Can you find the angle X? | (Justify your answer) | #math #maths | #geometry

Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!

Can you find the angle X? | (Justify your answer) | #math #maths | #geometry

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

#SolveForX #Angle #Angles #ExteriorAngleTheorem #Quadrilateral #GeometryMath #PythagoreanTheorem #MathOlympiad #CongruentTriangles #ThalesTheorem #RightTriangle #RightTriangles #SimilarTriangles #EquilateralTriangles
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CollegeEntranceExam
#blackpenredpen #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #Angles #Height #ComplementaryAngles #TwoTangentTheorem
#MathematicalOlympiad #OlympiadMathematics #CompetitiveExams #CompetitiveExam
#IsoscelesTriangles
How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
Pre Math
Olympiad Mathematics
Two Methods to Solve System of Exponential of Equations
Olympiad Question
Find Area of the Shaded Triangle in a Rectangle
Geometry
Geometry math
Geometry skills
Right triangles
imo
Competitive Exams
Competitive Exam
Find the length m
Pythagorean Theorem
Right triangles
Similar triangles
Congruent Triangles
Exterior angle theorem
Isosceles triangles
coolmath
my maths
mathpapa
mymaths
cymath
sumdog
multiplication
ixl math
deltamath
reflex math
math genie
math way
math for fun

Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

[SkinnerRobot]

You make it look so easy. Thank you, PreMath.

[engralsaffar]

Angle CEB=135 therfore angle B=30 degrees.
Then, connect a line from D to O perpendicular to CB, and draw a cemicircle to with D, O and B at its circumference, from the circle theorem this makes DE, BE and OE radii in the circle.
Angle ODE=60 degrees (30-60-90), and angle DOE is also 60, therefore triangle DOE is an equilateral triangle, and angle OED=60. Therefore angle OEC=15, and OCE is an isosceles triangle with OC=OE, and it is also equal to OD, therefore ODC is also an isosceles triangle with a 90 angle,

Therefore the bases are equal to 45 degrees, therefore x+15=45
x=30 degrees

[jimlocke]

As I noted yesterday, the 15°-75°-90° triangle, while not considered "special" in geometry, appears quite frequently in problems. Since we are allowed to use our notes when solving these problems, I recommend that everyone add the properties of the 15°-75°-90° triangle to their notes (or just memorize the properties)!

Ratio of sides short:long:hypotenuse (√3 - 1):(√3 + 1):2√2
Ratio short/long = (√3 - 1)/(√3 + 1) = (2 - √3)/1
Ratio long/short = (√3 + 1)/(√3 - 1) = (2 + √3)/1

In this problem, we see AD having length (2 - √3) and AC length 1, so the ratio AD/AC = (2 - √3)/1 and ΔDAC is a 15°-75°-90° right triangle. The short side is opposite the 15° angle.

[quigonkenny]

As ∠DEC is an exterior angle to ∆CEB at E, ∠DEC = ∠BCE+∠EBC.

∠DEC = ∠BCE + ∠EBC
45° = 15° + ∠EBC
∠EBC = 45° - 15° = 30°

Draw DF, where F is the point on BC where DF and BC are perpendicular. As ∠DBF = 30° and ∠BFD = 90°, ∠FDB must equal 60° and ∆BFD is a 30-60-90 special right triangle. If DF = a, then DB = 2a and BF = √3a.

As DB = 2a, DE = EB = a. Draw EF. As DF = DE = a and ∠FDE = 60°, then ∆FDE is an equilateral triangle, ∠DEF = ∠EFD = 60°, and EF = DF = DE = a.

As ∠DEF = 60° and ∠DEC = 45°, ∠CEF = 60°-45° = 15°. As ∠FCE = 15° as well, ∆EFC is an isosceles triangle and FC = EF = a.

As FC = DF = a and ∠DFC = 90°, then ∆DFC is an isosceles right triangle, and ∠CDF = ∠FCD = (180°-90°)/2 = 45°. As ∠FCD = 45° and ∠FCE = 15°, ∠ECD = x = 45°-15° = 30°.

[uwelinzbauer]

Hallo Professor,
I exactly used the same approach as shown in the video.
Thanks for the interesting geometric puzzle!
I wish a happy 😊 weekend to you and the channel visitors!

