Math Olympiad | Find angle X in the triangle | Important Geometry skills explained step by step

Подписаться 405 тыс.

Просмотров 48 тыс.

50% 1

Learn how to find the angle X in the triangle. Important Geometry and Algebra skills are also explained: Pythagorean Theorem, Congruent Triangles Theorem; Isosceles triangles. Step-by-step tutorial by PreMath.com
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!
Step-by-step tutorial by PreMath.com
• Math Olympiad | Find a...
Need help with solving this Math Olympiad Question? You're in the right place!
I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at
/ premath
Math Olympiad | Find angle X in the triangle | Important Geometry skills explained step by step
Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!
#FindAngleX #GeometryMath #PythagoreanTheorem #CongruentTriangles
#MathOlympiad #ThalesTheorem #RightTriangle #RightTriangles #SimilarTriangles
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox? #FillInTheBoxes #GeometryMath #Geometry #RightTriangles
#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 height h
Pythagorean Theorem
Right triangles
Similar triangles
Congruent Triangles
Two Tangent Theorem
Isosceles triangles
Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

Хобби

Опубликовано:

21 июл 2023

Ссылка:

Скачать:

Готовим ссылку...

Добавить в:

Мой плейлист

Посмотреть позже

Комментарии : 64

@MateusmabialaMathews-bx8pc 11 месяцев назад

Because of your videos i am learning more ,so thank you master

@PreMath 11 месяцев назад

Glad to hear that. Excellent! Thank you! Cheers! 😀

@j.r.1210 11 месяцев назад

My method: Let z = base of triangle. Then tan x = 2y/z and tan 2x = z/3y. Use the double angle formula to convert tan 2x to [2(2y/z]/[1 - (2y/z)^2]. Equate all this to z/3y, and simplify. Eventually, you get 4y = z. Put this back in my first equation to get tan x = 2y/4y = 1/2. So x = arctan 1/2 = 26.565.

@anthonycheng1765 11 месяцев назад

i use this method

@bekaluu1 11 месяцев назад

Me too. Double angle formula for Tan and substitution did the trick

@soli9mana-soli4953 11 месяцев назад

me too

@duckymomo7935 9 месяцев назад

Using trig is much faster than whatever op did lol

@abcdefgq1816 9 месяцев назад

Of course ! Way more elegant imho...

@murdock5537 11 месяцев назад

Nice, many thanks! AB = k; tan⁡(x) = 2y/k tan⁡(2x) = k/3y = (2tan⁡(x))/(1 - tan^2(x)) = (4y/k)/(1/k^2 )(k^2 - 4y^2) = 4yk/(k^2 - 4y^2 ) → 16y^2 = k^2 → k = 4y → BD = 2y√5 sin⁡(x) = AD/BD = √5/5 → x ≈ 26,565°

@PreMath 11 месяцев назад

Excellent! Thank you! Cheers! 😀

@jimlocke9320 11 месяцев назад

Well done! Constructing ΔAEB congruent to ΔADB with shared side AB and determining that ΔECB is isosceles is the key to PreMath's streamlined solution to the problem. A minor change at the end: PreMath obtained the value of 2x from trigonometry and divided by 2. You can compute x directly. length AD = 2y and length AB = 4y, so tan(x) = (2y)/(4y) = 0.5. When I compute arctan(0.5) to 2 decimal places, I get x = 26.57°, as PreMath did.

@AnishJha-ds3ih 5 месяцев назад

My method: Side opposite to x = 2y .•. Side opposite to 2x =4y .•. AB = 4y In Triangle DAB Tan theta = Opposite side/Adjacent side .•. tan x = 2y/4y .•. tan x = 1/2 If tan theta = 1/2 then theta = 26.57° .•.x = 26.57°

@ghhdcdvv5069 11 месяцев назад

تمرين جيد جميل. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم. تحياتنا لكم من غزة فلسطين .

@PreMath 11 месяцев назад

Thanks dear❤️

@Ahmed-mn1ri 11 месяцев назад

Nice way

@slomojohnjoshi5990 9 месяцев назад

My method: Let base = h, where h=3y•tan2x Then one hypotenuse will be √(h²+9y²) and another one will be √(h²+4y²). Consider the scalene triangle in the upper part. Area of the triangle is ½•(product of two sides)•(sine of the angle between them). Take two products, that is ½y√(h²+9y²)sin2x and ½y√(h²+4y²)sin(90+x), which give the area of the scalene triangle. Then equate them and and simplify (using 1+tan²θ=sec²θ, quadratic equation, etc). Finally we will get x=½cos-¹(3/5)≈26.57°

@SuperMtheory 11 месяцев назад

I liked your approach. Thanks!

