You can use this method too. Make an ABFD square. And then look at triangle FBC, which is an equilateral triangle. After look FDC triangle it is isosceles triangle. Thats why FDC angle is 15 degree. Therefore x is 75 degrees.
An interesting problem and a very nice solution. One more easier and simpler solution: Draw two lines from points B and D such that ABED is a square,. Contact points C and E. Now BCE is an isosceles triangle Where BC=BE=AB (sides of a square) Angle EBC=60 degree(150-90) So angleBCE= angleBEC=60 degree Now we know that BCE is an equilateral triangle SoEC=BC=DE। (sides of a square) Now CED is an isosceles triangle Angle DEC=150 degrees(90+60) So angleDCE=angle CDE=15 degrees Angle X= angle ADE- angleCDE 90-15=75
Awesome, many thanks, Sir. You are great. A slightly different approach: r = a = radius from point B = AB = AD = BC = BG (= BE + EG) = AF = BF → r = a = radius from Point A → F = crossing point of the two circles → ∆ABF = equilateral triangle → EBC = θ = 30° → FAB = ABF = BFA = 2θ = 60° → DAF = 30° → ADF = AFD = x = 75° 🙂
You haven't shown that F lies on the line DC. You have just assumed that it does, ie from the last part of your argument you claim ADF=x, how is that true? The angle x is ADC, not ADF. Again you need to show that F lies on DC.
Really great explanation, although I’m curious to know why you didn’t stop at 5:09, as the line FC seems like a wasted step. Basically, once the distance from EC was proven to be half the length of AD then the triangle ACD must be isosceles, meaning angle DAC must equal angle x. Since we know angle DAC is 75 degrees then angle x must be 75 degrees… or am I missing something?
Almost exactly my reasoning. First was to connect AC. With both smaller legs (AB BC) being same length then, and 150° in ∠B, only (180 - 150) = 30° gets divided in 2 for the deep corner angles. Since ∠A is 90°, take away 15°, and the remainder is 75° Next extending B→E allows the △BEC to be recognized as a 30-60-90 △ I figured if the height EC was '1', then the 3 congruent angles were each '2'. Again, extending a horizontal line from AD→C establishes a 15-75-90△ with small side [1] The upper part, also goes to point C, and thus is congruent. Therefore ∠𝒙 must be 75° Nice.
Once lines AC, CE and CF are drawn, and the simple angles and length CE have been established :- Tan 75 = AE/CE. AE = tan75 x CE = 3.732. (CE = 1). FC = AE. FD = 2 - 1 = 1. Triangle DFC. Tan X = 3.732 / 1. Arctan, X = 75.
I'm an old land surveyor, and I'm not sure how, but after looking at this for a few seconds I somehow new the correct answer without the geometric calisthenics. This solution didn't rely solely on geometry though. Knowing that the sine of 30° is 0.5 is technically trigonometry.
Depends upon how one defines sine and cosine. If using similar (right) triangles as a basis for them, then it's geometry. If using a unit circle at the origin of a Cartesian coordinate system, then it's trigonometry. In any case the two definitions are consistent with each other. Also sines and cosines are not necessary for showing that the side opposite the 30 degree angle of a right triangle is half the length of the hypotinuse, just geometry.
Yes, the solution should be accessible to the younger pupils, it should be as easy as possible. The method with the square and the equilateral triangle is more accessible than the one with the sine.
At 2:47 you were describing a special triangle and allocating unit lengths of 1 and 2. I don't understand how you can do this or why this triangle is special?
we solved the same question earlier by seeing the square, then the equilateral triangle now we have seen the solution by another method thanks PreMath🌟
Place point E on the segment DC such that AB = AE = BE. Triangle ABE is an equilateral triangle, triangle BCE is a right isosceles triangle, and triangle AED is an isosceles triangle. Therefore, x = 180 - 60 - 45=75.
This could be constructed by making a regular triangle ACE towards the D side. Then you find that angle DAE = 15 = angle BAC, hence triangle DAE is congruent to BAC by SAS. Then you get DE = BC = BA = DA. With CA = CE you get AE is symmetrical about CD. Hence angle ADC = angle ADE/2 = angle ABC /2 = 75 degree.
I found an easier approach Draw a line segment/bisector BD then draw segments BE and EC as seen in the video. We now have 3 triangles rt∆BAD, rt∆BEC and ∆CBD Let's focus on rt∆BAD. Notice that it forms a 45-45-90 triangle. So, angle ADB= 45° Due to bisector BD angle CBD= 105° Now, let's focus on rt∆BEC. Since angles ABD, DBC, and EBC form a straight line. Angle EBC = 30° Thus forming a 30-60-90 angle So angle BCE = 60°. Now let's focus on the last triangle ∆CBD Since the sides AD, AB, and, BC are congruent, their corresponding angles are also congruent (ADB=ABD=BCD). So this makes angle BCD = 45° Now for the final step, let's continue focusing ∆CBD Let us add up all the angles formed. But first notice that segment BD also bisects ADB and BDC So angle BDC = x-45° We can now solve for x by using ∆BDC 105+x-45+45= 180° x+105°= 180° x= 75° And this is our final answer. God bless from the Philippines! 🇵🇭
Why is that easier?. I looked at it, made an educated guess, and got the right answer. That was the easiest method of all. (I'm an old surveyor who did a lot of construction layout by "double-chaining," right triangles, so perhaps I had an unfair advantage.) Of course afterwards I sat there and stared at the diagram trying to figure out how to prove it.
