Find the angle X in the Circle | Learn how to Solve this Tricky Geometry problem Quickly

Learn how to find the angle X in the given diagram. Solve this tricky geometry problem by using Thales's Theorem, Central Angle Theorem, Triangle Sum Theorem, and Isosceles Triangles properties.

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Find the angle X in the Circle | Learn how to Solve this Tricky Geometry problem Quickly

#FindAngleX #Geometry #GeometryMath
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#IsoscelesTriangle #IsoscelesTriangles #Isosceles #Triangle #Triangles #ThalesTheorem
#CongruentTriangles #CentralAngleTheorem

Olympiad Mathematics
pre math
Po Shen Loh
Learn how to find the angle X
How to Solve this Tricky Geometry problem Quickly
Exterior Angle Theorem
Alternate interior angles
premath
premaths
circle
Radius
Diameter
Radii
Isosceles triangles
Congruent Triangles
Thales's Theorem
Thales' Theorem
Central Angle Theorem
Triangle Sum Theorem

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[mtr-math]

Out of all the educational videos I have seen, I like yours the most.nice problem.

[anatoliy]

Happy New Year, sir! The year is new, whereas the math is old as the world! All the best to you, dear Mr PreMath

[HappyFamilyOnline]

Very well explained👍
Thanks for sharing😊😊

[Copernicusfreud]

Happy New Year. I solved the problem a long and messy way. I did not know the Central Angle Theorem. I did not know Thale's Theorem.

Step 1: I drew a radius line segment from point O to point B. The line segments OA = OB = OD = OC, because they are all a radius of the circle.

Step 2: Triangle OAB is an isosceles triangle. If angle OAB = x, then angle OBA = x also. Angle OBA = x.

Step 3: Angle AOB = 180 - 2x. Angle BOP = Angle BOC = 180 - (180 - 2x)= 2x. Also angle DOP = 180 - 112 = 68.

Step 4: Triangle BOD is also an isosceles triangle, and angle ODB = angle OBD. All three angles = 180. Angle BOD = 68 + 2x.
(68 + 2x) + (angle OBD) + (angle ODB) = (68 + 2x) + (angle OBD) + (angle OBD) = (68 + 2x) + 2 * (angle OBD) = 180. And 2 * (angle OBD) = 180 - (68 + 2x). 2 * (angle OBD) = 112 - 2x. Angle OBD = 56 - x. Angle ODB = 56 - x.

Step 5: Angle OPD = 180 - 68 - (56 - x). Angle OPD = 112 - (56 - x). Angle OPD = 112 - 56 + x. Angle OPD = 56 + x.
Angle OPD and Angle BPC are vertical angles, so those angles are congruent. Angle BPC = 56 + x.

Step 6: Triangle BPC is also an isosceles triangle. If angle BPC = 56 + x, then angle BCP = 56 + x. Angle BCP = Angle OCB. Angle OCB = 56 + x.

Step 7: Triangle OBC is also an isosceles triangle. From step 3, Angle BOP = 2x. Angle BOC = Angle BOP. Angle BOC = 2x. Angle OCB = Angle OBC. 2x + Angle OCB + Angle OBC = 180. 2x + Angle OCB + Angle OCB = 180. 2x + 2 * (Angle OCB) = 180.
2 * (x + Angle OCB) = 180. (x + Angle OCB) = 90. Angle OCB = 90 - x

Step 8: From step 6, Angle OCB = 56 + x. From step 7, Angle OCB = 90 - x. Angle OCB = Angle OCB.
56 + x = 90 - x. x + x = 90 - 56. 2x = 34. x = 17.

[murdock]

Happy New Year, Professor! Awesome question, excellent way solving the problem, many thanks!
AOD = 2φ = 112° → DBA = φ = 56° → ABC = 90° (Thales!) →
90° - φ = γ = 34° → 180° - γ = 2β → β = 73° → 90° - β = 17° = x 🙂

[bigm]

Another step wise explanation. Very good!👍😀

[alster]

Amazing solution.

Happy New Year 2023, PreMath!

[Nayeem_]

Sir I have a prove that type question of Recurrence Relation that I am unable to solve, can u solve it and make a video on it. Question is :
Given,
F1 = 0
F2 = 1
F3 = 3
F4 = 4
Fn = Fn-1 + Fn-3 + Fn-4, for any natural number n >= 5
For example, F5 = F5-1 + F5-3 + F5-4 = F4 + F2 + F1 = 4 + 1 + 0 = 5
Prove that sqrt(F2p) = sqrt(F2p-2) + sqrt(F2p-4) for any natural number p >=3 .
Can u please prove this, u can verify the question by checking the values till around F16, they will always follow this pattern.

[tombufford]

At a quick glance: Angle DAO is (180-112)/2=34. from Central angle theorem, angle DBA = 112/2 = 56. Angle POD = 180-112 = 68, DPC = 112. Thales theorem, ABC =90, PBC = 90-56 =34 and BPC, BCP = (180 -34)/2= 73. BPA = 180 -73 = 107. x= 180-107-56= 17 degrees.

[imonkalyanbarua]

A very interesting question sir and very well solved! 😀👏👏👏 Wishing you a Very Happy New Year! 💐

[fadimakky]

Well done. angle ABC=90, so angle PCB= angle CPB= 90-X, PBC = 180-(90-X)-(90-X)= 2X.
angle CPB=angle OPD=90-X, and angle DOP=180-112=68, so in triangle DOP angle PDO = 180- (90-X)-68=X+22.
triangle DOB is an isosceles triangle OD=OB radius, so angle OBP =ODP=x+22.
triangle AOB is an isosceles Triangle aswell, OA=OB radius than angle CAB= angle ABO =X.
angle ABC =90= angle ABO+ angle OBP+ angle PBC= X+(22+X)+(2X)=90 SO 4X=68, X=17

[Hebrews.]

I got to x=17 by connecting O with B and working out the angles from those isosceles triangles.

[syedraufurraheemjaffrey]

A very simple circle theorem question, these kind of questions are given in grade 9 when IGCSE student start to learn the circle theorem first time.

[sameerqureshi-khcc]

Happy New Year to our dear professor of mathematics 😊👍

[yuusufliibaan]

Good lesson thank you very much dear teacher

[alexniklas]

Angle COD=180-112=68°
Angle CBD=COD/2= 34°
Angle ACB= (180-34)/2= 73°
Angle x= 90-73= 17°

[rudychan]

PCB = (180-34)°/2 = 90°-17°
A + C* = 90°
A = 90°-C = 90°- (90°-17°) = 17°
Easy• AC is diameter, so B = 90° 🙂

[thembamabuya]

Well explained as usual. Happy new year.

[APUS_NUNN]

Ein gutes neues Jahr und weiterhin viele schöne gute Ideen wünscht
Apus Nunn

[rukaiyashukra]

I tried to solve first before watching the video but I couldn't but just after you said about the central angle theory I solved it answer 17 came out