@PreMath 11 месяцев назад

Glad it was helpful! You are very welcome! Thank you! Cheers! 😀

@soniamariadasilveira7003 11 месяцев назад

Gostei muito!

@PreMath 11 месяцев назад

Excellent! So nice of you. Thank you! Cheers! 😀

@harikatragadda 11 месяцев назад

If AB = a, TanX = 2y/a Tan2X = a/3y TanX*Tan2X = 2/3 = 2Tan²X/(1 - Tan²X) TanX = 1/2

@PreMath 11 месяцев назад

Thank you! Cheers! 😀

@syedmdabid7191 4 месяца назад

3y = 3y cos 2x or cos 2x=1 or 2x= nπ or x= nπ/2

@williamwingo4740 11 месяцев назад

Here's a trigonometric solution: Let AB = a, as you did. then tan x = 2y/a and tan 2x = a/3y. But tan 2x = (2tanx)/(1 -- tan^2 x) so (2)(2y/a)/(1 -- 4y^2/a^2) = a/3y; cross-multiplying, taking common denominators, and simplifying (details omitted), we eventually get a = 4y, as you did. Then x = arctan (2y/4y) = arctan (1/2) = 26.57 degrees. Cheers. 🤠

@HappyFamilyOnline 11 месяцев назад

Great explanation👍 Thanks for sharing😊

@PreMath 11 месяцев назад

Glad you liked it

@misterenter-iz7rz 11 месяцев назад

Wow 😮. I attempt, but unsure to solve it. tan x/tan (90-2x)=2/3, or tan x/cot 2x=2/3, tan x tan 2x=2/3, 2tan^2x /1-tan^2=2/3, 6tan^2x=2-2tan^2x, 8tan^2x=2, tan^2x=1/4, tanx=1/2, x=arc tan 1/2=26.565 approximately. 😅

@PreMath 11 месяцев назад

Excellent! Thank you! Cheers! 😀

@khalidayubi01 11 месяцев назад

Nice working sir..Every point explained very well..

@PreMath 11 месяцев назад

Thanks and welcome

@ravenheartFF 11 месяцев назад

This is how I did it: a equals the length of segment AB. 2y/a=tan(x) 2y=a(tan(x)) y=a(tan(x))/2 a/3y=tan(2x) 3y/a=1/tan(2x) 3y=a/tan(2x) y=a/3tan(2x) a(tan(x))/2)=a/3tan(2x) tan(x)/2=1/3tan(2x) tan(x)=2/3tan(2x) tan(x)tan(2x)=2/3 tan(2x)=(2/3)*(1/tan(x)) tan(2x)=(2/3)*(cos(x)/sin(x)) tan(2x)=2cos(x)/3sin(x) sin(2x)/cos(2x)=2cos(x)/3sin(x) 2sin(x)cos(x)/cos(2x)=2cos(x)/3sin(x) sin(x)/cos(2x)=1/3sin(x) sin(x)=cos(2x)/3sin(x) 3sin^2(x)=cos(2x) 5sin^2(x)-2sin^2(x)=cos(2x) 5sin^2(x)-2sin^2(x)=1-2sin^2(x) 5sin^2(x)=1 sin^2(x)=1/5 2sin^2(x)=2/5 -2sin^2(x)=-2/5 1-2sin^2(x)=3/5 cos(2x)=3/5 2x=cos^-1(3/5) x=cos^-1(3/5)/2 x=26.565deg

@Jop_pop 11 месяцев назад

Another solution: Since no length is given, assume y=1. Let z=AB. Then ztan(x)=2 z/tan(2x)=3 Divide these to get tan(x)tan(2x)=2/3 And note that tan(2x)=2tan(x)/(1-tan^2(x)). Let t stand for tan(x). Then 2t^2/(1-t^2)=2/3. So 3t^2=1-t^2, so 4t^2=1 and thus t=0.5 So x=arctan(0.5) because t=tan(x)

@SarvarbekB-bd8qv 6 месяцев назад

x=1/2*arccos3/5 Sir, is this answer correct?

@wackojacko3962 11 месяцев назад

SOH CAH TOA ! One of the finest acronyms known to man!