@@jhandle4196 If not easier, then that would be a faster technique for me, forming special right triangles namely 45-45-90 (Isosceles right triangle) and 30-60-90 (equilateral equiangular triangle with a central altitude forming 2 right triangles)
It seems there is a much easier way. Draw a straight line perpendicular to AB from B and Draw a straight line from D perpendicular to AD, then If H is the intersection of the two lines, triangle ABC and triangle DHC are congruent. At this time, since the rectangle ABHD is a square, the angle of ADC is the angle of ADH minus the angle of CDH. Angle of ADH = 90° Angle of CDH = 1/2 (180°-150°) = 15° So 90° - 15° = 75°.
I disagree. Approximately at 3:00, you evaluate that Angle CBA is 30 Deg and Angle BCE is 60 Deg in the right angle triangle ECB. Therefore, the side BE (in front of 60 Deg angle) would be twice than the side EC (in front of the 60 Deg angle), AND NOT the Side BC, the hypotenuse in the right angle triangle. In fact the hypotenuse, BC, would be Square root of 5. Please correct me, if I am wrong! Thanks, Padmakar Srivastava (Ex IITian), PhD, PE Member ASCE, Member & Reviewer ASTM Retired US Army USA
A BETTER way to solve is lline segment DB is the hypotnuse to right isosceles triangle ABD which means it has length Ysqrt(2), Y=DA=AB=BC. Then you use law of cosines to find side DC 1.93185...*Y. Then use law of cosines again to find angle BDC = 30 degrees. X = 45 + 30 = 75 degrees.
Interesting problem. I was trying to visualize the diagram before watching the video and was on the right path but just didn't make the final connection to solve it until you drew the line FC. As soon as I saw that line actually drawn in the figure, I instantly recognized the final steps to match the two angles and get the result.
Nice! I resorted to trig, deriving quickly that tan x = 2 + sqrt(3) and sin 2x = 1/2; and since x clearly greater than 15 and less than 90 degrees, it's 75 degrees.
@@jhandle4196 LOL On what basis do you NOT regard solving a problem that's ostensibly geometry, using instead trigonometry, as "thinking outside the box"? Much of the beauty of mathematics is its unity across sub-disciplines.
I did he same thing by expanding things out; once I saw the 75° answer I kept going out of the box to find the geometric solution too. Sad it took me so much thinking and paper as I used to be so much quicker in math. I think regardless of how you get there the fact you know enough how to solve and can solve it is the point.
I couldn't quite figure it out using geometry. I was thinking maybe I could put it into a coordinate system and find out the expressions of all the lines. Whilst I was doing that, I realised that the slope of BC is simply sin(30 degrees) and figured out x using trig as well.
i creat a new point ABDE is square, Angle EBC is 60, side EB=BC=CE, so angle DEC is 150, because side DE=CE, angle EDC + angle ECD+150=180 , so angle EDC = 15, x=90-15, x =75, thanks
when you said outside the box thinking, I took a sheet of paper and traced the shape, I then realised you can make a hexagon, and then it was obvious angle x had to be half of 150= 75
I did a completely different route. AB=DA=BC=Y BE=Ycos30=(2+root(3))/2Y CE=Ysin30=1/2Y Extend a line from point D parallel to line AE to point C, call that point F, second right triangle forms. AE=DF=Y+Yroot(3)/2=(2+root(3)/2)Y CF=DA-CE=1/2Y Angle FDC=Tan^-1 (1/2Y/(2+root(3)/2)Y)=15 degrees Angle ADF is a 90 degree angle, 90 degree-15 degree=75 degrees.
Found arctan(2+sqrt(3)). Let E be a point of AD such the CE is perpendicular to AD. We have that angle BCE is 30°. Let's say that length of AD is 2. Then we have that length of AE = 2 * sin(30°) = 1, also length of CE = 2 + 2 * cos(30°) = 2 + sqrt(3). We can deduce the length of DE, which is one. By definition of tan, tan x = 2 + sqrt 3, so x = arctan(2 + sqrt(3))
PreMath, All exercises are very promising to improve geometrical thinking. I use NEBO for drawing graphically. But as I realized, it is not so good as yours. Could you write me please the name of this application?
