In step 5, you just guess values for x and y. Your guess happens to be right, but there are other guesses that would have been wrong. You have shown how to check whether a guess is right, but not how to make the right guess.
No need to guess X and Y, the problem itself gives a unique solution. Extend AB by a and DB by b to meet the circumference. Using the Intersecting Chords Theorem, we get 2*b=6*a and from R²= ((6-a)/2)²+((b+2)/2)² a=4, b=12 BC= √26
My sollution: Extended DB and AB, making 2 chords. Called DB's extention as x and AB's extension as y. DB*x=AB*y 2x=6y x=3y (1) Then, by 4r²=DB²+x²+AB²+y² found that x²+y²=160 (2) Then, inserting (1) into (2) found that x=12 and y=4. Then I extended BC, calling one part as √50+BC and the other as √50-BC. Then, picking one of the other two chords, found BC 2*12=(√50+BC)(√50-BC) BC=√26
@@soumyaroy5796 It's Intersecting Chords Theorem. When you extended BC, one part os the chord is √50+BC and the other is √50-BC, becouse the given radius is √50.
Thanks William for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed my friend😃
tan β = 6/2 β = 71,565° Labelling point 'F' to the extension of '2' intersecting the circumference DF = DB + BF = 2 + a ACF is isosceles triangle R sin β = ½ FA FA = 2 R sin β FA = 2 √50 √0,9 FA = √180 = 13,416 cm FA² = 6² + a² a = 12 cm Intersecting chords theorem: (R+z)(R-z) = 2 . a R²-z² = 2 . 12 z² = R² - 2 . 12 z² = 50 - 24 = 26 z = √26 cm ( Solved √ )
As many have stated here... guessing and then checking afterwards is NOT the right way to go. It could be that a problem has more than one solution for the length x, e.g. The easiest way to proceed from @8:00 (still only using Pythagoras' theorem to solve this problem) is as follows: Consider ΔACE: - AE = 6 - y - EC = x = 8 - 3y - AC = √50 Applying Pythagoras' theorem in this triangle (AC^2 = EC^2 + AE^2) gives: 50 = (8 - 3y)^2 + (6 - y)^2 Expanding the squares and simplifying the resulting expression yields: y^2 - 6*y + 5 = 0 This quadratic equation has two solutions: y=1 and y=5. Since x = 8 - 3y, this leads to x = 5 or x = -7. Since x represents a length (and must therefore be positive), we can ignore the solution y = 5. Therefore (x, y) = (5, 1) is the only solution. Now proceed with this result at @10:06.
tan α = 2/6 α = 18,435° Labelling point 'E' to the extension of '6' intersecting the circumference AE = AB + BE = 6 + b R sin α = ½ ED ED = 2 R sin α ED = 2 √50 √0,1 ED = √20 = 4,472 cm ED² = 2² + b² b = 4 cm Intersecting chords theorem: (R+z)(R-z)=6.b R²-z² = 6.b z² = R² -6.b z² = 50 - 6 . 4 = 26 z = √26 cm ( Solved √ )
Use the sketch at 3:46, r=sqrt(50), Take the Intersecting Chord Theorem, About F, (x+2)^2=(r-y)(r+y) x^2+4x+4= r^2-y^2=50-y^2 x^2+y^2=46-4x.............................(1) About E, (6-y)^2=(r-x)(r+x) y^2-12y+36=r^2-x^2 y^2+x^2=14+12y..........................(2) Equate equations (1) and (2), 46-4x=14+12y 12y+4x=32,3y+x=8 y= (8-x)/3.....................................(3) Use the triangle DFC, DF^2+FC^2=r^2 (x+2)^2+y^2=(sqrt(50))^2 (x+2)^2+((8-x)/3}^2=50 (substitute y from (3)), x^2+4x+4+(x^2-16x+64)/9=50 9x^2+36x+36+x^2-16x+64=450 10x^2+20x+100=450 x^2+2x+10=45 x^2+2x-35=0 (x-5)(x+7)=0 x=5, reject the -ve root. y=(8-x)/3 from (3), =(8-5)/3,y=1. From sketch, BC^2=x^2+y^2=46-4*5=26, from (1) BC=sqrt(26) units and that is our answer. This is a good place to stop. Thanks for the problem PreMath.