[solimana-soli]

since this problem is substantially equivalent to the one previously posted, without doing any calculations I would say that x = 30°

[alexj]

Make point O on AB such as EO=EB. Angle EOB=EBO=30. Therefore angle OEB=30. Therefore OC=EO=DE. Connect O with D, we’ve got equilateral triangle DOE. therefore DOB=90. DO=CO and COD=90. DO=CO, DCO=45 therefore X=45-15=30

[magamoodley]

Use midpoint and parallel lines Mid point theorem. Try neat solution

[PrithwirajSen-njqq]

BE=DE= m AD=n
AB= 2m +n
AC =AE = m +n (as 🔺 ACE is 90-45-45 triangle )
Ang ADC =45+x
tan (45+x)=AC /AD =(m+n)/n=m/n. +1
Ang CBD=45-15=30 degs
tan 30=AC/AB= 1/√3 = (m+n)/(2m+n)
m√3 -2m= n-n√3
m(√3-2)=n(1-√3)
m/n=(1-√3)/(√3-2)
Hence tan (45+x)
=[(1-√3)/(√3-2)] +1
= 2+√3= tan 75 deg
Then x =75-45=30 degs
Note -- angles measured in degrees in all cases

[CliffordMorris-lslc]

Basic angles if a triangle equals 180 was all you needed to see ACB was 60degrees and ACE was 45 degrees

[fantasticyt]

As DE=EB
Considering triangle BCD,
We can say that CE is the median of triangle BCD
Hence according to triangle laws,
x=2 x 15
= 30 degrees

[marcelowanderleycorreia]

Drop a perpendicular line from point D to line BC - DP lenght. Lenght CP = a, since CPB is a special 30 - 90 - 60, PD is also = a, hence x + 15 must be iqual to 45 degrees, x = 45 - 15 = 30 degrees. I spent all day to view this...

[cyruschang]

The big triangle is a 30° - 60° - 90° triangle with the base (y + x + x) and height (x + y)
2x + y = (x + y)✓3
(2 - ✓3)x = (✓3 - 1)y
x = (✓3 - 1)y/(2 - ✓3) = (✓3 + 1)y
(x + y)/y = ✓3 + 2
y/(x + y) = 2 - ✓3
x = 60° - 15° - arctan(2 - ✓3) = 45° - arctan(2 - ✓3)

[NinaDolgan]

This supplement has made my nervous system function more steadily.

[marcgriselhubert]

Let's use an orthonormal center A and first axis (AB). We have A(0; 0) E(c; 0) C(0; c) with c = AE =AC
The equation of (CB) is y - c = -tan(30°).x or y = (-sqrt(3)/3).x. The intersection of (CB) with (AB) is then B((sqrt(3).c; 0) and EB = (sqrt(3) -1).c
As EB = DE, we then have DE = (sqrt(3) -1).c and D((2 -sqrt(3)).c; 0)
Now we have VectorCD((2 -sqrt(3)).c; -c) and CD^2 = (c^2).(4+3-4.sqrt(3) +1) = (c^2).(8 -4.sqrt(3)) and CD = c.sqrt(2).(sqrt(3) -1)
Also we have VectorCE(c; -c) and CE^2 = 2.(c^2) and CE = c.sqrt(2).
In triangle CDE we have DE^2 = CD^2 + CE^2 - 2.CD.CE.cos(x), so: (c^2).(3+1-2.sqrt(3)) = (c^2).(8 -4.sqrt(3)) + 2.(c^2) -2.c.sqrt(2).(sqrt(3) -1).sqrt(2).c.cos(x)
We simplify: 4.(sqrt(3) -1).cos(x) = 6 -2.sqrt(3) or cos(x) = (3 -sqrt(3))/(2.(sqrt(3) -1)) = sqrt(3)/2, which gives that x = 30°
(In fact the length c has no effect, we could have choosen c = 1 at the beginning without changing the generality of the problem.
This problem is quite similar to the one from yesterday)

[imetroangola]

*Solução:*

Seja DE=BE=a. Além disso, o ângulo B= 45° - 15°= 30°. Aplicando a Lei dos senos no ∆CED:

sen x/DE = sen 45°/CD

*sen x/a = sen 45°/CD (1)*

Aplicando a Lei dos senos no ∆CDB:

DB/sen (x + 15°) = CD/sen B

*2a/sen (x + 15°) = CD/sen 30° (2)*

Multiplica (1) e (2):

2 sen x/sen (x + 15°) = sen 45°/sen 30°

sen x/sen (x + 15°) = (sen 45°/sen 30°)×1/2

sen x/sen (x + 15°)=(sen 45°/sen 30°)×sen 30°

sen x/sen (x + 15°) = sen 45°

sen x/sen (x + 15°) = 1/√2

√2 sen x = sen (x + 15°)

Facilmente, *x=30°* satisfaz está última expressão.

[michaeldoerr]

The answer x = 30°. I think that this application is very similar to one of Math Booster favorite constructions. And furthermore I think that this easy enough process is ONLY applicable to 30°-60°-90° triangles.

[giuseppemalaguti]

t=CE, a=DE..t/sin(135-x)=a/sinx...t/sin30=a/sin15...divido rimane

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال
DB=1
AD=a
ACD=y
x+y=45

tany=tan15
y=15
x=30

[黃羅賓]

It has a easier way to solve this problem, make a line from E to CB