This one cracks open easily once you find the trick. My solve process was roughly: 1 - Triangle composed of ABC is given isosceles -> angle BAC = angle BCA = 15 degrees -> angle DAC = (90 - 15) = 75 degrees 2 - The angle of vector BC relative to vector AD is necessarily -60 degrees -> BC projected along AD yields cos(60) = 0.5x of vector AD -> triangle CAD is also isosceles -> angle ADC = angle DAC = 75 degrees
Triangle ABC is isosceles so angle CAB is 15 and angle CAD is complementary to angle CAB so it is 75 From sine law in triangle ABC calculate side length of AC in terms of AB From cosine law in triangle ADC calculate length of CD in terms of AB first then also from cosine law calculate cosine of x This last cosine law would not be necessary if we compare side length of AC and CD
Respected Sir I want to use all geometry in share market forex and comodity trading plz hint..I eagerly waiting for next video using on price and time reversal or retracement of price and dates on financial market
I haven't done these problems for sixty years, so it took me quite a long time to manage a method. Not as simple as yours. 1. Draw line AC. ABC is an isosceles triangle, so angle BAC is 15. 2. Draw line DF, equal to line AD, inclined towards C. Make angle ADF 150. 3. Draw line AF. ADF is an isosceles triangle, so angle DAF is 15. 4. Angle CAF is 90 - (BAC15 + DAF15) = 60 5. Draw line CF. 6. CA and AF are equal, so CAF is an equilateral triangle . 7. Therefore, CA =CF. 8. CA = AF. AD = DF. CD is common. 9. Therefore CAD and CFD are congruent. 10. Therefore, angle ADC = CDF. 11. Therefore ADC = half 150 = 75
Utilizzando il teorema dei seni per i 2 triangoli abbiamo, dopo vari calcoli, l'equazione sinx/sin150=sin(x+75)/sin15....e dopo altri calcoli semplici... X=75
I drew a II segment starting from B going to CD where I Got a F point in this segment and drew an another segment starting from B direct to D This gave me a isosceles triangle ABD and a Isoceles triangle BFC Another segment was drew going from A to F, and AFB is equilateral triangle… Equilateral triangle has 60° each angle… so A point is now divided in 60° plus 30° and AF and AD has the same lenght so ADF is isoceles with A 30°… so 180 - 30 = 150… and 150 / 2 is 75° Am I right?
Legends trick: Draw BD.then ABD is isosceles traingle(since AD=AB). then angle ABD=75. Similarly angle ADB=75(sice opposite angle in isosceles triangle are equal).🤨. Agree?
I solved it via a somewhat more difficult route, I showed that 2 circles of radius AD centered at A and B and the line DC are concurrent. From that the angle x=75 pops out.
@@jeffreygreen7860 To understand your statement I had tried to draw the figure accordingly but didn't understand about the radius AD. Maybe you can construct properly as per your imagination.
@@Giveup00 Sorry about the confusion. I'm not sure if I had labeled my diagram differently from that in the video or that I just had the labels switched in my head when I wrote the original comment (
@Awakened1729 uhhh, just saw that I had said the centers were on A&B. Now I'm not sure where your confusion lies. Could you elaborate a bit about the problem you're having?
Let x = AD = AB = BC Draw BD. ADB = 45 and DB = sqrt(2)x by Pythagorus DBC = 105 degrees Using cosine law on DBC DC = sqrt((sqrt(2)x)^2 + x^2 - 2(sqrt(2)x)(x)cos105) = xsqrt(3-2sqrt(2)cos(105)) = xsqrt(3-2sqrt(2)(1-sqrt(3))/(2sqrt(2)))) = xsqrt(2+sqrt(3)) Draw AC, use cosine law on ABC AC = sqrt(x^2 + x^2 - 2(x)(x)cos(150)) = xsqrt(2 - 2cos(150)) = xsqrt(2 - 2(-sqrt(3)/2)) = xsqrt(2 + sqrt(3)) So DC = AC = xsqrt(2+sqrt(3)) that is DCA isosceles Also ABC isosceles CDA = CAD = 90 - BAC = 90 - (180-150)/2 = 75 degrees
I took the view that as ADB is an isosceles triangle then angle ADB is 45 deg. Angle EBC is 30 deg. and by adding the two together, I got ADC to be 75 deg. I'm sure my logic is faulty but I got the right answer. Maybe this was just a coincidence?!
A given for the problem. Start by drawing the line AB. At A draw the perpendicular. Set the lengths AD and AB, you choose. At B draw the line 30 degrees up from AB and away from A. Set the length BC=AB (just draw a circle at B of radius AB when setting the lengths AD and AB). Draw the line DC.
