Challenge Question: Can you find the Angle X?| Step-by-Step Explanation
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- Опубликовано: 18 фев 2022
- Learn how to find angle x in the given compound triangle by using the exterior angle theorem. Quick and simple explanation by PreMath.com
#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam 
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Thank you for step by step analysis that widen our perspection.
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It can be solved by a easier way...
The triangle is isosceles so to of it's sides that is angle A and angle C =x so 2x +20⁰=180 =>x=80... This is easier
Dang. I love when you use auxiliary lines and move triangles to find compound angles. Very creative problem solving. Great job!
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It is called synthetic geometry, and I love it too! This guy's channel is great, no b.s. and a good problem almost every day.
That “Therefore” is trademark of the house 😃 Great video as usual. Cheers!!!
That's why Premath rock & roll express moves with the speed of light 😊👍
Thank you Sameer! Cheers!
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Thanks drawing the congruent triangle was a great move
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You are awesome Pranav 😀
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if I don't know the answer,how on earth we know we need to draw an identical triangle in order to solve it?
Awesome, many thanks!
Thank you for your detailed and nice solution.
My aproach is a bit different.
1. Step 3: because the triangle CDB is an isosceles one, just draw the bisector DH from D to BC. From C, draw a circle whose radius is equal to CB. By that way this circle cuts DH(extended) at point P and then PC cuts AB at point I. Now we notice that newly formed triangle CPB is an equilateral one---->the angle ICD=IDC= 40 degrees ----> the triangle CID is a isosceles one -----> CI=ID; AI= PI-----> ACDP is an isosceles trapezoid ------> the angle x= the angle CPH=30 degrees.
Answer x=30 degrees
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I did it slightly differently. I flipped the CDB triangle so C was at D and vice versa. That results in an equilateral triangle ABD. Triangle ABC is the same as triangle ADC which leads to 2x=60.
Your solutions are amazing!
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Thank you for sharing this informative video!
Another solution (but same approach) is copying the ACD triangle so that CD and BD overlap each other. Therefore we have the new ABC triangle, which is an equilateral triangle. Then by connecting the point D with the new point A, we can see that the new angle A is combined by 2 angles "x" (it can be easily demonstrated).
Btw, nice solution from you 👍
pl put a small fig.not clear for me.
I figured out that CDB is an isosceles triangle and that DB=CD, but I didn‘t see the trick to attach the triangle to the lower part of the triangle ADC. Once I saw this step in the video, it became clear that the new quadralateral shape is composed of an equilateral triangle and an isosceles triangle, with angles 50-50-80. Thus x = 30 degrees!
A very tricky problem and a very creative solution!
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wrong answer do sum of triangle for check the answer
sum of triangle 180
Great question with a sweet solution!
Beautifully explained!
Awesome👍
Thanks for sharing😊
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I found that if I use "The Law of Sines", x is about 26.343 degree."Out of box" is really a better idea for avoiding deviation !!!
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You have some error in your calculations. Law of sines + Pythagorean => arccot(v3) = 30°.
x=70 not 30
Beautiful explanation sir really i like it
Beautiful problem!!
V nice tips ofsolving this multiple step question l had to watch it two times to understand it but it turned out veasy brilliantly explained
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4:21 After getting A,C and D are on the circumference, we can conclude ∠CAD = 60°/2 = 30°.
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Thanks for video. Good luck!!!!
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Mathematics just a good activity as your explanation is encouraging
Thank you for sharing this math🇧🇩
Let's draw an equilateral triangle DBE and connect C and E. Then ∠CBE is equal to 60-20=40 degr., ∠DEB=60 d., ∠EDB also 60d. ∠CDE is equal to 180-60-40=80 d. Notice that triangle CDE is isosceles since ED=DB=CD. We get ∠DEC=∠DCE=(180-80)/2=50 d. and ∠ECB=50-20=30 d. Now lets concider triangles ADC and CBE. AD=CB, DC=BE and ∠ADC=∠CBE=40 d. Thus, tr. ADC= tr. CBE and ∠CAD=∠ECB=30d.
Nice video, well presented
Wow thank you this so Amajing
Awesome! 🌸💗
Very nice solution. Thank you.
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Extremely beautiful ❤️
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I somehow manage to complete this question and get the answer in my head 😮
من المغرب شكرا لكم على المجهودات
نستعمل علاقة sin في المثلثACD وصيغة cos3x بدلالة cosx نجد sinx=0,5 ايx=30°
Amazing
So nice of you.
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This is why trigonometry was developed! :)
Really very nice solution! it is surprising how, after a series of intractable angles, an equilateral triangle magically emerges, which is the right key to solve the problem! Congratulations!