Hello but I’m the last why did you pick whole numbers? It could have also been irrational or numbers that are not integers and that was not mentioned in the question
Here is another solution. Extend AB and let E be the point where it intersects the circle. From the angle of circumference theorem, angle DCE is twice angle DAE. Also, CD and CE are equal in length, so triangle CDE is an isosceles triangle. If we draw a line bisecting angle DCE and set F as the intersection with DE, then angle DFC is a right angle. Therefore, triangle ABD and triangle CFD are similar. From the similarity ratio, the length of DF is the sqrt 5. The length of DE is twice that of DF. Finally, extend BC in both directions, and let G and H be the intersections with the circle, respectively. the length of BG is the radius-x. the length of BH is the radius+x. Therefore 4*6=(x-radius)(x+radius). 24=x^2-50. x^2=26. x=sqrt 26.
Guessing and checking is no solution. To solve, say x,y is remaining lengths of two given intersecting cords having 6,2 as part lengths respectively. This amount to y=3x as products of segments of intersecting cords are equal. Now draw diameter parallel to any cord and find x using Pythagorean triplet relation. This would give x=4 and thus y=12. Then find BC=√26, again using Pythagorean triplet relation. This is the correct way.
I understood how to find the correct solution. The two equations obtained by pitagoras are equations of circunference so you need to draw both grafics. Then you find the interception between both circunferences points (5,1) and (-7,5). The equation x+3y=8 is going to intercept both points. The point (-7,5) has only physical meaning if we use a system of coordenate relative instead of distances.
I used Trig. 1- Pythag AD, then Trig to calculate angle DAB. 2- Connect CA & CD to get triangle CAD, with all known side values. Law of Cos to calculate angle CAD. 3- Now look at triangle CAB. Angle CAB = (angle CAD - angle DAB) 4- Now triangle CAB has 2 known sides and middle angle. Law of Cos to calculate distance CB, our answer.
Great 👍 Thanks Abhishek for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
3:34 This problem is meant to be solved in trigonomeetry to avoid trial and error. ∠DAB = artan(2/6) = 18.43° ∠DAC = 63.43° (sine rule from isoseles △DCA with both radius =√50 and hypotenuse of △DBA = √40) Notice that ∠BAC = 45°!!!~ ∠ECA = 45° Therefore, EA = EC = 5 (Pythagorean Theorem of △EAC given AC = √50) BE = BA - EA BE = 6 - 5 = 1 So, BC = √26 = 5.099
Sometimes is it allowed to "guess" and check later? In my check (2,2) didn't seem to work with equations (1) & (2). Also x=-7, y=5 would work in eq (3), but might be with wrong geometry, of a kind... To avoid "guessing", one could try solve coordinate points for A as (-5,-5) and D as (-7,1) and then point B at (-7+2,-5+6)=(-5, 1) and C was at (0,0). Then distance from B to C is sqrt((0-(-5))^2+(0-1)^2)=sqrt(26).
Extend DBto the far side of the circle , calling it H, and extend AB. to the top of the circle, calling it G.lNote that if you use the values 1 and 5 that you have in your solution that DBxBH does not equal ABxBG. If you use this chord-chord product theorem, together with the Pythagoras applications in your solution, you can find that BC=2sqrt2.
I agree that there is a issue with this solution when you say...to make things simple let take interget numbers for x e y ...there is no constraint in the exervise that allows you to say that x and y are intergers so this solution is not really proved..
Could also use the chords theorem and solve a quadratic equation in x (or y!) No need to guess then verify! But I have to admit that I struggled a bit before solving this one! EUREKA! 😂😂
Dear Javan, thanks for sharing. Elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
I found other way to solve . You need to combine the equation of x+3y =8 with one of the circunference equations then you will obtain one equation of second degree where the roots are 5 and 1 then you use these roots in the equation of x+3y=8 to find the value of the other variable.
Hi, great - this was a hard nut to crack ! If found the solution with some trigonomertry. First I got the angle BCA with the law of cosines, the same with angle DAC - after I got the distance DA . Now I calculated angle DAB and found angle BAC . Now the law of cosines again gives the distance BC….