Hola no soy de usar geometría porque me llega costar que hacer así que recurro a trigonometria en mi caso lo que hice fue trazar una diagonal BD así que sus ángulos son 45° así que como ya tenemos eso y a la diferencia entre x y 45° la llamamos theta continuando sabemos también por Pitágoras que la diagonal es raiz cuadrada de 2 por a ahora también sabemos que en el triángulo oblicuangulo que se formo sabemos que 2 de sus lados miden raíz cuadrada de a y a y que uno de sus ángulos es 105 así que usando teorema del seno y coseno obtenemos que theta vale 30° así que solamente sumamos 45+3o que es 75 así que el valor de x es 75°
∠CBE=180º -150º=30º → CE=BC/2=DA/2 →→ AB=BC→∠CAE=∠CBE/2=30º/2=15º → ∠DAC=90º -15º=75º = ∠ADC= X, by symmetry with respect to the FC axis. Thanks and best regards
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Excellent sir
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You are awesome. Keep it up 👍
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You can use this method too.
Make an ABFD square. And then look at triangle FBC, which is an equilateral triangle.
After look FDC triangle it is isosceles triangle. Thats why FDC angle is 15 degree.
Therefore x is 75 degrees.
Did the same. More simple and thus elegant solution.
There's more than one way to skin a cat.
@@jhandle4196 Yes you can hold your left ear with your right hand many ways but one of them is more simple.
I solved it this way also, and I like it more as it is not implying trigonometry. Just geometry.
The best answer IMO
An interesting problem and a very nice solution.
One more easier and simpler solution:
Draw two lines from points B and D such that ABED is a square,.
Contact points C and E.
Now BCE is an isosceles triangle
Where BC=BE=AB (sides of a square)
Angle EBC=60 degree(150-90)
So angleBCE= angleBEC=60 degree
Now we know that BCE is an equilateral triangle
SoEC=BC=DE। (sides of a square)
Now CED is an isosceles triangle
Angle DEC=150 degrees(90+60)
So angleDCE=angle CDE=15 degrees
Angle X= angle ADE- angleCDE
90-15=75
Awesome, many thanks, Sir. You are great.
A slightly different approach:
r = a = radius from point B = AB = AD = BC = BG (= BE + EG) = AF = BF →
r = a = radius from Point A → F = crossing point of the two circles → ∆ABF = equilateral triangle →
EBC = θ = 30° → FAB = ABF = BFA = 2θ = 60° → DAF = 30° → ADF = AFD = x = 75° 🙂
You need to show that the 2 circles and the top line of the quadrilateral are concurrent first.
@@jeffreygreen7860 The congruent circles probably weren't drawn simultaneously, so they are not exactly concurrent.
You haven't shown that F lies on the line DC. You have just assumed that it does, ie from the last part of your argument you claim ADF=x, how is that true? The angle x is ADC, not ADF. Again you need to show that F lies on DC.
Oops, different person. Murdock5537 needs to show that F lies on DC.
@@jeffreygreen7860
Thanks for your comments. Solving the problem without circles:
ABC = φ = 150° → θ = 180° - φ
AD = AB = BC ∶= k → AF = BF = k → FAB = ABF = BFA = 2θ →
FBC = 3θ → sin(3θ) = 1 → DAF = θ → DF = DR + FR; DR = FR →
ADF = DFA = φ/2 = x
Btw:
sin(θ) = 1/2 → cos(θ) = √3/2 →
sin(θ/2) = √(1 - cos(θ)/2) = (1/2)√(2 - √3) = DR/k → DR = (k/2)√(2 - √3) →
DF = k√(2 - √3) → cos(x) = DR/k = sin(θ/2) → x = 75°
or: (DF)^2 = 2k^2(1 - cos(θ)) = k^2 (2 - √3) → DF = k√(2 - √3)
and:
CBE = θ → CAB = BCA = θ/2 → DAC = x = 5θ/2 → AC = CD
sin(θ) = 1/2 → cos(θ) = √3/2 → sin(θ/2) = (1/2)√(2 - √3) = (k/2)AC →
AC = DC = k√(2 - √3)(2 + √3) → (AC)^2 = k^2(2 + √3) →
AD = AP + DP; AP = DP →
CP = √((AC)^2 - (k/2)^2) = (1/2)k√(7 + 4√3)
Really great explanation, although I’m curious to know why you didn’t stop at 5:09, as the line FC seems like a wasted step. Basically, once the distance from EC was proven to be half the length of AD then the triangle ACD must be isosceles, meaning angle DAC must equal angle x. Since we know angle DAC is 75 degrees then angle x must be 75 degrees… or am I missing something?
تمرين جميل جيد. رسم واضح مرتب. شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا .تحياتنا لكم من غزة فلسطين .
I am 64 and I enjoy these problems, thank you .
Wonderful!
You are very welcome!
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I am now just addictive with this channel!
Excellent!
Glad to hear that!
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@@PreMath This is my mom's I'd .My real name is Aryan Raj☺️
My friend recommended this to me.
Special thanks to him 😌
@@vandanakumari1634 Thank you Raj. You are the best!
What are the withdrawal symptoms?