Sir namaskar..I find your videos very exciting..if you could make some videos on trigonometry std 9 and coordinate geometry,slope etc. Thank you sir.....
Please make more this kind of hard geometry problems
well explained!
x=30 answer
let label CB as 4 then for the isosceles triangle, 20, 20, 140 we have an Angle Side Ange with sides 2.124, 2.124 and 4. Hence the large
triangle ABC side is 6.124 , angle 20 degrees, and side 4, thus a Side Angle Side. Hence angles are 130, 20, and 30. x=30 answer
Thanks sir
Great method
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Thank u soooo much mister please which program u r useing for explaning the problems??
شكرا كثيرا على الحل كما يقول رسول الإسلام :( من لم يشكر الناس لم يشكر الله ) و شكرا مرة أخرى
well explain
Будет проще, если построить правильный треугольник, переложив АСД, поменяв местами точки С и Д. Тогда А1Д будет серединным перпендикуляром в правильном треугольнике и х=60:2=30.
Лихо!
Very nice.
After getting angle of 80°
80° is a vertex of isosceles triangle
So x+20=180-80/2=50
So x=30°
Awesome
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Let a = AD and b = DB. Then, BC = a and DC = b. *Now use the sine law of Trigonometry on 2 triangles to get 2 independent equations in a and b*
sin(140-x) / sin x = a / b = sin(140)/sin(20) = sin(180-140) / sin(20) = sin 40 / sin 20 = 2 sin 20 cos 20 / sin 20 = 2 cos 20
So, 2 cos 20 = sin (140-x) / sin x = sin (40 + x)/sin x = sin 40 cos x / sin x + cos 40, which implies cot x = (2 cos 20 - cos 40) / sin 40 = csc 20 - cot 40 = √3 = cot 30
Therefore, x = 30 deg.
Is this the kind of problem that works well only for certain values of angle measurements but would be a mess if the given angles were not 40 and 20, but let's say 37 and 23 degrees? I solved the problem with the Law of Sines and some manipulations with basic trig identities. My method did not involve much mess, and it didn't require any auxiliary lines, so it is pretty simple, but it still wouldn't work if triangle CBT wasn't isosceles. So, in summary, I think that the author's method only works for these particular angle measurements, while mine is good for more generic situation as long as the first angle is twice the second. Am I correct? Any corrections or comments would be greatly appreciated.
Excellent
Brilliant
Excelentes
very well explained, thanks for sharing this compound triangle problem
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D is the Circumcentre of Tr. ABC. So x = < CDB /2 = 60/2 = 30
Nice🙏🙏
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Too easy no need for that excessive maths - that’s 20 degrees 👍👨🏭👨🏭
Lol you are wrong
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👏👏👏
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You are awesome, Atiqa 😀Keep it up 👍
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Good grief!I found the triangles,I found the equilateral, but putting the triangle adjacent to the other one was beyond me.I must do a lot more of these problems.
I will try to know clear about this exam .
Angle d is 140i triangle adc angle cis100
Hence angle a is 40
Little construction and then applied SAS similarity to solve this question easily
Very well done Vats
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Lets see my different solution here, using trigonometry and the sine law :
We apply the sine law in the triangle CAD : (*) CD/sinx = CA/sin40 = AD/sin(ACD) and since the angle ACD is 140-x we can write the last equality as AD/sin(140-x)=AD/sin(40+x)
Now we apply the sine law to the triangle CBD : (**) CD/sin20 = CB/sin140=BD/sin20 .
Now we take the first and the third ratio of the (*) and the first and the second ratio of the (**) and divide them by parts and we get : sin20/sinx = sin140/sin(40+x) . But sin140=sin40 so finally we get : sin20/sinx=sin40/sin(40+x) which is the same analogy as sinx/sin(40+x) = sin20/sin40 = 1/2cos20 . This is equivalent to : sin(40+x)=2 sinx cos20 .
Using the standard trigonometric indentities of the sine of a sum, we get easily that tanx=sin40/(2cos20-cos40) . Here is the tricky part : we calculate the sin and cos of 20 and 40 using the angle of 30 degrees. We have that tan(x)=(sin30*cos10+sin10*cos30)/(cos30*cos10+3*sin30*sin10). Here we divide both the numerator and the denominator by cos30*cos10 , hence we get that : tan(x)=(tan30+tan10)/(1+3*tan30*tan10)=(sqrt(3)/3+tan10)/(1+3*sqrt(3)/3*tan10)=(sqrt(3)+3*tan10)/(3+3*sqrt(3)*tan10)=1/sqrt(3)=tan30. Hence x=30 .
please professor what program u use to explan the lessons??
