First comment and second like,you can pin it?

@PreMath

20 күн бұрын

You are pinned😀

@Imran-tc6sn

20 күн бұрын

Thankyou sir

Thanku sir

@PreMath

20 күн бұрын

You are very welcome! Thanks ❤️

y=6, CD=7\/3, (4х-1)/(х+2)=14/6; 10х=34; х=3,4!

@PreMath

20 күн бұрын

Excellent! Thanks for sharing ❤️

Validity check: After computing x and y, we should check that the resulting figure is valid. First check: Is BC longer than CD? We note that BC is the hypotenuse of ΔBCD, so must be longer than either side. Length BC = 4x - 1 = 4(3.4) - 1 = 12.6 and, from CD being the long side of a 30°-60°-90° special right ΔACD with short side AD = 7, has length 7√3 or approximately 12.12. So, BC is longer than CD. Second check: Is point E between points B and D, as shown in the diagram? We can compute length BD from the Pythagorean theorem as √(BC² - CD²) = √((12.6)² - (7√3)²) = √(158.76 - 147) = √(11.76) or approximately 3.43. AB = AD + BD = 7 + 3.43 = 10.43. From similar triangles, BE is 3/7 of AB = 4.47. So, point E lies between A and D, not between B and D as shown in the diagram. We are given that the diagram may not be 100% true to the scale, but does that mean that a point shown as between B and D need not be between those points?

@PreMath

20 күн бұрын

Bravo! Thanks for sharing ❤️

@jimlocke9320

20 күн бұрын

@@PreMath You're welcome! Giving the problem more thought, let's say that CF were given a length of 4x - 3 instead of 3x -3. Length BC becomes 5x - 1. Skip ahead to about 6:50 where the formula BC/BF = AC/EF is presented. The formula is still valid but becomes (5x - 1)/(x + 2) = 7/3. Doing the algebra, we find x = 17/8. We might think that we are done. However, if we substitute 17/8 for x in BC = 5x - 1, we find that BC = 9.625. So, BC, the hypotenuse of right ΔBCD, is shorter than CD, a side equal to 7√3, or about 12.12. There is no valid solution. The problem might appear on an exam erroneously. On the other hand, a professor might deliberately put the problem on an exam to see how many students catch the error, perhaps offering partial credit for a solution of x = 17/8.

All 30 60 90 degree triangles in radians have sides with basic ratios : pi/6-pi/3-pi/2. The shortcuts used to solve this problem along with corresponding angles of parallel lines and side length proportional ratios of similar triangles is fun. 🙂

@PreMath

20 күн бұрын

Excellent! Thanks for the feedback ❤️

This math problem is very easy. Got it in less than a minute.

Thank you!

@PreMath

20 күн бұрын

You're welcome!❤️

△ADC is a special 30°-60°-90° right triangle. CD = (AD)√3 = 7√3 AC = 2(AD) y + 8 = 2(7) = 14 y = 6 Draw a segment thru F and a point G on segment AB such that the segments are perpendicular. By the Corresponding Angles Theorem, ∠A ≅ ∠BEF. So, m∠BEF = 60°. So, △ADC ~ △EGF by AA. FG/CD = EF/AC FG/(7√3) = 6/14 FG/(7√3) = 3/7 FG = [3(7√3)]/7 = (21√3)/7 = 3√3 △BDC & △BGF share ∠B. So, △BDC ~ △BGF by AA. BC/BF = CD/FG (4x - 1)/(x + 2) = (7√3)/(3√3) (4x - 1)/(x + 2) = 7/3 12x - 3 = 7x + 14 5x - 3 = 14 5x = 17 x = 17/5 = 3.4 So, the value of x is 3.4, and the value of y is 6.

Similar triangles BEF and ABC Then (3x -3 + x +2)/(x+2) =(y + 8)/y (4x - 1)/(x +2) =(y+8)/y It is a relation between y and x From triangle ACD we say y +8=7*2 y=6 Putting this value in eq (1) we may get the value of x

In ∆ACD Cos(6o)=AD)AC=7/y+8 1/2=7/y+8 y+8=14 so y=14-8=6 AC=6+8=14 CD=√14^2-7^2=7√3 ∆ ABC ~ ∆ FBE 6/14=(x+2)/4x-1 so x=17/5 Hence x=17/5 ; y=6

@PreMath

20 күн бұрын

Thanks for sharing ❤️

VERY simple. In triangle ADC we have AD/AC = cos(60°) =1/2, so 7/AC = 1/2 and AC = 14, so y +8 = 14 and y =6. Triangles ABC and EBF are similar, so BF/BC = EF/AC, so (x+2)/(4.x-1) = y/(y+8) = 6/14 = 3/7, so 7.x+14 = 12.x -3, so 5.x = 17 and x =17/5. That's all.