Almost exactly my reasoning. First was to connect AC. With both smaller legs (AB BC) being same length then, and 150° in ∠B, only (180 - 150) = 30° gets divided in 2 for the deep corner angles. Since ∠A is 90°, take away 15°, and the remainder is 75°
Next extending B→E allows the △BEC to be recognized as a 30-60-90 △ I figured if the height EC was '1', then the 3 congruent angles were each '2'. Again, extending a horizontal line from AD→C establishes a 15-75-90△ with small side [1] The upper part, also goes to point C, and thus is congruent.
Therefore ∠𝒙 must be 75°
Nice.
Very detailed description. It is useful for everyone to understand easily. Congratulations brother.🌷🌷
The way you pronounce degrees is hypnotic
Once lines AC, CE and CF are drawn, and the simple angles and length CE have been established :-
Tan 75 = AE/CE.
AE = tan75 x CE = 3.732. (CE = 1).
FC = AE.
FD = 2 - 1 = 1.
Triangle DFC.
Tan X = 3.732 / 1.
Arctan, X = 75.
It was suchhhh an INTERESTING question.....nice video!!
Great example, ...
Note to self !... Remember to think outside the box
It is much easier to add BD and use trigonometry to get BDC
Math Magic!!! ❤️💯😉🌟🌟🌟🌟🌟👍👍👍👍👍
I'm an old land surveyor, and I'm not sure how, but after looking at this for a few seconds I somehow new the correct answer without the geometric calisthenics.
This solution didn't rely solely on geometry though. Knowing that the sine of 30° is 0.5 is technically trigonometry.
Depends upon how one defines sine and cosine. If using similar (right) triangles as a basis for them, then it's geometry. If using a unit circle at the origin of a Cartesian coordinate system, then it's trigonometry. In any case the two definitions are consistent with each other. Also sines and cosines are not necessary for showing that the side opposite the 30 degree angle of a right triangle is half the length of the hypotinuse, just geometry.
Yes, the solution should be accessible to the younger pupils, it should be as easy as possible. The method with the square and the equilateral triangle is more accessible than the one with the sine.
At 2:47 you were describing a special triangle and allocating unit lengths of 1 and 2. I don't understand how you can do this or why this triangle is special?
we solved the same question earlier by seeing the square, then the equilateral triangle
now we have seen the solution by another method
thanks PreMath🌟
Great job
Thanks for your feedback! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Place point E on the segment DC such that AB = AE = BE.
Triangle ABE is an equilateral triangle, triangle BCE is a right isosceles triangle, and triangle AED is an isosceles triangle.
Therefore, x = 180 - 60 - 45=75.
This could be constructed by making a regular triangle ACE towards the D side. Then you find that angle DAE = 15 = angle BAC, hence triangle DAE is congruent to BAC by SAS. Then you get DE = BC = BA = DA. With CA = CE you get AE is symmetrical about CD. Hence angle ADC = angle ADE/2 = angle ABC /2 = 75 degree.
I thought he will use sin and cos thetas to solve this....this is a good idea 👌
Ah! I couldn't figure out the trick of adding the interior line AC. It all fell into place with that segment! Nice overview! 👍
Thanks for video.Good luck sir!!!!!!!!!!!!
I found an easier approach
Draw a line segment/bisector BD then draw segments BE and EC as seen in the video.
We now have 3 triangles
rt∆BAD, rt∆BEC and ∆CBD
Let's focus on
rt∆BAD.
Notice that it forms a 45-45-90 triangle.
So, angle ADB= 45°
Due to bisector BD
angle CBD= 105°
Now, let's focus on rt∆BEC.
Since angles ABD, DBC, and EBC form a straight line.
Angle EBC = 30°
Thus forming a 30-60-90 angle
So angle BCE = 60°.
Now let's focus on the last triangle
∆CBD
Since the sides AD, AB, and, BC are congruent, their corresponding angles are also congruent (ADB=ABD=BCD).
So this makes angle BCD = 45°
Now for the final step, let's continue focusing ∆CBD
Let us add up all the angles formed.
But first notice that segment BD also bisects ADB and BDC
So angle BDC = x-45°
We can now solve for x by using ∆BDC
105+x-45+45= 180°
x+105°= 180°
x= 75°
And this is our final answer.
God bless from the Philippines! 🇵🇭
Why is that easier?. I looked at it, made an educated guess, and got the right answer. That was the easiest method of all.
(I'm an old surveyor who did a lot of construction layout by "double-chaining," right triangles, so perhaps I had an unfair advantage.)
Of course afterwards I sat there and stared at the diagram trying to figure out how to prove it.
@@jhandle4196 If not easier, then that would be a faster technique for me, forming special right triangles namely 45-45-90 (Isosceles right triangle) and 30-60-90 (equilateral equiangular triangle with a central altitude forming 2 right triangles)
It seems there is a much easier way.
Draw a straight line perpendicular to AB from B and
Draw a straight line from D perpendicular to AD, then
If H is the intersection of the two lines, triangle ABC and triangle DHC are congruent.
