.9452 area
I solved the problem using your method:
x= 13/6; Р=7,5; S= a×b×sin(60°)/2=
=(3/2×5/2×√3/2)/2=15√3/16
Thanks sir!❤
Let's do it again with another method:
.
..
...
....
.....
First of all we add point D such that ACD is a right triangle and B is located on CD. Then we have ∠ADB=90° and ∠ABD=180°−120°=60°. Therefore ABD is a 30°-60°-90° triangle and we can conclude:
BD = AB/2 = (3x − 5)/2
AD = √3*BD = (√3/2)(3x − 5)
Now we can apply the Pythagorean theorem to the right triangle ACD:
AC = 3x − 3 = y + 2
AD = (√3/2)(3x − 5) = (√3/2)y
CD = BC + BD = (3x − 4) + [(3x − 5)/2] = (y + 1) + y/2 = 3y/2 + 1
AC² = AD² + CD²
(y + 2)² = [(√3/2)y]² + (3y/2 + 1)²
y² + 4y + 4 = 3y²/4 + 9y²/4 + 3y + 1
0 = 2y² − y − 3
... (see my first comment) 🙂
Let's face this challenge:
.
..
...
....
.....
We should be able to find the value of x by applying the law of cosines. With y=3x−5 we obtain:
AC² = AB² + BC² − 2*AB*BC*cos(∠ABC)
(3x − 3)² = (3x − 5)² + (3x − 4)² − 2*(3x − 5)*(3x − 4)*cos(120°)
(y + 2)² = y² + (y + 1)² − 2*y*(y + 1)*(−1/2)
(y + 2)² = y² + (y + 1)² + y*(y + 1)
y² + 4y + 4 = y² + y² + 2y + 1 + y² + y
0 = 2y² − y − 3
0 = 2y² + 2y − 3y − 3
0 = 2y(y + 1) − 3(y + 1)
0 = (2y − 3)(y + 1)
First solution:
y + 1 = 0
⇒ y = −1
⇒ x = (y + 5)/3 = (−1 + 5)/3 = 4/3
⇒ AC = 3x − 3 = 3*(4/3) − 3 = 4 − 3 = 1
⇒ AB = 3x − 5 = 3*(4/3) − 5 = 4 − 5 = −1
⇒ BC = 3x − 4 = 3*(4/3) − 4 = 4 − 4 = 0
This is not a valid solution.
Second solution:
2y − 3 = 0
⇒ y = 3/2
⇒ x = (y + 5)/3 = (3/2 + 5)/3 = (3/2 + 10/2)/3 = (13/2)/3 = 13/6
⇒ AC = 3x − 3 = 3*(13/6) − 3 = 13/2 − 3 = 13/2 − 6/2 = 7/2
⇒ AB = 3x − 5 = 3*(13/6) − 5 = 13/2 − 5 = 13/2 − 10/2 = 3/2
⇒ BC = 3x − 4 = 3*(13/6) − 4 = 13/2 − 4 = 13/2 − 8/2 = 5/2
Now we are able to calculate the area and the perimeter of the triangle:
A(ABC) = (1/2)*AB*BC*sin(∠ABC) = (1/2)*(3/2)*(5/2)*sin(120°) = (15/8)*(√2/2) = (15/16)√2
P(ABC) = AB + BC + AC = 3/2 + 5/2 + 7/2 = 15/2
Best regards from Germany
Yhank you. I thought that x had to be intyeger.
Acertei !!!!!
Excellent!
Glad to hear that!
Thanks for the feedback ❤️
*Solução:*
Seja y = 3x - 5. Daí,
BC= y + 1 e AC = y + 2. Você pode usar a lei dos cossenos, porém , vamos construir uma perpendicular AD em relação ao lado BC. Assim,
O ângulo ABD = 60° e, usando a definição de seno e cosseno no triângulo retângulo ∆ABD, temos:
BD = y/2 e AD=(y√3)/2. Assim, DC = y/2 + y+1 = (3y+2)/2. Por Pitágoras no ∆ACD:
(y+2)² = [(y√3)/2]² + [(3y+2)/2]²
y²+4y+4 = 3y²/4+(9y²+12y+4)/4
y²+4y+4 = (12y²+12y+4)/4
y²+4y+4 = 3y²+3y+1
2y² - y -3 = 0, com y > 0. Resolvendo essas equação do 2° grau, obtemos y = 3/2, logo:
AB=3/2, BC=5/2, AC=7/2 e AD=3√3/4. Temos:
*_Perímetro=_* 3/2 + 5/2 + 7/2 = *15/2 U*
*_Área=_* AD×BC/2 =
= 5/2 × 3√3/8 = *15√3/16 U.Q*
Col teorema del coseno risulta 18x^2-63x+52=0..x=39/18,x=4/3(no)..per cui i lati sono 7/2,3/2,5/2...A=(1/2)(3/2)(5/2)sin120=15√3/16..P=15/2
First and foremost as a US citizen I follow the Constitution of the United States of America. If anyone is offended by that, I don't care! Be offended...🤣. @ 7:43 , never say never! Things get very strange very fast in this 4-dimensional space-time continuum.😊
😀
Thanks for sharing ❤️