Thanks Gowri for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
My solution. First, join C, D and C, A join A, D furthermore. Drop perpendicular from C on AD. Let the foot of the perpendicular be F. CA =CD=root 50 and AD=root over(36+4)=2 (root 10),now Cos(angle CAD) =1/root5 and sin(angle CAD) =2/root5. See, sin(angle BAD ) =1/root10 and Cos(angle BAD ) =3/root10. Now sin(angle BAC)=sin(angle CAD - angle BAD)=(2/root5)(3/root10) - (1/root5) (1/root10) =1/root2=sin45. Drop perpendicular at E on AB, from C. Traingle CEA will be isosceles with EA=EC=root over(50/2)=5 (because angle BAE=45), now BE=6-5=1 and CE=5 therefore BC=root ovet(25+1)=(root 26). This is our ans. I could do it with cosine rule but I preferred doing it step by step by precise construction
Many values of x and y satisfies the equation. Take x=2, y =2 this also does satisfy. But this does not satisfy equation 1 and equation 2. So, it is defined. But too long.
I have a genuine solution. Dont use trial and error method. If x and y are not integers then what you do? I have simple and clear solution but i cant explain in comment box.
Hello sir,love from Bangladesh 🇧🇩🇧🇩🇧🇩
Thank you so much Partho for your continued love and support. Take care dear and stay blessed😃 You are awesome. Keep smiling😊
@@PreMath, sir would you like to find the radius if the value of BC is given. This will be another fun.
In step 5, you just guess values for x and y. Your guess happens to be right, but there are other guesses that would have been wrong. You have shown how to check whether a guess is right, but not how to make the right guess.
True what you said. In fact, the right answer is 5.1 haha and Mr Phisterer blundered!
Sorry, I didn't follow the explanation to the end, Sir PreMath. Bear with me. However, I was sidetracked by the initial result attempts.
Possible explanation if condition; x and y is natural number
No need to guess X and Y, the problem itself gives a unique solution.
Extend AB by a and DB by b to meet the circumference.
Using the Intersecting Chords Theorem,
we get 2*b=6*a
and from R²= ((6-a)/2)²+((b+2)/2)²
a=4, b=12
BC= √26
My sollution:
Extended DB and AB, making 2 chords.
Called DB's extention as x and AB's extension as y.
DB*x=AB*y
2x=6y
x=3y (1)
Then, by 4r²=DB²+x²+AB²+y² found that
x²+y²=160 (2)
Then, inserting (1) into (2) found that
x=12 and y=4.
Then I extended BC, calling one part as √50+BC and the other as √50-BC.
Then, picking one of the other two chords, found BC
2*12=(√50+BC)(√50-BC)
BC=√26
Why 2*12 =( BC+√50)(√50-BC)?? Is there any relation fir using this?
@@soumyaroy5796 It's Intersecting Chords Theorem. When you extended BC, one part os the chord is √50+BC and the other is √50-BC, becouse the given radius is √50.
@@alancs85 got the clear picture. Thanks!
8:18 Can x and y be picked freely as long as making x+3y=8? No. It has to be obtained by solving either Eq. 1 or 2.
I have been out of school for a long time.
This man would have been a great teacher for math class 👏
Thanks William for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed my friend😃
tan β = 6/2
β = 71,565°
Labelling point 'F' to the extension of '2' intersecting the circumference
DF = DB + BF = 2 + a
ACF is isosceles triangle
R sin β = ½ FA
FA = 2 R sin β
FA = 2 √50 √0,9
FA = √180 = 13,416 cm
FA² = 6² + a²
a = 12 cm
Intersecting chords theorem:
(R+z)(R-z) = 2 . a
R²-z² = 2 . 12
z² = R² - 2 . 12
z² = 50 - 24 = 26
z = √26 cm ( Solved √ )
Great challenge, awesome way to solve it - especially "let's pick some x and y values..."
(2 + a)^2 + (6 - b)^2 = a^2 + b^2 →
a = 3b - 10 → (b - 3)^2 = 4 → b1 = 5; b2 = 1 → a = 3b - 10 > 0 → b = b1 = 5 →
a = 15 - 10 = 5 → BC = √26
obviously:
a > 2
b < 6
and:
a^2 + b^2 = 50 → 50 = 2(25) = 2(5)^2 → a = b = 5
or:
tan(θ) = AB/BD = 3 → θ = 71,565°
AF = DF = AD/2 = √10 → tan(φ) = √10/√40 = √(1/4) = 1/2 → φ = 26,565°→
γ = DAC = 90° - φ = 63,435°→
θ + γ = 135° → AR = a = CR = b = √50/√2 = √25 = 5 = a = b → 25 + 1 = (BC)^2
(AR = parallel to DF; CR = parallel to AB)
Another approach:
tan(α) = BD/AB = 2/6 = 1/3 → α = 26,57°
(AD)² = 4 + 36 = 40 → AD = 2√10.