@PreMath 11 месяцев назад

Thank you! Cheers! 😀

@honestadministrator 9 месяцев назад

AB = AC tan (2 x) AB = AD cot (x) 3 tan ( 2 x) = 2 / tan ( x) 3 tan^2 ( x) = 1 - tan^2 ( x) tan ( x) = 1/2 x = arc tan (1/2)

@rajendraameta7993 11 месяцев назад

In the last,we can find x from tanx=2y÷4y=0.5

@ybodoN 11 месяцев назад

In this kind of case, 3:4:5 and 1:2:√5 special right triangles are the usual suspects 🧐

@PreMath 11 месяцев назад

Excellent! Thank you! Cheers! 😀

@user-qs3tz6hh5g 11 месяцев назад

Triangle ADB: tan x=AD:AB=(2y):(4y)=1/2. So, x=arctan (1/2)

@PreMath 11 месяцев назад

Thank you! Cheers! 😀

@businesswalks8301 11 месяцев назад

why isn't 4y the adjacent? how do you know when to label them correctly for SOHCAHTOA

@user-nm1cp3ex6i 11 месяцев назад

abc est un triangle rectangle égale à 180degre begale à 90 degré a égale à 2x begale à 90 et c égale à x donc:2xplus xplus 90 égale à 180 ce qui veut dire que x égale à 30degre

@andirijal9033 11 месяцев назад

tan (x) and tan (2x) corelation

@pralhadraochavan5179 11 месяцев назад

Good evening sir

@PreMath 11 месяцев назад

Hello dear

@anestismoutafidis4575 11 месяцев назад

x= 30° 2x=60°

@jamesrocket5616 2 месяца назад

x=26.5651°

@MarieAnne. 11 месяцев назад

Here is my solution: In △ABC, tan 2x = AB/AC = AB/(3y) → AB/y = 3 tan 2x In △ABD, tan x = AD/AB = 2y/AB → AB/y = 2 / tan x 3 tan 2x = 2 / tan x 3 * 2 tan x / (1 − tan²x) = 2 / tan x 3 tan x / (1 − tan²x) = 1 / tan x 3 tan²x = 1 − tan²x 4 tan²x = 1 tan²x = 1/4 tan x = 1/2 x ≈ 26.565°

@skywang-io5cc 9 месяцев назад

is there any other ways to solve it with only that triangle itself? plz

@Abby-hi4sf 3 месяца назад

with trig yes, let AB = a; then tan⁡(x) = 2y/a , also tan⁡(2x) = a/3y The Tan(2x ) identity = 2tan⁡(x))/(1 - tan^2(x)) = You can use tan (2x) formula and find the segment AB

@giuseppemalaguti435 11 месяцев назад

sinx=rad(1/5)

@PreMath 11 месяцев назад

Thank you! Cheers! 😀

@andrewskipwith9401 11 месяцев назад

what is this?

@mohammedtarteer9439 11 месяцев назад

CB=6Y but CE=5Y

@bigm383 11 месяцев назад

😀🥂❤️

@PreMath 11 месяцев назад

Excellent! So nice of you. Thank you! Cheers! 😀

@bigm383 11 месяцев назад

@@PreMath 👍

@mahouddeye8923 11 месяцев назад

I want someone to help me in maths i have exams really soon 😢

@racquelsabesaje4562 9 месяцев назад

math

@boudjemaabouattou1546 11 месяцев назад

I tried another method and I get x=16

@neverythingk3270 10 месяцев назад

dont you feel its way too complex explanation. Its show easy to prove but you made it so tough.

@quigonkenny Месяц назад

sin(x) = AD/DB sin(x) = 2y/DB DB = 2y/sin(x) cos(x) = BA/DB BA = DBcos(x) cos(2x) = AC/CB cos(2x) = 3y/CB CB = 3y/cos(2x) sin(2x) = BA/CB sin(2x) = DBcos(x)/CB sin(2x) = (2y/sin(x))cos(x)/CB sin(2x) = 2y/CBtan(x) sin(2x) = 2y/(3y/cos(2x))tan(x) sin(2x) = 2cos(2x)/3tan(x) tan(2x) = 2/3tan(x) tan(x) = 2/3tan(2x) tan(x) = 2(1-tan²(x))/3(2tan(x)) 3tan²(x) = 1 - tan²(x) 4tan²(x) = 1 tan²(x) = 1/4 tan(x) = √(1/4) = 1/2 x = tan⁻¹(1/2) ≈ 26.565°