At this time, since the rectangle ABHD is a square, the angle of ADC is the angle of ADH minus the angle of CDH.
Angle of ADH = 90°
Angle of CDH = 1/2 (180°-150°) = 15°
So 90° - 15° = 75°.
well done👍👏
سلام
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متشکرم.
I disagree. Approximately at 3:00, you evaluate that Angle CBA is 30 Deg and Angle BCE is 60 Deg in the right angle triangle ECB. Therefore, the side BE (in front of 60 Deg angle) would be twice than the side EC (in front of the 60 Deg angle), AND NOT the Side BC, the hypotenuse in the right angle triangle. In fact the hypotenuse, BC, would be Square root of 5. Please correct me, if I am wrong!
Thanks,
Padmakar Srivastava (Ex IITian), PhD, PE
Member ASCE, Member & Reviewer ASTM
Retired US Army
USA
I agree with you, this was very confusing. The side opposite to 30° angle in a right-angled triangle is half of the hypotenuse.
Great video. I really enjoyed it. I enjoy a good puzzle.
Glad you enjoyed it!
Thanks for your feedback! Cheers!
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A BETTER way to solve is lline segment DB is the hypotnuse to right isosceles triangle ABD which means it has length Ysqrt(2), Y=DA=AB=BC. Then you use law of cosines to find side DC 1.93185...*Y. Then use law of cosines again to find angle BDC = 30 degrees. X = 45 + 30 = 75 degrees.
Interesting problem. I was trying to visualize the diagram before watching the video and was on the right path but just didn't make the final connection to solve it until you drew the line FC. As soon as I saw that line actually drawn in the figure, I instantly recognized the final steps to match the two angles and get the result.
Yes. It's pretty simple once you know the answer.
Thank you very much ❤❤
Nice! I resorted to trig, deriving quickly that tan x = 2 + sqrt(3) and sin 2x = 1/2; and since x clearly greater than 15 and less than 90 degrees, it's 75 degrees.
messy. It's a geometry problem though. Not trigonometry.
@@jhandle4196 LOL On what basis do you NOT regard solving a problem that's ostensibly geometry, using instead trigonometry, as "thinking outside the box"?
Much of the beauty of mathematics is its unity across sub-disciplines.
I did he same thing by expanding things out; once I saw the 75° answer I kept going out of the box to find the geometric solution too. Sad it took me so much thinking and paper as I used to be so much quicker in math. I think regardless of how you get there the fact you know enough how to solve and can solve it is the point.
I couldn't quite figure it out using geometry. I was thinking maybe I could put it into a coordinate system and find out the expressions of all the lines. Whilst I was doing that, I realised that the slope of BC is simply sin(30 degrees) and figured out x using trig as well.
i creat a new point ABDE is square, Angle EBC is 60, side EB=BC=CE, so angle DEC is 150, because side DE=CE, angle EDC + angle ECD+150=180 , so angle EDC = 15, x=90-15, x =75, thanks
Excellent sirr
Glad you think so!
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You are awesome, Sahil. Keep it up 👍
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Are u from USA??
@@sahilchaudhary2606 Yes!
I used the sin law to calculate length of AC, then the cosine law to calculate CD, then the sine law again to get A.
Aslam u Alaikum
With the help of your video
We can learn easily Math
From the diagram, we have |AB| = |AC| = |AD| = 2 units. Then |DF| = |AF| = |CE| = 1 unit bec. AECF is a rectangle. Now
|BE| = |BC|.sin(∠CBE) = 2.(√3)/2 = √3. So tan(∠ADC) = |FC|/|DF| = (|AB|+|BE|)/|DF| = 2+√3. So ∠ADC = tan⁻¹(2+√3) = 75°.
Note tan(75°) = tan(45°+30°) = {tan(45°) + tan(30°)} / {1 - tan(45°).tan(30°) = {1+ 1/√3}/{1- 1.1/√3} = {√3 +1}/{√3 -1}
= {√3 +1}.{√3 +1} / {3 -1} = (4+ 2√3)/2 = 2+√3. Trigonometry is king !
when you said outside the box thinking, I took a sheet of paper and traced the shape, I then realised you can make a hexagon, and then it was obvious angle x had to be half of 150= 75
Excellent!
I did a completely different route.
AB=DA=BC=Y
BE=Ycos30=(2+root(3))/2Y
CE=Ysin30=1/2Y
Extend a line from point D parallel to line AE to point C, call that point F, second right triangle forms.
AE=DF=Y+Yroot(3)/2=(2+root(3)/2)Y
CF=DA-CE=1/2Y
Angle FDC=Tan^-1 (1/2Y/(2+root(3)/2)Y)=15 degrees
Angle ADF is a 90 degree angle, 90 degree-15 degree=75 degrees.