AD/2 = AF = DF = √10 → (CF)^2 = (AC)^2 - (AF)^2 = 50 - 10 = 40 →
CF = 2√10 = 2AF = 2DF = AD
Draw x- and y-axis:
x = horizontal through A;
y = AB → find the coordinates of Point F →
F cuts AD in half → A(0; 0) → D (-2; 6) → F(-1; 3)
two equations to solve the problem:
f(x) = g(x) in F(-1; 3)
f(x) = AD = -3x
g(x) = (x/3) + k → g(x) = 3 = (-1)/3 + k → k = 10/3 → g(x) = (x/3) + (10/3)
f(x) = g(x) → crossing in Point F
FC' = parallel to DB → CC' = parallel to AB → (CC')/(FC' ) = 1/3 = tan α →
(FC)^2 = (FC' )^2 +(CC') = 40 = a^2 + (3a)^2 = 10a^2 → a^2 = 4 → a = 2 →
CC' = 2 → FC' = 6 → (BC)^2 = √(25 + 1) = √(26) 🙂
or:
tan(δ) = BD/AB = -3 → tan(α) = (CC')/(FC' ) = 1/3 →
CF = √40 = 2√10 → CC' = k → (CF)^2 = 40 = 10(k)^2 →
k = √4 → k = 2 (s.t.: k > 0) →
CC' = k = 2 → FC' = 6 → B'C' = a = 6 - 1 = 5 = b → BC = √26
Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Stay blessed 😀
As many have stated here... guessing and then checking afterwards is NOT the right way to go. It could be that a problem has more than one solution for the length x, e.g.
The easiest way to proceed from @8:00 (still only using Pythagoras' theorem to solve this problem) is as follows:
Consider ΔACE:
- AE = 6 - y
- EC = x = 8 - 3y
- AC = √50
Applying Pythagoras' theorem in this triangle (AC^2 = EC^2 + AE^2) gives:
50 = (8 - 3y)^2 + (6 - y)^2
Expanding the squares and simplifying the resulting expression yields:
y^2 - 6*y + 5 = 0
This quadratic equation has two solutions: y=1 and y=5.
Since x = 8 - 3y, this leads to x = 5 or x = -7.
Since x represents a length (and must therefore be positive), we can ignore the solution y = 5.
Therefore (x, y) = (5, 1) is the only solution.
Now proceed with this result at @10:06.
tan α = 2/6
α = 18,435°
Labelling point 'E' to the extension of '6' intersecting the circumference
AE = AB + BE = 6 + b
R sin α = ½ ED
ED = 2 R sin α
ED = 2 √50 √0,1
ED = √20 = 4,472 cm
ED² = 2² + b²
b = 4 cm
Intersecting chords theorem:
(R+z)(R-z)=6.b
R²-z² = 6.b
z² = R² -6.b
z² = 50 - 6 . 4 = 26
z = √26 cm ( Solved √ )
Use the sketch at 3:46,
r=sqrt(50),
Take the Intersecting Chord Theorem,
About F,
(x+2)^2=(r-y)(r+y)
x^2+4x+4= r^2-y^2=50-y^2
x^2+y^2=46-4x.............................(1)
About E,
(6-y)^2=(r-x)(r+x)
y^2-12y+36=r^2-x^2
y^2+x^2=14+12y..........................(2)
Equate equations (1) and (2),
46-4x=14+12y
12y+4x=32,3y+x=8
y= (8-x)/3.....................................(3)
Use the triangle DFC,
DF^2+FC^2=r^2
(x+2)^2+y^2=(sqrt(50))^2
(x+2)^2+((8-x)/3}^2=50 (substitute y from (3)),
x^2+4x+4+(x^2-16x+64)/9=50
9x^2+36x+36+x^2-16x+64=450
10x^2+20x+100=450
x^2+2x+10=45
x^2+2x-35=0
(x-5)(x+7)=0
x=5, reject the -ve root.
y=(8-x)/3 from (3),
=(8-5)/3,y=1.
From sketch, BC^2=x^2+y^2=46-4*5=26, from (1)
BC=sqrt(26) units and that is our answer.
This is a good place to stop.
Thanks for the problem PreMath.
Hello but I’m the last why did you pick whole numbers? It could have also been irrational or numbers that are not integers and that was not mentioned in the question
Wow.. this was tougher than it looked at first blush! Thx.
