Can you find Angle X? Justify your answer! | Quick & Simple Explanation
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- Опубликовано: 28 фев 2022
- Learn how to find the unknown angle in the given triangle. Use the straight angle property! Step-by-step tutorial by PreMath.com
#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam 
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Pin me or I won't solve this
You have a great sense of humor😀
I'm sorry I don't get it...would you please enlighten me...I'd love to laugh with you .🙂
45and135
I’m retired and enjoy trying your problems so that I can preserve whatever brain cells are left. For this problem, I discovered what was most likely the most difficult solution. I created equations for line segments AC and CB with point C located at coordinate (0,0). Then I drew a line perpendicular to AB that intersected point C. The point where it intersected AB I labeled as point Z. To simplify the problem as much as possible I set the length for segment DZ as 1 unit. Without going into horrendous detail, I used the line equations to determine that segment CZ was also 1 unit long thus making the angle 45 degrees. Thanks for your challenging videos!
You are very welcome my dear friend.
Glad to hear that!
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Keep it up 😀
Love and prayers from the USA!
I am also a retired person. Often I watch his videos to brush up my memories.
Liked your solution as well 👍
@fevengr keep at it! I'm retired for 15 years and 79 years old. See my simple solution using the law of sines.
Let BD = y and AD = BD = z. Angle BCD = 150-x and angle ACD = x-15. Thus
sin15/y = sin(150-x)/z and sin30/y = sin(150-x)/z. Eliminate y and z to obtain
tanx = (2sin150sin15+sin15)/(2cos150sin15+cos15)=1 and x=45°.
I just did this after spending an hour on the treadmill to get the blood circulating through my brain.
X=45,X=135
Your explanations always make to so obvious (afterwards!). I get cross with myself for so rarely spotting the solution! Thank you.
You are very welcome.
So nice of you.
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You are awesome David 😀
Love and prayers from the USA!
I’ve watched all the way through and absolutely love the way you got to the answer through higher level/simple angles and properties.
With that said, 4:30 is where I’d have pulled a ratio of distance and run a trig function to solve.
I’be downloaded this to show my youngest when he gets to geometry. Great job!
Did not think to this kind of point of view at all...! Thanks !
Nice challenging puzzle and well explained 👍
We can let |AD| = |DB| = 1 unit. Also let |CD| = b units, angle ACD = y, & angle BCD = z. Then y + 15° = x, so y = x - 15° and sin(z) = sin(180° - 30° - x) = sin(x+30°). Now by the Sine Rule, sin(x-15°)/1 = sin(15°)/b & sin(x+30°)/1 = sin(30°)/b. So from the 2nd equation, 1/b = sin(x+30°)/sin(30°). Substituting in the 1st eq., we get sin(x-15°) = sin(15°).sin(x+30°)/ sin(30°). So sin(30°).sin(x-15°) = sin(15°).sin(x+30°). Expanding both sides we get,
sin(30°). [sin(x).cos(15°) - sin(15°).cos(x) ] = sin(15°). [sin(x).cos(30°) + cos(x).sin(30°)]. So [sin(30°).cos(15°) - sin(15°).cos(30°)].sin(x) = [sin(30°).sin(15°) + sin(30°).sin(15°)].cos(x). Thus sin(x)/cos(x) = 2.sin(30°).sin(15°)/sin (30°-15°) = 2.(1/2).sin(15°)/sin (15°) = 1. So tan(x)=1 & hence x=45°. In case, they're needed: y= x-15°= 30°, z= 180°-30°-x =105°, & b= sin(15°)/sin(x-15°) = sin(15°)/sin(30°) = sin(15°)/(1/2) = (√6 - √2)/2. Trig. is king!
.
I do like this problem, and I love your solution. There is a sort of shortcut that makes use of the interaction between a square and an equilateral triangle external to it and sharing a side with it. Obviously, the internal angles of the triangle are each 60°. Less obviously, if T is the vertex of the triangle not touching the square, and line segments are drawn from T to each of the two corners farthest from it, then those line segments subtend an angle of 30°.
Excellent solution! Hats off!👌
I love these as I have never learned any of this at school. My first action was to create a regular quadrilateral starting at point C to include point D as the opposite corner. This gave an angle of 45. I then watched the video to see how to do it correctly. )))
Very nice.... Again recollected school/college time... Geometry.. Thanks sir
Thank you for a great math lesson.
Wow that was amazing!
So creative and cool!
Great leson!
Another great question from PreMath,thank you.
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You are the best Metehan
Keep it up 😀
that is cool thank you.
Beautiful solution! Thanks!
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You are awesome Sergio 😀
Very nice explanation👍
Thanks for sharing🎉
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Keep it up 😀
Very nice solution. Easy to understand. Thank you Sir.
You are very welcome.
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You are awesome Luis😀
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“… and here’s our much nicer looking diagram!”
I love it👍
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You are awesome Tc 😀
very well explained, thanks for sharing
Glad it was helpful!
