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!
@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.
This is a very cool question and the added "think outside the box" construction lines are clever! Reminds us to consider completing a 30-60-90 right triangle when we see a length 2k side adjacent to a 30° angle. But I wanted to see if I could find an answer by "thinking inside the box" instead. (Reminder: I'm 53 years old, have no natural talent for math or logic, and my brain is very slow). Here's the answer I came up with: 1. For ease of convenience, label ∠BCD as ∠y and label ∠ACD as ∠z. Also, let's call the length AD=DB=a. 2. By Law of Sines we find sin y/a = sin 30°/DC and sin z/a = sin 15°/DC. 3. Combining these equations produces sin y/sin z = sin 30°/sin 15°. From this we might guess y=30° and z=15° but that's not the correct answer since we know y+z=135° 4. Applying the double angle for sine, we can rewrite the equation as sin y/sin z = 2⋅cos 15° (Note that cos 15° = cos(45° - 30° ) so not too hard to compute with cosine angle difference formula.) 5. Recalling that y+z=135°, the equation can be further rewritten as sin (135°-z)/sin z = 2⋅cos 15°. Applying sine difference formula to the denominator gives (sin 135°⋅cos z - cos 135°⋅sin z)/sin z = 2⋅cos 15°. 6. Now sin 135° = 1/√2 and cos 135° = -1/√2, so doing some algebra on the equation and multiplying both sides by √2 we find (cos z + sin z)/sin z = 2⋅cos 15°⋅√2 7. Separating the left side of the equation and calculating out the right side of the equation gives cot z + 1 = √3 + 1, hence z = arccot(√3) = 30°. From this we can find y=105° and then x=45°. Note that solving for z is easier than solving for y, since the expression one ends up with at the end of an analogous calculation does not obviously resolve to y=105°, unlike arccot(√3) that is more easily seen to be 30°. The corollary that sin 105° ⋅ sin 15° = (sin 30°)² seems surprising at least to me.
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!
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°.
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. )))
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°.
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°
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 ✓ ≡=-⋅
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!
@@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.
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
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 😀
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 🙏🙏
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.
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. 🙂.
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 …….. "?
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
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
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!
Thank you for your feedback! Cheers!
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
Did not think to this kind of point of view at all...! Thanks !
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.
Glad to hear that!
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You are awesome David 😀
Love and prayers from the USA!
This is a very cool question and the added "think outside the box" construction lines are clever! Reminds us to consider completing a 30-60-90 right triangle when we see a length 2k side adjacent to a 30° angle.
But I wanted to see if I could find an answer by "thinking inside the box" instead. (Reminder: I'm 53 years old, have no natural talent for math or logic, and my brain is very slow). Here's the answer I came up with:
1. For ease of convenience, label ∠BCD as ∠y and label ∠ACD as ∠z. Also, let's call the length AD=DB=a.
2. By Law of Sines we find sin y/a = sin 30°/DC and sin z/a = sin 15°/DC.
3. Combining these equations produces sin y/sin z = sin 30°/sin 15°. From this we might guess y=30° and z=15° but that's not the correct answer since we know y+z=135°
4. Applying the double angle for sine, we can rewrite the equation as sin y/sin z = 2⋅cos 15° (Note that cos 15° = cos(45° - 30° ) so not too hard to compute with cosine angle difference formula.)
5. Recalling that y+z=135°, the equation can be further rewritten as sin (135°-z)/sin z = 2⋅cos 15°. Applying sine difference formula to the denominator gives (sin 135°⋅cos z - cos 135°⋅sin z)/sin z = 2⋅cos 15°.
6. Now sin 135° = 1/√2 and cos 135° = -1/√2, so doing some algebra on the equation and multiplying both sides by √2 we find (cos z + sin z)/sin z = 2⋅cos 15°⋅√2
7. Separating the left side of the equation and calculating out the right side of the equation gives cot z + 1 = √3 + 1, hence z = arccot(√3) = 30°. From this we can find y=105° and then x=45°.
Note that solving for z is easier than solving for y, since the expression one ends up with at the end of an analogous calculation does not obviously resolve to y=105°, unlike arccot(√3) that is more easily seen to be 30°. The corollary that sin 105° ⋅ sin 15° = (sin 30°)² seems surprising at least to me.
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!
Looked easy at first then realized it was a nightmare to solve 😂
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°.
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. )))
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°.
