Can you calculate the value of angle X? | (Quadrilateral) | #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 calculate the value of angle X? | (Quadrilateral) | #math #maths | #geometry

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

#SolveForX #Angle #Quadrilateral #Triangles #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.

[shashijoshi]

An alternate simpler approach will be: after getting equation 1, consider triangle CFB, and get the equation 2 as,
alpha + 2x = 90 deg.. then solve these two simultaneous equations.🙏🏼

[jmcxzcs]

Put point P on line AB so that
CP bisects the angle C.
Then there triangles ADP DCP
CBP are congruent.
Therefore angle C is 6x and
x is 24

[aminimam]

sin(5x-90)=a/2a=1/2
5x-90=30
x=120/5=24

[waheisel]

Hello, I drew auxiliary line DE so that DE equals AD (and DC and CB) with E on line AB. THen you can show that angle CEB is 2x, therefore triangle CEB is isosceles and so EC=CB (and DE and DC). Thus triangle DEC is equilateral. You can then show that 5x=120.

[Z-eng]

I was able to solve it with sine law and trig lots of manipulations, but creating a triangle around the 4x angle and bisecting it to get 2 congruent triangles with each being congruent to a triangle that can be formed around the non-bisected 2x angle was truly a genius idea.
(Even if you didn't fully imagine what can happen when you go after bisecting the 4x angle, the part about closing 1 isosceles triangle was obvious, just not obvious which one it should be, you bisect the 4x angle which got you a best case scenario of 2 right triangles, from there the next step of closing a right triangle around the 2x angle was obvious, but ending up with a 30-60-90 trianglewas such a beautiful coincidence, truely a beautiful solution for a beautiful puzzle)

[uwelinzbauer]

Difficult for me, I had to watch the video.
The proceeding shown is great 👍
Best wishes!

[mirbekmuratov]

Another way A=90-2x, A=3x-30 then 90-2x=3x-30, x=24

[Eror]

No hablo inglés. Pero explicas tan bien, que te entendí todo . 👍

[wackojacko]

@ 2:35, droppin that perpendicular auxiliary is genius! ...imoh 🙂.

[countysecession]

sin(3x)-sin(7x)-sin(2x)=0
Not sure of any good way to solve that, but it turns out to be correct.

[MeenakshiSinha-dcuu]

Sir after getting that AC = 2CF .then Sin 30 = CF / ACthat is 1/2 we can obtain the equation 30 + alpha = 3x
we can get the 2 nd easiest equation from any of the triangle AED, DEC, Or CFB as every triangle is rectangular triangle hence alpha + 2x will always be 90 .
I think it is more easier to calculate the value of X, with the help of these two equation by replacing alpha The answer will be 5x=120 i.e x = 24

[egillandersson]

It needs only elementary skills, but the way to find how combine them is not easy at all. I didn't find this time. Thank your for this nice puzzle !

[alster]

The aux lines were tricky but the algebra part became easy

[windy]

When you know that angle C is 2x angle A, so C=6x. So A+B+C+D = 2x180 degrés (2 triangles) = 15 x. So x in 24 degrees…

[giuseppemalaguti]

Per il teorema dei seni risulta lato uguale che si semplifica, rimane l'equazione in incognita x..risulta cos5x=(-1/2).x=24

[unknownidentity]

Let's find x:
.
..
...
....


First of all we add point E on AB such that ADE is an isosceles triangle (AD=DE). Then we can conclude:

∠AED = ∠DAE = 3x
⇒ ∠ADE = 180° − ∠DAE − ∠AED = 180° − 3x − 3x = 180° − 6x
⇒ ∠CDE = ∠ADC − ∠ADE = 4x − (180° − 6x) = 4x − 180° + 6x = 10x − 180°

Since AD=DE and AD=CD, the triangle CDE is also an isosceles triangle (CD=DE), so we obtain:

∠DCE = ∠CED = (180° − ∠CDE)/2 = [180° − (10x − 180°)]/2 = (180° − 10x + 180°)/2 = (360° − 10x)/2 = 180° − 5x
⇒ ∠BEC = 180° − ∠AED − ∠CED = 180° − 3x − (180° − 5x) = 180° − 3x − 180° + 5x = 2x = ∠CBE

The triangle BCE is an isosceles triangle as well and we finally obtain:

CE = BC = CD = AD = DE ⇒ CD = CE = DE

Therefore the triangle CED is not only an isosceles triangle, it is an equilateral triangle and we can conclude:

∠DCE = ∠CED = 180° − 5x = 60°
⇒ 120° = 5x
⇒ x = 24°

∠CDE = 10x − 180° = 60°
⇒ 240° = 10x
⇒ x = 24° ✓

Best regards from Germany