Math Olympiad | Find angle X in the triangle | Important Geometry skills explained step by step
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- เผยแพร่เมื่อ 21 ก.ค. 2023
- 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
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Step-by-step tutorial by PreMath.com
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Math Olympiad | Find angle X in the triangle | Important Geometry skills explained step by step
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Because of your videos i am learning more ,so thank you master
Glad to hear that.
Excellent!
Thank you! Cheers! 😀
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.
i use this method
Me too. Double angle formula for Tan and substitution did the trick
me too
Using trig is much faster than whatever op did lol
Of course ! Way more elegant imho...
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°
Excellent!
Thank you! Cheers! 😀
تمرين جيد جميل. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم. تحياتنا لكم من غزة فلسطين .
Thanks dear❤️
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.
I liked your approach. Thanks!
Glad it was helpful!
You are very welcome!
Thank you! Cheers! 😀
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°
Nice way
why isn't 4y the adjacent? how do you know when to label them correctly for SOHCAHTOA
Great explanation👍
Thanks for sharing😊
Glad you liked it
Nice working sir..Every point explained very well..
Thanks and welcome
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. 😅
Excellent!
Thank you! Cheers! 😀
Gostei muito!
Excellent!
So nice of you.
Thank you! Cheers! 😀
In this kind of case, 3:4:5 and 1:2:√5 special right triangles are the usual suspects 🧐
Excellent!
Thank you! Cheers! 😀
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°
SOH CAH TOA ! One of the finest acronyms known to man!
Thank you! Cheers! 😀
x=1/2*arccos3/5 Sir, is this answer correct?
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. 🤠
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
Triangle ADB: tan x=AD:AB=(2y):(4y)=1/2. So, x=arctan (1/2)
Thank you! Cheers! 😀
If AB = a,
TanX = 2y/a
Tan2X = a/3y
TanX*Tan2X = 2/3
= 2Tan²X/(1 - Tan²X)
TanX = 1/2
Thank you! Cheers! 😀
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)
is there any other ways to solve it with only that triangle itself? plz
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
3y = 3y cos 2x or cos 2x=1 or 2x= nπ or x= nπ/2
In the last,we can find x from tanx=2y÷4y=0.5
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
Good evening sir
Hello dear
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)
tan (x) and tan (2x) corelation
x= 30° 2x=60°
x=26.5651°
😀🥂❤️
Excellent!
So nice of you.
Thank you! Cheers! 😀
@@PreMath 👍
sinx=rad(1/5)
Thank you! Cheers! 😀
what is this?
I want someone to help me in maths i have exams really soon 😢
CB=6Y but CE=5Y
I tried another method and I get x=16
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°
dont you feel its way too complex explanation. Its show easy to prove but you made it so tough.
math
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°