Very nice.... Again recollected school/college time... Geometry.. Thanks sir
Very nice solution. Easy to understand. Thank you Sir.
You are very welcome.
Glad to hear that!
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You are awesome Luis😀
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Excellent solution! Hats off!👌
Great lessons sir!!!
Glad you like them!
Thank you! Cheers! 😀
“… and here’s our much nicer looking diagram!”
I love it👍
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You are awesome Tc 😀
Wow that was amazing!
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!
Glad to hear that!
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You are awesome 😀
Great leson!
this is really good ! good job sir !
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You are the best Haofeng
Keep it up 😀
Beautiful solution! Thanks!
You are very welcome.
Glad to hear that!
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You are awesome Sergio 😀
What tools are you using to make this video? Could you please tell me?
Very nice explanation👍
Thanks for sharing🎉
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Keep it up 😀
I wait for new problem which is more difficult than one. Thaks for all
Sir , have you created this question your own ??
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 ✓ ≡=-⋅
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!
Thank you for your feedback! Cheers!
Keep it up Philip 😀
Wow 😳 sir ❤️🙏🙏🙏🙏 superb creation 🙏🙏🙏❤️
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You are awesome 😀
Love and prayers from the USA!
Nice explanation 🙂
Glad it was helpful!
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Keep it up Mahalakshmi 😀
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 🙏🙏
Another great question from PreMath,thank you.
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You are the best Metehan
Keep it up 😀
Nice challenging puzzle and well explained 👍
Can you find x on law sine ?
I dont think anyone can solve this as so many trick reconstructions. Any easier solutions without reconstruction
Thank you for a great math lesson.
Thanks for video. Good luck!!!!!!!!
Thanks for watching!
You are very welcome.
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very well explained, thanks for sharing
Glad it was helpful!
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Keep it up 😀
So creative and cool!
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
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
Great solution
Glad to hear that!
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You are awesome Gurwindar😀
45 and 135
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 😀
Some people just have a talent for seeing what others miss. That's what IQ tests measure. But more practice seeing such videos as this can help.
Good very good🌹🌹🙏🙏
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Keep it up Engineer 😀
Solved very easily sir
Good job Vats
Excellent
Thank you so much 😀 Cheers!
Keep it up John 😀
that is cool thank you.
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
"if CD is the median then it must bisect ∠ACB"
Not necessarily.
Simply just Put AD/DC = BD/DC = Sine of Opposite angles
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Keep it up Pankaj 😀
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.
Very tough question !
good. aproa.
but we can solve this by using quadrilateral , parallagram properties
Thnku
You are very welcome.
You are awesome Pranav 😀
Sin15/sinx=sin30/sin(135-x). If you expand no calculator needed x=45 finished.
Cool
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🙏🙏
The shading was helpful to focus attention on a particular part of the diagram at the 03:20 time stamp and elsewhere.
Very good question
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Keep it up 😀
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.
Thank you for your feedback! Cheers!
Keep it up 😀
Nice solution. I first solved it using trigonometry. Then I did with circumscribed circle. Will post solutions on my channel shortly.
Thank you for your feedback! Cheers!
You are awesome 😀
Link to geometric solution: ruclips.net/video/usE2jUsXrOQ/видео.html
Link to trigonometric solution: ruclips.net/video/kihsNLXUM9M/видео.html
The best
Whoa! It looks so complicated 😆😆😆
Yes! It's quite challenging.
Thank you for your feedback! Cheers!
You are awesome 😀
It was impossible to solve angle x
It is not an easy one.
Thank you for your feedback! Cheers!
You are awesome Mmatlou 😀
Extremely difficult problem
suy luận tìm góc rất tuyệt.
شكرا لكم
لدينا حلان X=45،X=135
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. 🙂.
Thank you for your wise feedback! Cheers!
You are the best
Keep it up 😀
Significiant solvement
45 degree
"Quick & Simple Explanation" is very funny. Especially quick...
X = 45°
X=45
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 …….. "?
good
I got the radius of the small circle correctly (1 and 25 as well), however, forgot to calculate the areas... :(
Then third angle should be 180-30-45=105. Not seems like this
X=30
يا ليت الشرح باللغه العربيه حتى نتابعك
Thinks
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
What the hell this construction is
How yo think this bro
Can you please tell
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
X== 50
🖤
:)!!!
Your presentation is very slow.
Thank you
👏👏👏👏👏 👏
You're welcome 😊
You are awesome Algeria 😀