So what it comes down too, is not the value of X but the relationship of X in the transformational ether...,that's what I'm thinking. 🙂
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Let BC be 2a, then DB=CD=a/cos 20, CE=CDsin 40=a sin 40/cos 20=2a sin 20, AE=2a-CDcos 40=2a-a cos 40/cos 20=a(2-cos 40/cos 20), tan x=2sin 20/(2-cos 40/cos 20), x=30.😮
97.3°
Damn!! Nice problem
x=arctg(sin40/(2*cos20-cos40))
I don't know why I get such a rush solving this type of problems in my head
Did this myself
.
Excellent!
You are awesome Ryuga 😀
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What approach should we follow in these type of questions???
#Plz-reply.
These are trick questions. And most of these require making magic isosceles and equilateral triangle.
Dear Manoj, these problems little different. So we have to little creative in solving these problems. Please keep watching and you'll get the feelings how to handle such problems. Practice, perseverance, and experience are the essence!
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Triangle ACD has been proved to be an isosceles triangle with sides AD =CD and therefore angleCAD=angleDCA=x+20. because angleCDA=80°, x+20 must be 50°hence x=30°.to me it is a better explanation.
Side AD is NOT equal to side CD
The given is
AD = BC
If CD = AD
Then CD = CB, making CDB an Isosceles, with sides CD & CB & base DB.
But side angles are not equal, therefore it is not an Isosceles.
The triangle BCD is Isosceles, BUT with sides CD & BD & Base CB.
Angle CDB, by inspection, is 180-40 = 140
Therefore,
Angle BCD, by inspection, is (140+20+x=180) x=20
Therefore, side angles are equal.
Another ‘angle’ 😎
Your given is
X=30
ADC=40
Therefore
DCA=110
Therefore the three angles are different
Therefore Triangle ACD is NOT isosceles
AD is equal to BC? what does that mean?
I cheated. I just drew a perpendicular from point C to the line segment AB (point E) and then used the trig functions of sin( angle ) = opp/hyp and tan (angle ) = opp/adj. I also randomly picked a number like BC = AD = 10. CE = 10 * sin (20) = approximately 3.420201. Using the trig functions and the value of 10 for AD, I was able to use a calculator to get the values of BE = 4.076037, and AE = 10 - 4.076031 = 5.923963. tan x = CE/AE and then x = 30.
I wonder why it you called it cheating. Seems ok to me.
@@gol-r5836 I needed the calculator.
No worries Mark
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It looks like ray optics theorem,I think it's snells law.
nice problem
120
Suggestions is wroung
80
X 20 degree
30
This sum is no where near as easy as it looks and I got the solution using trigonometry
I have a general solution. Be the unknown angle = x. Solution tan x = tan 40 * sin 20/(tan 40-sin 20). Let's make the angle 20 = a, 40 = b.
General solution: tan x = tan b*sin a/(tan b-sin a), so x = arctan(tan b * sin a / (tan b - sin a)). This is for at any angle such as 21 & 42, 19.9 & 39.8
I am 70 years old, hold a Chemistry Ph.D, but I have studied Physics and Math for my private hobby.
Sir can you please tell me what this identity called?
↓
1/n(n+1) = 1/n - 1/n+1
I didn't understand your question! It has been decomposed into two fractions (Partial Fraction Decomposition). It's a telescoping series that converges to 1. All the best!
@@PreMath Thank you Sir I just needed that term “Partial Fraction Decomposition”
@@experimentingalgorithm1546 My pleasure!
You are the best. Keep iy up dear
100 degree
Super question. But this time I have failed.
No worries Mustafiz
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How are you supposed to figure that out?
We have to think out-side-the-box to solve these challenging problems! Thanks for asking.
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@@PreMath thx
Most students will not copy the isosceles triangle to the right over to be under the other triangle...
Un système d'équations 3 avec trois inconnus et c'est tout
There was no construction needed here
I did it in mind calculations
Maybe you should have mentioned at the start that A D B is a straight line? (
Из-за такого обозначения можно запутаться
Screen goes blank!! Can’t follow!
No But x=80 degrees
I doubt if the question is correctc
you needed 8 min. to fid it out ? I needed 3 secs. 🙂
40° + 20° + X = 90° . . . . . 30° !
?
With trigonometry and a calculator it is being solved much easier. No living being can invent this auxiliary triangle, if this being does not know the answer beforehand. Thus, these auxiliaries are COMPLETELY USELESS, long live trigonometry!