First comment and first like,can you pin it?

@PreMath

Ай бұрын

Done!❤️

@inyomansetiasa

Ай бұрын

​@@PreMaththank you

Adding the extra two parallel lines made me confused. Here is an approach without extra lines: 1) Due to parallel lines, the angle from downward L1 to CG is the same as the angle from upward L2 to GC. 2) Set a equal to the angle at CED (also the angle at GEF) 3) Based on the ABC triangle, the angle between CD and L1 is 30 degrees. 4) Based on isosceles triangle, angle DCE is 180 - 2a. 5) Based on second isosceles triangle, EFG and EGF are both equal to 90 - a/2 6) From #1 and the other angle expressions: 30 + 180 - 2a = 18 + 90 - a/2 7) Rearranging: 210 - 2a = 108 - a/2 or 102 = 3a/2 8) a = 68. 9) x = 180 - 18 - EGF, or x = 180 - 18 - 90 + a/2 => x = 106

@mannlahoti

Ай бұрын

This is much better way of solving without the need of drawing auxiliary lines. Not everybody can think of drawing imagibary lines. But anybody can solve by this method. I was about to write the same and thought that there could be others who would have thought like me before me 😂

@kassuskassus6263

6 күн бұрын

Good job. That's my method !

Label a point on l1 above C on the diagram as point H and a point below G on l2 as point J. As found in the video,

I think I get the roundabout solution award, for making this problem much more difficult than you showed it was. I came out with four unknowns and four equations, based on the sum of interior angles for any triangle being 180. To solve that I produced an equation with a 4x4 matrix multiplied by a 4X1 matrix of unknowns to come out with a 4X1 result. I multiplied both sides by the inverse 4x4 matrix and came out with the solution, 106 for x, and 68, 56, and 44 for the others. I guess the most satisfying part of this was showing it actually worked.

Very cool puzzle! Congratulations teacher!!

When you started the problem, l1 and l2 were not necessarily parallel, then magically they became parallel at 0:53 in the video. Makes it hard for people trying to do the problem before watching your solution ;) Also, if you label the top of the leftmost line "H" and the bottom of the rightmost line, "I", then ∠HCG and ∠CGI are equal and the solution is a bit more simple, no need for auxiliary lines. If you label ∠DEC as α then you get 30+180-2α = (180-α)/2+18 and from there you can easily solve for X.

@egillandersson1780

Ай бұрын

At the beginning, at 0:19, he says the lines are parallel

Due to the parallel lines we have: angle(DCE) + 30° = angle(EGF) + 18° However, we also have that: angle(DCE) = 4*angle(EGF) - 180° (due to the two isosceles) This can be solved, so: angle(EGF) = 56° and: X = 180° - 18° - 56° = 106°

How did you come up with a fantastic way of adding 2 lines in the middle?😭 Like, why can't I find the treasure of solution? Can you give some tips on that to me?

@mannlahoti

Ай бұрын

Check the other comments, there are solutions without drawing the lines

Auxiliary lines!

After seeing 56+18 = 74, I paused the video and got the answer immediately

Let angle egf= beta and angle ced = alpha. Angle gef=180-2 beta so angle dec also=180-2 beta degrees. This means that angle DCE =4beta-180. The angle acb = 30 so the vertically opposite angle also =30. So 4beta-180 +30=18+1beta. Then 3beta=168 degrees so beta =56. Therefore x= 180- 56-18=106 degrees

Let's find x: . .. ... .... ..... Since the triangle ABC is a right triangle, we obtain: sin(∠ACB) = AB/AC = 1/2 ⇒ ∠ACB = 30° From the sketch we can conclude that ∠BCG=x. Now we can proceed: ∠DCE = 180° − ∠BCG − ∠ACB = 180° − x − 30° = 150° − x The triangle CDE is an isosceles triangle, so we can conclude: ∠CDE = ∠CED = (180° − ∠DCE)/2 = [180° − (150° − x)]/2 = (30° + x)/2 = 15° + x/2 From the sketch we can conclude that ∠CED=∠FEG. Since the triangle EFG is also an isosceles triangle, we obtain: ∠EFG = ∠EGF = (180° − ∠FEG)/2 = [180° − (15° + x/2)]/2 = (165° − x/2)/2 = 165°/2 − x/4 Now we are able to calculate the value of x: 180° = x + ∠EGF + 18° 180° = x + 165°/2 − x/4 + 18° 79.5° = (3/4)*x ⇒ x = 4*79.5°/3 = 106° Best regards from Germany

106°

Uglily difficult 😢, angle E=90-a/2, angle F,G=(180-(90-a/2))/2=90+a/4, 18+90+a/4+x=180, x+a/4=72, 180-x=30+a, a=150-x, x+(150-x)/4=72, 3/4 x=72-150/4=(288-150)/4=138/4, 3x=138, x=46.😅

Do it again, 180-x=30+c, c=150-x, e=90-(150-x)/2=90-75+x/2=15+x/2, g=90-(15+x/2)/2=90-15/2-x/4=165/2-x/4, 180=x+165/2-x/4+18, 3/4 x=180-165/2-18=162-165/2=(324-165)/2=159/2, x=159/2 × 4/3=53×2=106.😅😅😅😅😅😅😅😅

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asnwer=153 isit