Calculate the angle X and justify | Learn how to Solve the Geometry problem Quickly

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.

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Calculate the angle X and justify | Learn how to Solve the Geometry problem Quickly

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Olympiad Mathematics
pre math
Po Shen Loh
Learn how to find the angle X
Isosceles Triangles
Triangle
premath
premaths
Exterior Angle Theorem

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[solimana-soli]

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

[alster]

Tricky at first but after seeing the Angle Addition Postulate and the substitution technique, the problem became a piece of cake.


God bless from The Philippines 🇵🇭

[bigm]

Thanks Professor, you crushed those triangles in your vicelike grip!🥂

[dye]

Let us split E into <AEL=<LED=a, x, <BER=<REF=b, where L on AC and R on BC, this makes <ELC=<ERC=90. Now we have <REL=127=a+b+x, and also 2a+2b+x=180, so a+b=53, x=74

[theoyanto]

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

[lindafromcalifornia]

You are such a blessing to so many. Keep up your great work. 👍

[MrPhilippos]

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 😊

[jasonk]

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.

[johnspathonis]

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.

[jim]

Excellent! Love the logical application of geometric theorems to solve the problem.

[annatygrys]

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

[fried_ady]

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.

[roycuyler]

You are a very good instructor, sir. Thank you.

[vandanakumari]

Yayyy New video!
Thank you 💕 professor

[murdock]

Absolutely stunning, awesome, many thanks, Sir!

[honestadministrator]

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°

[santiagoarosam]

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
Thanks for the curious problem and best regards

[sngmn]

Ninty nine percent of all your videos are EASY? but easy for me butnot for my scoolers. You are great, thanx u

[abdulwahab_]

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

[sameralkhabaz]

Thanks professor, Which program do you use to draw diagram?