Here is another solution.
Extend AB and let E be the point where it intersects the circle. From the angle of circumference theorem, angle DCE is twice angle DAE. Also, CD and CE are equal in length, so triangle CDE is an isosceles triangle. If we draw a line bisecting angle DCE and set F as the intersection with DE, then angle DFC is a right angle. Therefore, triangle ABD and triangle CFD are similar. From the similarity ratio, the length of DF is the sqrt 5. The length of DE is twice that of DF. Finally, extend BC in both directions, and let G and H be the intersections with the circle, respectively. the length of BG is the radius-x. the length of BH is the radius+x. Therefore 4*6=(x-radius)(x+radius). 24=x^2-50. x^2=26. x=sqrt 26.
The explanation was missing, so I'll add it.
Using the Three Square Theorem, DE^2-DB^2=BE^2. Therefore BE^2=20-4=16. Therefore BE=4.
説明内容に従って絵を描きました.理解しました。すごい
@@안다순-x5g ありがとう！
@@ebi2ch And also extend BC not BE
@@hrisheekeshbhave5607 Your explanation is correct. This was a simple typo on my part. Thank you for pointing it out.
First calculate AD, then
My way too
Guessing and checking is no solution. To solve, say x,y is remaining lengths of two given intersecting cords having 6,2 as part lengths respectively. This amount to y=3x as products of segments of intersecting cords are equal. Now draw diameter parallel to any cord and find x using Pythagorean triplet relation. This would give x=4 and thus y=12. Then find BC=√26, again using Pythagorean triplet relation. This is the correct way.
I understood how to find the correct solution.
The two equations obtained by pitagoras are equations of circunference so you need to draw both grafics.
Then you find the interception between both circunferences points (5,1) and (-7,5). The equation x+3y=8 is going to intercept both points. The point (-7,5) has only physical meaning if we use a system of coordenate relative instead of distances.
Sir I want to send you some problems where do I send?
I used Trig.
1- Pythag AD, then Trig to calculate angle DAB.
2- Connect CA & CD to get triangle CAD, with all known side values. Law of Cos to calculate angle CAD.
3- Now look at triangle CAB. Angle CAB = (angle CAD - angle DAB)
4- Now triangle CAB has 2 known sides and middle angle.
Law of Cos to calculate distance CB, our answer.
Beautifully explained Sir
Awesome video solution sir...I understood everything.. thank you very much 👍👍
Great 👍
Thanks Abhishek for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
Sir can you please solve this ....
Decompose to partial fractions :
1) 2(x^2-3x+2) / x^3+2x-3
2) 54 / x^3-21x+20
photo math ，the app on phone ，inside have calculater
In step 5, x+3y=8 , let y=t , x=8-3t ( t>0 , 8-3t>0) = ( 0
Theorem of Chords:
x • z = w • y
2 • z = 6 • y
z = 3y (1)
Perpendicular Chords:
4r² = x² + y² + w² + z²
4(√50)² = 2² + y² + 6² + z²
200 = 4 + y² + 36 + z²
200 = 40 + y² + z²
160 = y² + z² (2)
(1) & (2) ==> 160 = y² + (3y)²
160 = y² + 9y²
160 = 10y²
16 = y²
==> y = 4 (3)
Theorem of Chords:
6 • y = (√50 + BC) • (√50 - BC)
6y = (√50)² - (BC)²
(BC)² = 50 - 6y (4)
(3) & (4) ==> (BC)² = 50 - 6 • 4
(BC)² = 50 - 24
(BC)² = 26
BC = √26
Used different construction and used cosine rule three times giving me square root of 26
Incredible John! Smart move.
Thanks dear for sharing. You are awesome 👍 Take care dear and stay blessed😃
Can also be done using cosine rule three times
3:34 This problem is meant to be solved in trigonomeetry to avoid trial and error.
∠DAB = artan(2/6) = 18.43°
∠DAC = 63.43° (sine rule from isoseles △DCA with both radius =√50 and hypotenuse of △DBA = √40)
Notice that ∠BAC = 45°!!!~
∠ECA = 45°
Therefore,
EA = EC = 5 (Pythagorean Theorem of △EAC given AC = √50)
BE = BA - EA
BE = 6 - 5 = 1
So,
BC = √26 = 5.099
In the equation x + 3y = 8, another possible solution is (2,2) which was not checked.
