As an industrial electrician who use 30 and 45 special triangle formula to help bend conduit I find these videos very useful. Although most electricians use multiplier formulas. I like to think of bends as triangles and use formulas like offset height divided by the sine of any degrees. This. Give me my hypotenuse or space in-between bends. So for students who doesn't think math is important, it is because without it I could not put food on my table for my children. They would have no phone to even watch RUclips videos. Lol. So ingrain these formulas theorems and tricks in your head so you can provide for your family in the future
Third Method: * drop line from C to intersect line AB extended at point D, making the corner of right triangle CDB. * CBD must be 75 degrees and then BCD is 15 degrees. Since BC is given as √2, then CD is √2(sin 75) and can be rewritten: 1/2*(√3+1) * looking at the right triangle CDA, angle ACD is 60 degrees (45+15) and so length AC is: (1/2*(√3+1))/cos(60) = √3+1
Reflect a copy of ABC along AC. You get a combination of an equilateral triangle of side 2 and an Isosceles Right triangle with an altitude 1. Hence AC = 1+✓3
I think this is more of a mental math problem for anyone who remembers that sin 45 deg = cos 45 deg = 1/sqrt2, and sin 30 deg = 1/2 and cos 30 deg = sqrt3/2. These are easily remembered by visualising the unit square and unit equilateral triangle respectively. So the moment you realise the unknown angle is 45degs and one side is sqrt2, drop a perpendicular from the 105deg angle to the side opp it. Then you can easily work out the unknown sides. Visualising the basic figues -- unit square, unit equilateral triangle, and one or two more figures -- and knowing their proportions is good lifelong.
Draw perpendicular from B on AC. Half done. Now let the perpendicular touch AC at O. BO =1 CO=1 because it forms a right isosceles. BD =1/tan30 =root3. AC=1+root3
At a quick glance , Extend AB to D, the intersection of a perpendicular to C. The angle BCD is 15 degrees and side AC = sqrt(2) * cos(15)/sin(30) = 2.732
Great stuff. I actually used a different method from both those you showed. To the right of the main triangle I constructed a narrow, imaginary triangle with angles of 90, 75, and 15 degrees. I won't go into all the details but I figured out at AC = (sin(75) x sqrt(2)) / sin(30). I arrived at the same answer as you.
i use * for degrees as i have no degree thing ob my phone if you make a perpendicular line from B to AC, and call the point D, angle BDC is a 90* angle. in ABC, you know angle A=30* and angle B=105* so angle C=45*. this means triangle BCD is a special 1:1:sqrt2 triangle, the hypotenus (BC) is sqrt2 so the 2 other sides (BD,CD) are 1. since angle D is a perpendicular, triabgle ABD has angle A =30* and angle D=90*, so angle B=60*. this means we have another special triangle with 1:sqrt3:2=BD:AD:AB. we know that BD is 1 so we can substitute: 1:sqrt3:2=1:AD:AB so it is a 1x multiplication which means AD=sqrt3 (AB is irrelevant). recall we want to know AC, that is also equal to AD+CD=sqrt3+1 and that’s the awnser edit: there is probably an easier way with like the cos rule but i dont know that (is it a^2=b^2+c^2-2ab•cos(α)?)
Draw the perpendicular BH to AC. We now have 2 triangles: BHC and BHA. The triangle BHC is a right isosceles in which the hyposetenuse BC =sqrt of 2 so the sides BH=CH = 1. The triangle BHA is half of a equilateral triangle in which AH is the height and AB is the side, BH half of the side, so AB= 2 BH= 2, and AH= 1/2 ABX sqrt of 3, thus AH= 1/2X2X sqrt of 3= sqrt of 3. Thus AC=AH+CH= sqrt of 3+1.
Answer 1+ sqrt 3 =2.732 Since the triangle angles are 30, 45, and 105, with one side = sqrt 2, two triangles,30-60-90 and 45, 90 45 can be formed by drawing a perpendicular line from AC to angle 105. let's label the point on AC as P, then PA =sqrt 3 and PC =1. Therefore AC = 1 +sqrt 3 = 2.732.
We make a perpendicular line BM on AC, where M is on AC. If A is 30 and B is 105 then C is 45 so it means triangle MBC is isosceles right triangle so BM will be equal to sqrt of 2 then we calculate AM which will be 2 times BM as the 30 angle is there so AM will be 2 times sqrt of 2 then use Pythagorean theorem in triangle MBC to get MC then do the sum and u will get AC Edit: I was almost right as the hypotenuse of BMC will be BC
Got it in one, with the law of sines (which is good exercise because I never learned it in school). Nice exercise! 👍 (BTW, I'm glad you agree the use of standard triangles makes life easier 😉) ps If you want the last one exactly, the sine of 105 is (1+sqrt(3))/(2*sqrt(2).
sin105=sin(180-75)=sin75=(sqrt2+sqrt6)/4 Use sine rule, sin30/sqrt2=sin75/AC AC=(sqrt2*(sqrt2+sqrt6)/4)/(1/2)=2*sqrt2(sqrt2+sqrt6)/4 =(sqrt2+sqrt6)/sqrt2=(1+sqrt3) AC=(1+sqr3) units and that is our answer.
Well done Shadrana - but it would have been even nicer if you had included the line below. Jai Bhaarat Maa. sin(75°) = sin(45°+30°) = sin(45°).cos(30°) + cos(45°).sin(30°) = (√2/2).(√3/2 + 1/2 ) = (√2).(1+√3)/4 = (√2+√6)/4. .