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Keep it up 😀
I wait for new problem which is more difficult than one. Thaks for all
Great lessons sir!!!
Glad you like them!
Thank you! Cheers! 😀
Wow 😳 sir ❤️🙏🙏🙏🙏 superb creation 🙏🙏🙏❤️
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You are awesome 😀
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this is really good ! good job sir !
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You are the best Haofeng
Keep it up 😀
Thanks for video. Good luck!!!!!!!!
Thanks for watching!
You are very welcome.
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Simple solution using law of sines. Let BD = y and AD = BD = z. Angle BCD = 150-x and angle ACD = x-15. Thus
sin15/y = sin(150-x)/z and sin30/y = sin(150-x)/z. Eliminate y and z to obtain
tanx = (2sin150sin15+sin15)/(2cos150sin15+cos15)=1 and x=45°.
What tools are you using to make this video? Could you please tell me?
suy luận tìm góc rất tuyệt.
angle en CAB on a les deux angles de la base du triangle abc et la somme des angles est de 180° angle y cab = 180 - 15 - 30° = 180 - 45 = 135° comme AD = DB l'angle CDB = CAB / 2 = 135/2 et x + 135°/2 + 30° = 180° => x =
Can u tell me answer of the question in which interior angle is x and other two r only exterior angles y and z and i need to find the value of angle y which is exterior. No specific measurements are given...pl help ur explanation r very nice so i need ur help
Your step by step explanation plus your construction part is simply superb. Thank you alot
You are very welcome.
So nice of you.
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You are awesome Niru dear 😀
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@@PreMath 🙏🙏
Sir , have you created this question your own ??
Great solution
Glad to hear that!
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You are awesome Gurwindar😀
Can you find x on law sine ?
Nice explanation 🙂
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Keep it up Mahalakshmi 😀
Thank you
👏👏👏👏👏 👏
You're welcome 😊
You are awesome Algeria 😀
Nice solution 👍 I have another one that is the same approach as yours.
I doubled the two angles ∠A and ∠B, so that I make a right triangle △ABF, whose ∠A = 30° and ∠B = 60°.
Next we have 2 congruent triangles *△BCD ≅ △BCF* (because of BD = BF; ∠CBD = ∠CBF = 30° and they have the common side BC).
The point C has to be the center of the inscribed circle of *△ABF* (intersection of two bisectors AC and BC), then we have ∠BFC = 90°/2 = 45°.
Finally the value of x is equal to ∠BFC = 45°
Excellent!
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You are awesome 😀
Excellent
Thank you so much 😀 Cheers!
Keep it up John 😀
45 and 135
good. aproa.
but we can solve this by using quadrilateral , parallagram properties
The best
Looked easy at first then realized it was a nightmare to solve 😂
Solved very easily sir
Good job Vats
I have to admit I couldn‘t solve it - but the approach was very creative and incorporated many relevant aspects of triangles. Very clever! Liked it a lot!
Glad it helped!
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Keep it up Philip 😀
Thnku
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You are awesome Pranav 😀
45 degree
Very good question
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Keep it up 😀
A good task. Just how to learn how to make the right additional constructions?👍
Glad it was helpful!
As stated in the video, I knew that one of the angle is 30, so we'd construct a 30-60-90 triangle and then go with the flow. In mathematics, practice and perseverance are name of the game!
Keep it up 😀
Thinks
Very tough question !
Nice solution. I first solved it using trigonometry. Then I did with circumscribed circle. Will post solutions on my channel shortly.
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You are awesome 😀
Link to geometric solution: ruclips.net/video/usE2jUsXrOQ/видео.html
Link to trigonometric solution: ruclips.net/video/kihsNLXUM9M/видео.html
Good very good🌹🌹🙏🙏
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Keep it up Engineer 😀
But sir it can be done from another method
Since AD = DB it means CD is the median of the given ∆ABC if CD is the median then it must bisect ∠ACB.
now ..... ∠ACB + ∠CAB + ∠CBA = 180° (ASP of ∆'s)
=>. From here we get ∠ACB as 135°
As CD bisects ∠ACB it means ∠DCB = ∠DCA = ½∠ACB or ∠DCB =∠DCA = 67.5°
Now in ∆DCB ...
∠DCB + ∠CBD + ∠BDC = 180° (ASP of ∆'s)
From here we get ∠BDC = x = 82.5°
Hence x = 82.5°
Sir if I am wrong then please correct me 🙏🙏
Median jabse bisect karne lag gyi
X = 45°
good
I solved it using exyerior angle : 180 - 15 =165 ; 180 - 30= 150. And then algebra. Thanks for the lessons. Ciao
Che procedimento hai fatto con l'algebra?
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You are awesome
Keep it up DR 😀
@@matteoieluzzi4133 Oh, I've been around the world, but I have a mitigating factor. At university I studied Ancient Languages and only a few months ago I started repeating math. I find these lessons relaxing. The prof is very clear in the explanations and carries out all the steps.