Sometimes is it allowed to "guess" and check later?
In my check (2,2) didn't seem to work with equations (1) & (2).
Also x=-7, y=5 would work in eq (3), but might be with wrong geometry, of a kind...
To avoid "guessing", one could try solve coordinate points for A as (-5,-5) and D as (-7,1) and then point B at (-7+2,-5+6)=(-5, 1) and C was at (0,0).
Then distance from B to C is sqrt((0-(-5))^2+(0-1)^2)=sqrt(26).
x=8-3y => x2+(y-6)2=50 => y2-6y+5=0 => y = 1 (true) and y = -5 (wrong).
I think it's better than guess.
Extend DBto the far side of the circle , calling it H, and extend AB. to the top of the circle, calling it G.lNote that if you use the values 1 and 5 that you have in your solution that DBxBH does not equal ABxBG. If you use this chord-chord product theorem, together with the Pythagoras applications in your solution,
you can find that BC=2sqrt2.
It is good for 1,5. Check again.
I agree that there is a issue with this solution when you say...to make things simple let take interget numbers for x e y ...there is no constraint in the exervise that allows you to say that x and y are intergers so this solution is not really proved..
Sir please make a video on this question next
Integral from 0 to 2pi(1/(3+2sin(x))dx
Using contour integrals
Sure Aditya!
Pretty soon I'll do that. Thanks for asking.
Take care dear and stay blessed😃
Thanks sir
You are best
Amazing! ♥️🔥
Greetings and Prayers from India! ❤️🇮🇳
@Freefire RUclipsr??
actually we can put x=8-3y into equation (1) to solve x and y, no need to guess
Could also use the chords theorem and solve a quadratic equation in x (or y!) No need to guess then verify! But I have to admit that I struggled a bit before solving this one! EUREKA! 😂😂
Dear Javan, thanks for sharing. Elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
How?
I found other way to solve . You need to combine the equation of x+3y =8 with one of the circunference equations then you will obtain one equation of second degree where the roots are 5 and 1 then you use these roots in the equation of x+3y=8 to find the value of the other variable.
Attention the values 5 and 1 are for y in tne equation of x+3y=8
Thank you for premath
Довольно интересное решение! Respect!
Большое спасибо, Владимир, за вашу неизменную любовь и поддержку. Береги себя, дорогая, и оставайся счастливым😃 Ты потрясающий. Продолжай улыбаться😊
Hi, great - this was a hard nut to crack !
If found the solution with some trigonomertry.
First I got the angle BCA with the law of cosines, the same with angle DAC - after I got the distance DA .
Now I calculated angle DAB and found angle BAC . Now the law of cosines again gives the distance BC….
excelente profe !
Thank you sir
Thanks Gowri for such an elegant feedback. Thank you so much for taking the time to leave this comment. You are awesome 👍 Take care dear and stay blessed😃
My solution. First, join C, D and C, A join A, D furthermore. Drop perpendicular from C on AD. Let the foot of the perpendicular be F. CA =CD=root 50 and AD=root over(36+4)=2 (root 10),now Cos(angle CAD) =1/root5 and sin(angle CAD) =2/root5. See, sin(angle BAD ) =1/root10 and Cos(angle BAD ) =3/root10. Now sin(angle BAC)=sin(angle CAD - angle BAD)=(2/root5)(3/root10) - (1/root5) (1/root10) =1/root2=sin45. Drop perpendicular at E on AB, from C. Traingle CEA will be isosceles with EA=EC=root over(50/2)=5 (because angle BAE=45), now BE=6-5=1 and CE=5 therefore BC=root ovet(25+1)=(root 26). This is our ans. I could do it with cosine rule but I preferred doing it step by step by precise construction
Sans se fatiguer 50^.5 -2 est égal environ à 5,071 !!!!! Without getting tired 50^.5 -2 is approximately equal to 5.071!!!!!
But x = square(50) - 2 no?
Many values of x and y satisfies the equation. Take x=2, y =2 this also does satisfy. But this does not satisfy equation 1 and equation 2. So, it is defined. But too long.
That was a bit more challenging.
Right triangle
I used trigonometry to solve this. It was an easy one.
See
#Pythagoras #PythagoreanTheorem
I have a genuine solution. Dont use trial and error method. If x and y are not integers then what you do?
I have simple and clear solution but i cant explain in comment box.
Square root of 50
Square root of 26
Pythagorean Theorem
300 th like
Pythagoras