Calculate the angle X and justify | Learn how to Solve the Geometry problem Quickly
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- เผยแพร่เมื่อ 5 ก.ย. 2024
- Learn how to find the angle X in the given diagram. Solve this tricky geometry problem by using the isosceles triangle properties and Exterior Angle Theorem.
Today I will teach you basic tips and hacks to solve this tricky geometry problem in a simple and easy way. Step-by-step tutorial by PreMath.com
• Calculate the angle X ...
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Calculate the angle X and justify | Learn how to Solve the Geometry problem Quickly
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Learn how to find the angle X
Isosceles Triangles
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Exterior Angle Theorem
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Knowing that alfa+beta=127, for the reasons mentioned, I have considered the quadrilateral EFCD from which
X + 53 + 180 - alpha + 180 - beta = 360 that easily leads to x=74
Let us split E into
Tricky at first but after seeing the Angle Addition Postulate and the substitution technique, the problem became a piece of cake.
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Explaining is very nice
Thanks Professor, you crushed those triangles in your vicelike grip!🥂
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Consider the quadrilateral CDEF :
X+53 + y1 + y2 = 360 or
X + 53 + (180 - a) + (180 - b) = 360 or
X = ( a + b ) - 53 or ( exterior angle )
X = 127 - 53 = 74 😊
Yay! I solved it.
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I tackled this from a geometric direction. Since the question was general in nature then the answer must work for all positions of E along AB. Let CA equal CB in length and let E be located mid length.. The rest is just fill in the spaces.
We can see that 127 is an exterior angle to triangle CAB, so then it will be the sum of angle A and angle B (external angle postulate). That means angle A and angle B add up to 127. Now, we look at triangle AED and triangle BEF. Since they are both isoceles, angle A will equal angle EDA, and angle B will equal angle EBF. So, if we were to add these two triangles together, we would get a sum of 360 degrees. 360= angle A + angle A (EDA, reflexive property) + angle B + angle B (EFB, reflexive property) + angle DEA and angle FEB. Since there are two pairs of angle A + B, which add up to 127, we can substitute and simplify. Therefore, 360=254+angle EFB+ angle DEA. Therefore, angles EFB + DEA have a sum of 360-254=106. Then we can look at the straight angle E. It consists of X + DEA + EFB. Substituting the sum of DEA and EFB, we get x+106=180, thus x =74 degrees.
Superb!!!, more like this please...
I sure need the practice, you just seem to wave your hands and the solution appears, such great detailed explanation. Thanks again
Herein.
angle ABC + angle CAB = 127°
Hereby angle AED + angle BEF
= (180° - 2 angle DAE)
+ (180° - 2 angle FBE)
= 360° - 2 x 127° = 106°
Hereby x°
= 180° - (angle AED + angle BEF)
= 74°
Superb, sir! I choice "the long way around Robinhood's barn" to receive a system of 3 equations with three unknowns :x, a and b. The first step was obvious :The exterior angle 127=a +b. Thus a=127-b Later on I used two other exterior angles :180-2a +x =2b and 180-2b +x=2a As next I substituted 180 - 2(127-b)+x=2b 180-254+2b+x=2b (-74) =(-x) x=74
The conditions are not given but AC // EF and interior angle is 53 so according to corresponding angles angleF is 53 and since FBE is isosceles so angle B is also 53 and angle FEB will be 74. According to corresponding angles angle A is 74 also and angle D is also 74 since ADE is isosceles and angle E will be 32. So x will be 180 - 74 -32 = 74
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Wow. You walked us through that like it was a walk in the park! You made it look so easy. That was kind of a fun journey.
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∠DCF = 180 - 127 = 53
Let ∠DAE = y.
Then ∠EBF = 127 - y (or 180 - 53 - y)
Because AE = DE, ∠ADE = y, so ∠CDE = 180 - y.
Because BE = BF, ∠EFB = 127 - y, so ∠CFE = 180 - (127 - y) = 53 + y.
So the interior angles of CDEF = 53 + (53 + a) + (180 - a) + x = 360
53 + 53 + a + 180 - a + x = 360
286 + x = 360
x = 74.
The initial conditions of the problem do not prevent me from considering AC=CB → ∆ACB is isosceles just like ∆ADE and ∆EFB → Drawing a parallel to CB through vertex A, it is verified that ∠A=127º/2 → ∠A=∠B =127º/2 → ∠AED=∠BEF=180º -(2*127º/2)=180º -127º=53º → ∠DEF=180º -53º -53º= 74º=X
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Great solution!
Excellent! Love the logical application of geometric theorems to solve the problem.
Ninty nine percent of all your videos are EASY? but easy for me butnot for my scoolers. You are great, thanx u
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Thanks professor, Which program do you use to draw diagram?
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This os absolutely beautiful love your channel
let angle DAE =a ant angle FBE= b
a +b =127
Let Andre DEF= m and angle FEB= n
Consider triangles ADE and FEB, the sum of their angles would be
2a+2b + m +n = 360
m + n = 360 - 2•127
= 106
Angie x = 180 -106
=74
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Thanks to Sir Master 🇧🇷
Very nice 👍
I love maths , bcuz of u
I looked at the 360 degrees of the rectangle CDEF containing x
Absolutely stunning, awesome, many thanks, Sir!
Sir I have 3 question maths . Can ask you personally any way
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Interesting
I solved it
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α + β = 127°
x = 180° - (360° - 2 * 127°) = 180° - 106° = 74°
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74
OMG I DID IT WITH DIFFERENT METHOD AND STILL GOT IT RIGHT
good
127=A+B, 2*127=2(A+B)=180+x, x=254-180=74, done.🙂
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SUPER!
360° -294° = 66° X=66°
There is no assumption that ABC is a triangle. -> how can you prove angle aeb is 180
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All i did was 360-(2x127) then subtract the difference from 180 to get 74 derees
Much easier to say that angles CDE +CFE = (360-127). Thus 53 + (233) + x = 360.
This is 6th 7th grade math in Asia, believe it or not. I remembered starting doing proof in 6th grade.
I am in 11th grade India. This problem took me 37 secs to solve mentally without pen/paper. (I usually time my solves)(my methods are different from the ones done here)
X=180°-(53×2)= 74
254-180=76
74
1st comment
Amazing job
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Let's go I solved it and is 72
Edit: it was 74 because I miscalculated 127×2 as 252 instead of 254
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74
74