150 =x +beta (opposite angles);
165 =delta +gamma (opposite angles);
(delta + x) - (x+beta) =180-150 =135=delta - beta;
If delta +beta =165 and delta - beta =30, gamma - beta =165-30=135;
180- 135= 45.
I think your answer is incomplete i cant see how you can solve it by just using exterior angles
Significiant solvement
Hello,☺
The laws of physic determine if there is a triangle with 30°, 15°, and (180-15+30=135)135°, angle X always remains zero, 0.
Simply just Put AD/DC = BD/DC = Sine of Opposite angles
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Keep it up Pankaj 😀
X=45
Sin15/sinx=sin30/sin(135-x). If you expand no calculator needed x=45 finished.
Cool
شكرا لكم
لدينا حلان X=45،X=135
X=30
The shading was helpful to focus attention on a particular part of the diagram at the 03:20 time stamp and elsewhere.
I dont think anyone can solve this as so many trick reconstructions. Any easier solutions without reconstruction
This is not the geometry problem exactly. It is triangle law problem. So , three construction and isoceles accidently doesn't work for any other problem like this. For this type of problem to solve by geometry, we have to see the angle relation like 30/2=15, 30+15=45*2=90. 15*4=60+30=90. like this type and then eventually we have to construct. It becomes quite complex. So triangle law is the best solution.
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Keep it up 😀
🖤
X== 50
Euler is smiling down on you Sir...but if he asks you in your dreams "what's the weather like down there?", and you reply "it's two o'clock "...he's gonna know something isn't quite right. 🙂.
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You are the best
Keep it up 😀
Whoa! It looks so complicated 😆😆😆
Yes! It's quite challenging.
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You are awesome 😀
"Quick & Simple Explanation" is very funny. Especially quick...
या तो सवाल आसान हो रहे हैं या मेरा गणित में दिमाग😍😍
Dear Dheeraj,
तुम बहुत होशियार और बुद्धिमान भी हो। इसे जारी रखो।👍
@@PreMath आपको देखकर सीख रहा हूँ मास्टरजी🙏🙏
Basta impostare l'equazione del teorema dei seni sui 2 triangoli, dove si semplificano i lati uguali e rimane x come incognita... Sono tutti uguali questi problemi ah ah
يا ليت الشرح باللغه العربيه حتى نتابعك
Funny but without a pen and paper it gets solved instantly using 2 tans, a dropped down RA has a 3rd tan of h/h, done. Providing that .999999999999 is a good enough approximation for one h.
"How _accurate_ does it …….. "?
What the hell this construction is
How yo think this bro
Can you please tell
It was impossible to solve angle x
It is not an easy one.
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You are awesome Mmatlou 😀
Extremely difficult problem
Then third angle should be 180-30-45=105. Not seems like this
I got the radius of the small circle correctly (1 and 25 as well), however, forgot to calculate the areas... :(
:)!!!
I played with cosine law in triangle ABC
to calculate length of |AC| and |BC| in terms of length BD
I played with cosine law in triangle BCD
to calculate length length CD in terms of length BD
Once more cosine law in triangle BCD to calculate cos(x)
this time length of BD will cancel
OK, your derivation is definitely BEAUTIFUL geometry! I however found it a trigonometric way.
Let
𝒉 = height of triangle.
𝒎 = length of one of the two congruent segments.
(𝒎 + 𝒂) is segment to left
(𝒎 - 𝒂) is segment to right
Seeing that the right triangle is a 30-60-90, then
𝒉 = (𝒎 - 𝒂)/√3
𝒉 = (𝒎 - 𝒂)*√⅓
The tangent of 15° = 0.267949 in general. This I will call T₁₅ for simplicity.
T₁₅(𝒎 + 𝒂) = (𝒎 - 𝒂)√⅓
Léts expand and flip the 𝒎's and 𝒂's around to isolate
T₁₅𝒎 + T₁₅𝒂 = √⅓𝒎 - √⅓𝒂 … rearrange
T₁₅𝒂 + √⅓𝒂 = √⅓𝒎 - T₁₅𝒎 … combines to
(T₁₅ + √⅓)𝒂 = (√⅓ - T₁₅)𝒎
Now, arbitrarily set (𝒎 = 1), in order to solve 𝒂
𝒂 = (√⅓ - T₁₅) / (√⅓ + T₁₅)
𝒂 = (0.577350 - 0.267949) / (0.577350 ⊕ 0.267949)
𝒂 = 0.366025
Armed with that, now figure height of triangle
𝒉 = √(⅓)(𝒎 - 𝒂)
𝒉 = 0.577350 (1 - 0.267949)
𝒉 = 0.366025
Well, that's the same as 𝒂, so the inscribed △ has ( height = base ), and thus must be a 45-45-90.
α = 45°
Tada.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
Your presentation is very slow.