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

[@MateusmabialaMathews-bx8pc]

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

[@PreMath]

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

[@j.r.1210]

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]

i use this method

[@bekaluu1]

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

[@soli9mana-soli4953]

me too

[@duckymomo7935]

Using trig is much faster than whatever op did lol

[@abcdefgq1816]

Of course ! Way more elegant imho...

[@murdock5537]

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]

Excellent! Thank you! Cheers! 😀

[@ghhdcdvv5069]

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

[@PreMath]

Thanks dear❤️

[@jimlocke9320]

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.

[@SarvarbekB-bd8qv]

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

[@AnishJha-ds3ih]

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°

[@SuperMtheory]

I liked your approach. Thanks!

[@PreMath]

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

[@misterenter-iz7rz]

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]

Excellent! Thank you! Cheers! 😀

[@khalidayubi01]

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

[@PreMath]

Thanks and welcome

[@slomojohnjoshi5990]

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°

[@ybodoN]

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

[@PreMath]

Excellent! Thank you! Cheers! 😀

[@wackojacko3962]

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

[@PreMath]

Thank you! Cheers! 😀

[@harikatragadda]

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

[@PreMath]

Thank you! Cheers! 😀

[@HappyFamilyOnline]

Great explanation👍 Thanks for sharing😊

[@PreMath]

Glad you liked it

[@williamwingo4740]

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. 🤠

[@ИванПоташов-о8ю]

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

[@PreMath]

Thank you! Cheers! 😀

[@businesswalks8301]

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

[@Jop_pop]

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)

[@ravenheartFF]

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

[@Samir-c8y]

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

[@soniamariadasilveira7003]

Gostei muito!

[@PreMath]

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

[@syedmdabid7191]

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

[@honestadministrator]

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)

[@skywang-io5cc]

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

[@Abby-hi4sf]

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

[@rajendraameta7993]

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

[@Ahmed-mn1ri]

Nice way

[@MarieAnne.]

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°

[@pralhadraochavan5179]

Good evening sir

[@PreMath]

Hello dear

[@giuseppemalaguti435]

sinx=rad(1/5)

[@PreMath]

Thank you! Cheers! 😀

[@andrewskipwith9401]

what is this?

[@andirijal9033]

tan (x) and tan (2x) corelation

[@anestismoutafidis4575]

x= 30° 2x=60°

[@jamesrocket5616]

x=26.5651°

[@mahouddeye8923]

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

[@quigonkenny]

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°

[@neverythingk3270]

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

[@mohammedtarteer9439]

CB=6Y but CE=5Y

[@boudjemaabouattou1546]

I tried another method and I get x=16

[@bigm383]

😀🥂❤️

[@PreMath]

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

[@bigm383]

@@PreMath 👍

[@racquelsabesaje4562]

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