Good question 😊
After 4:50 ACD is an isoceles triangle, so angle D= angle A=75
Bro how its become isoseles , tell me bro?
Amazing!
Found arctan(2+sqrt(3)).
Let E be a point of AD such the CE is perpendicular to AD. We have that angle BCE is 30°. Let's say that length of AD is 2. Then we have that length of AE = 2 * sin(30°) = 1, also length of CE = 2 + 2 * cos(30°) = 2 + sqrt(3). We can deduce the length of DE, which is one. By definition of tan, tan x = 2 + sqrt 3, so x = arctan(2 + sqrt(3))
I think inside the box. Make a point O, where OA=OD=AD, that makes angle AOC=DOC=150...now we have AC=DC, every angle can be calculated
Make ABCE square
ECB equilateral triangle
DE=EC
ECD=15° DCB=45°
CDA=75° the end
PreMath, All exercises are very promising to improve geometrical thinking. I use NEBO for drawing graphically. But as I realized, it is not so good as yours. Could you write me please the name of this application?
Very well explained.
Awesome!
How do you assume angel E is 90?
This one cracks open easily once you find the trick. My solve process was roughly:
1 - Triangle composed of ABC is given isosceles -> angle BAC = angle BCA = 15 degrees -> angle DAC = (90 - 15) = 75 degrees
2 - The angle of vector BC relative to vector AD is necessarily -60 degrees -> BC projected along AD yields cos(60) = 0.5x of vector AD -> triangle CAD is also isosceles -> angle ADC = angle DAC = 75 degrees
How would the solution change if the known angle was 160° instead of 150° ?
ABDを使った正方形を作図し、追加した頂点をEとする。
EBCは正三角形となる。
角DECが150° DECは二等辺三角形
角EDC＝15°
よって X＝75°
Sir why you assume length 1 or 2 units complicated no need please explain
Triangle ABC is isosceles so angle CAB is 15
and angle CAD is complementary to angle CAB so it is 75
From sine law in triangle ABC calculate side length of AC in terms of AB
From cosine law in triangle ADC calculate length of CD in terms of AB first
then also from cosine law calculate cosine of x
This last cosine law would not be necessary if we compare side length of AC and CD
good job bro
Allah bless u alot with success and happiness aameen 🌹💕
Ugh, it’s so easy once you explain it, I was going about it all wrong…
I love this more than oranges. and i like oranges A LOT, its really saying osmething
Respected Sir I want to use all geometry in share market forex and comodity trading plz hint..I eagerly waiting for next video using on price and time reversal or retracement of price and dates on financial market
Can i just... made a line between B and D, ad declare it "root of 2 length"?
Then i have a known angle (105*) between two known sides.
I solved it using Pythagoras, law of cosines and law of sines
Muy buen ejercicio, es importante saber pocas propiedades pero esenciales.
¡Excelente!
¡Gracias por tus comentarios! ¡Salud!
Usted es maravilloso. Sigue así 👍
¡Amor y oraciones desde los EE. UU.! 😀
I haven't done these problems for sixty years, so it took me quite a long time to manage a method. Not as simple as yours.
1. Draw line AC. ABC is an isosceles triangle, so angle BAC is 15.
2. Draw line DF, equal to line AD, inclined towards C. Make angle ADF 150.
3. Draw line AF. ADF is an isosceles triangle, so angle DAF is 15.
4. Angle CAF is 90 - (BAC15 + DAF15) = 60
5. Draw line CF.
6. CA and AF are equal, so CAF is an equilateral triangle .
7. Therefore, CA =CF.
8. CA = AF. AD = DF. CD is common.
9. Therefore CAD and CFD are congruent.
10. Therefore, angle ADC = CDF.
11. Therefore ADC = half 150 = 75
Nice.
Helpful
Utilizzando il teorema dei seni per i 2 triangoli abbiamo, dopo vari calcoli, l'equazione sinx/sin150=sin(x+75)/sin15....e dopo altri calcoli semplici... X=75
Eccellente!
Grazie per la condivisione! Saluti!
Sei fantastico. Continua così 👍
Amore e preghiere dagli Stati Uniti! 😀
@@PreMath thanks,sei molto gentile
from 4:50 onwards, triangle ADC is isosceles so therefore angle ADC is 75 degrees.
at that point you haven't really proved that triangle ADC is isosceles though as a result it is but you'll need to prove it first
EC is half AD, therefore isosceles by visual inspection
I drew a II segment starting from B going to CD where I Got a F point in this segment and drew an another segment starting from B direct to D
This gave me a isosceles triangle ABD and a Isoceles triangle BFC
Another segment was drew going from A to F, and AFB is equilateral triangle…
Equilateral triangle has 60° each angle… so A point is now divided in 60° plus 30° and AF and AD has the same lenght so ADF is isoceles with A 30°… so 180 - 30 = 150… and 150 / 2 is 75°
Am I right?
I just hoped that projection of C on AD would divide it in half and counted 75. I omitted the justification part though.