@MegaSuperEnrique

20 күн бұрын

You have either a math error or a typo

@marcgriselhubert3915

20 күн бұрын

@@MegaSuperEnrique Naturally!! y =6. That's rectified. Thanks.

cos 60º=7/(y+8) 1/2=7/(y+8) multiply both side by 2(y+8) to get y+8=14 y=6 in ACB and FEB angle B is common since AC is parallel to FE then angles A and C are congruent to angles FEB and EFB respectively we can conclude that the two triangles are similar then AC/CB=FE/FB (y+8)/(3x-3+x+2)=y/(x+2) since y=6 we can plug in 6 14/(4x-1)=6/(x+2) cross multiply and divide by 2 3(4x-1)=7(x+2) 12x-3=7x+14 add 3 and subtract 7x from both sides 12x-7x=14+3 5x=17 (divide by 5) x=17/5 x=3.4 so x=3.4 and y=6

Not difficult but need plenty computation, y+8=7×2=14, y=6, 6/(6+8)=3/7=(x+2)/(4x-1), 3(4x-1)=7(x+2), 5x=17, x=17/5.😅

@PreMath

20 күн бұрын

Excellent! Thanks for sharing ❤️

As ∆ADC is a right triangle and ∠CAD is 60°, ∆ADC is a 30-60-90 special right triangle. AC = 2•DA and CD = √3•DA. y + 8 = 2•DA = 2(7) y = 14 - 8 = 6 As AC and EF are parallel and both intersect CB and BA, ∠BAC = ∠BEF and ∠ACB = ∠EFB. As ∠B is also shared between them, triangles ∆ACB and ∆EFB are similar. (x+2)/y = (x+2+(3x-3))/(y+8) (x+2)/6 = (4x-1)/14 6(4x-1) = 14(x+2) 24x - 6 = 14x + 28 10x = 34 x = 34/10 = 17/5

@PreMath

20 күн бұрын

Excellent! Thanks for sharing ❤️

Let's find x and y: . .. ... .... ..... It is obvious that the triangle ACD is a 30°-60°-90° triangle. Since AD is the shorter cathetus, we can directly conclude: AC = 2AD y + 8 = 2*7 = 14 ⇒ y = 6 AC and EF are parallel to each other and the points B, C and F are located on the same line. Therefore the triangles ABC and BEF are similar. This helps us to calculate the value of x: BC/AC = BF/EF (BF + CF)/AC = BF/EF (BF + CF)/BF = AC/EF 1 + CF/BF = (y + 8)/y = (6 + 8)/6 = 14/6 = 7/3 CF/BF = 4/3 (3x − 3)/(x + 2) = 4/3 3(3x − 3) = 4(x + 2) 9x − 9 = 4x + 8 5x = 17 ⇒ x = 17/5 Best regards from Germany

@PreMath

20 күн бұрын

Excellent! Thanks for sharing ❤️🇺🇸

My way of solution is ➡ for the ΔCAD we can write, cos(60°)= AD/CA AD= 7 CA= y+8 cos(60°)= 0,5 ⇒ 0,5= 7/(y+8) 0,5y+4= 7 y= 6 FE // CA, so ∠ (EBF)= ∠ (ABC) ∠ (FEB)= ∠ (CAB) ∠ (BFE)= ∠ (BCA) ⇒ ΔEBF ~ ΔABC FE/CA= BF/BC= EB/AB if we consider the first two equations we get: FE= y FE= 6 CA= y+8 CA= 14 BF= x+2 BC= (3x-3)+(x+2) BC= 4x-1 ⇒ 6/14= (x+2)/(4x-1) 3/7= (x+2)/(4x-1) 12x-3= 7x+14 5x= 17 x= 17/5 x= 3,4 so, (x; y) = (3,4 ; 6)

How long did it take you to solve this just down comment and tell me(viewers)

@marcgriselhubert3915

20 күн бұрын

Just the time to write it on the computer!

@LuisdeBritoCamacho

20 күн бұрын

Some 24 hours of sleep!!

Let's get started!! Quest for y. 1) cos(60º) = 7 / (y + 8) ; 1 / 2 = 7 / (y + 8) ; y + 8 = 14 ; y = 6 Quest for x. 1) Thales's Theorem or Basic Proportionality Theorem (BPT) states that : "If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio." 2) BF / EF = BC / AC ; BF = (x + 2) ; EF = y = 6 ; BC = (x + 2) + (3x - 3) = (4x - 1) ; AC = y + 8 = 6 + 8 = 14 : 3) So : (x + 2) / 6 = (4x - 1) / 14 4) 14*(x + 2) = 6*(4x -1) ; 14x + 28 = 24x - 6 ; 10x = 34 ; x = 34/10 ; x = 17 / 5 5) Answer : x = 6 and y = 17/5. 6) Best Regards from the Kingdom of Portugal and the Algarves!!