Nada garantiza que trazar una bisectriz se forme un ángulo recto.
شكرا على المجهودات
نستعمل مبرهنة الكاشي في المثلثBCDنجدx=75
Legends trick:
Draw BD.then ABD is isosceles traingle(since AD=AB). then angle ABD=75.
Similarly angle ADB=75(sice opposite angle in isosceles triangle are equal).🤨.
Agree?
Thanks
I solved it via a somewhat more difficult route, I showed that 2 circles of radius AD centered at A and B and the line DC are concurrent. From that the angle x=75 pops out.
AD as a radius or diameter?
@@Giveup00 Radius...as in my statement. Problem??
@@jeffreygreen7860 To understand your statement I had tried to draw the figure accordingly but didn't understand about the radius AD. Maybe you can construct properly as per your imagination.
@@Giveup00 Sorry about the confusion. I'm not sure if I had labeled my diagram differently from that in the video or that I just had the labels switched in my head when I wrote the original comment (
@Awakened1729 uhhh, just saw that I had said the centers were on A&B. Now I'm not sure where your confusion lies. Could you elaborate a bit about the problem you're having?
I just made the wrong diagram
I did the same but thought that AECD is a rectangle and my x came 90
No worries!
Thanks for your feedback! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
Wow!
“Think outside the box.”
I see what you did there.
Let x = AD = AB = BC
Draw BD. ADB = 45 and DB = sqrt(2)x by Pythagorus
DBC = 105 degrees
Using cosine law on DBC
DC = sqrt((sqrt(2)x)^2 + x^2 - 2(sqrt(2)x)(x)cos105)
= xsqrt(3-2sqrt(2)cos(105))
= xsqrt(3-2sqrt(2)(1-sqrt(3))/(2sqrt(2))))
= xsqrt(2+sqrt(3))
Draw AC, use cosine law on ABC
AC = sqrt(x^2 + x^2 - 2(x)(x)cos(150))
= xsqrt(2 - 2cos(150))
= xsqrt(2 - 2(-sqrt(3)/2))
= xsqrt(2 + sqrt(3))
So DC = AC = xsqrt(2+sqrt(3)) that is DCA isosceles
Also ABC isosceles
CDA = CAD = 90 - BAC = 90 - (180-150)/2 = 75 degrees
Good morning sir
Using a combination of Pythagorean theorem and a law of cosines and sines yield a quicker correct answer
arctan((1+cos(30))/(1-sin(30)))=75
Proportional Reasoning:
x/120=150/240
How are AC and CD equal?
Because of SAS. See 6.39 of the video.
I took the view that as ADB is an isosceles triangle then angle ADB is 45 deg. Angle EBC is 30 deg. and by adding the two together, I got ADC to be 75 deg. I'm sure my logic is faulty but I got the right answer. Maybe this was just a coincidence?!
AC=sqrt(4+4-8cos(150°))=sqrt(8+4sqrt(3))
Or you could have said halfway through the video that we know that triangle ADC is isosceles because of the known 1/2 length side
'점C' 에서 내린 수선의 발이 '선분AD'를 2등분 하는 이유는 뭐죠?
아~ 특수삼각형을 찾아 1:2:루트2 비율로 표기를 한 것이군요.
_- Fing X!_
_- Here it is!_ [points to X]
I think that if you label the sides with x would be much better than 2
I agree, this was very confusing.
Actually I didnt understand how length of AB and AD BC equals BC ?
Haven't watched the video yet X=75°
How BC = AB = AD?
A given for the problem. Start by drawing the line AB. At A draw the perpendicular. Set the lengths AD and AB, you choose. At B draw the line 30 degrees up from AB and away from A. Set the length BC=AB (just draw a circle at B of radius AB when setting the lengths AD and AB). Draw the line DC.
👍👍
Hola no soy de usar geometría porque me llega costar que hacer así que recurro a trigonometria en mi caso lo que hice fue trazar una diagonal BD así que sus ángulos son 45° así que como ya tenemos eso y a la diferencia entre x y 45° la llamamos theta continuando sabemos también por Pitágoras que la diagonal es raiz cuadrada de 2 por a ahora también sabemos que en el triángulo oblicuangulo que se formo sabemos que 2 de sus lados miden raíz cuadrada de a y a y que uno de sus ángulos es 105 así que usando teorema del seno y coseno obtenemos que theta vale 30° así que solamente sumamos 45+3o que es 75 así que el valor de x es 75°
You didn’t need F, the large isosceles triangle gave you the angle measurement of angle C
This can be solved in a very easy way....
This was complex.
this is elementary school geometry in Taiwan..
∠CBE=180º -150º=30º → CE=BC/2=DA/2 →→ AB=BC→∠CAE=∠CBE/2=30º/2=15º → ∠DAC=90º -15º=75º = ∠ADC= X, by symmetry with respect to the FC axis.
Thanks and best regards