Draw a line from B perpendicular to AC intersection at D, ABD is 30 degree and 60 degree right triangle and BDC are 45-45 degree right triangle, BD=CD=1, AD=1.732, so AC=2.732
Before watching: AC is 1 + sqrt(3). Reason: Create a point D on AC, so that it creates half of a square (meaning BD=DC) Since BC is the diagonale of this square and is sqrt(2), we know that the sidelinght of this square must be 1. The remaining figure ABD is a 30/60/90 triangle. As the short side is 1 (side of square), we know that AB must be 2 and AD must be sqrt(3). Thus AC = 1 + sqrt(3). (under one minute). After watching: exactly what I did! (The law of sines method is less pretty and a bit messy, as you need a calculator to do it...)
Actually, students should be taught to think of sqrt2 as the diagonal of a unit square and sqrt3 as the longer diagonal of a rhombus created by putting together two unit equilateral triangles. Then they no longer need to memorize the basic trig values of 45deg, 60deg, 30 deg, and, if they have a little more ability to visualise, even that of 15degs.
Sir as you know I am also a youtuber. I am using Camtacia studio for recording as well as editing my videos before uploading. But I like this software you and Mr. presh talwalkar are using. Can I have this software name or downloading link plzzz....
Еще мне нравится первый способ! Однако я хочу поделиться обоими методами и позволить моей аудитории решать. Спасибо за ваш отзыв! Ваше здоровье! Ты классный 😀
Ponto H | AC | Que seja um ponto sobre a verdade! Se Abaixarmos uma Perpendicular do Ponto B com (105) º ao Ponto H, Dividindo o Ponto B com (105) º em Dois Ângulos Diferentes, Um com (45) º e o Outro (30) º (60) º Teremos um Triângulo direito com) º (90) º! BHC também é | BH | ^ 2 + | HC | ^ 2 = | BC | ^ 2 de | BH | = | HC | = x; => 2 * x ^ 2 = 2; => x ^ 2 = 1; x = 1; Será! Portanto, x = 1; x = | HC |; | HC | = 1; É possível! | BH | = 1; Oposto (30) º No Triângulo Den (30) º (60) º (90) º | BH | = 1; O oposto (60) º será | AH | = √3! E a resposta à sua pergunta | AC | Seu comprimento é | AC | = | AH | + | HC |; Será a soma de seus comprimentos! Isto também | AC | = √3 + 1 ; Vai estar em forma!
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Very interesting mathematics problems. Good luck!!!!!!
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As an industrial electrician who use 30 and 45 special triangle formula to help bend conduit I find these videos very useful. Although most electricians use multiplier formulas. I like to think of bends as triangles and use formulas like offset height divided by the sine of any degrees. This. Give me my hypotenuse or space in-between bends. So for students who doesn't think math is important, it is because without it I could not put food on my table for my children. They would have no phone to even watch RUclips videos. Lol. So ingrain these formulas theorems and tricks in your head so you can provide for your family in the future
Third Method:
* drop line from C to intersect line AB extended at point D, making the corner of right triangle CDB.
* CBD must be 75 degrees and then BCD is 15 degrees. Since BC is given as √2, then CD is √2(sin 75) and can be rewritten: 1/2*(√3+1)
* looking at the right triangle CDA, angle ACD is 60 degrees (45+15) and so length AC is: (1/2*(√3+1))/cos(60) = √3+1
This is more or less the way I did it.
Super
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Nancy same method I used💯
Reflect a copy of ABC along AC.
You get a combination of an equilateral triangle of side 2 and an Isosceles Right triangle with an altitude 1.
Hence AC = 1+✓3
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Using the half angle cosine formula sin(105)=cos(15)=cos(30/2) we get AC=1+sqrt(3). This solution has already been proposed by Simon Slade
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I think this is more of a mental math problem for anyone who remembers that sin 45 deg = cos 45 deg = 1/sqrt2, and sin 30 deg = 1/2 and cos 30 deg = sqrt3/2. These are easily remembered by visualising the unit square and unit equilateral triangle respectively. So the moment you realise the unknown angle is 45degs and one side is sqrt2, drop a perpendicular from the 105deg angle to the side opp it. Then you can easily work out the unknown sides. Visualising the basic figues -- unit square, unit equilateral triangle, and one or two more figures -- and knowing their proportions is good lifelong.
Draw perpendicular from B on AC. Half done. Now let the perpendicular touch AC at O. BO =1 CO=1 because it forms a right isosceles. BD =1/tan30 =root3. AC=1+root3
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@@PreMath I love solving your problems. I want you to post much more difficult sums
Because point C is a common height, you may wish to consider: AC.Sin(30)=BC.Sin(180-105) or AC=√2.Sin(75)/Sin(30)=2√2.Sin(75)
Thnx for the question it can also be written as 1/2√2=√3+1/AC2√2(i.e sin 75)=√3+1=2.732
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I have been learning about triangles, I love these brain teasers.
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At a quick glance , Extend AB to D, the intersection of a perpendicular to C. The angle BCD is 15 degrees and side AC = sqrt(2) * cos(15)/sin(30) = 2.732
Great stuff. I actually used a different method from both those you showed. To the right of the main triangle I constructed a narrow, imaginary triangle with angles of 90, 75, and 15 degrees. I won't go into all the details but I figured out at AC = (sin(75) x sqrt(2)) / sin(30). I arrived at the same answer as you.
Superbly maverick effort and execution sir ❤️🙏 without using law of sing and using law of sine extremely beautiful ❤️🙏❤️❤️🙏🙏❤️❤️🙏🙏🙏
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i use * for degrees as i have no degree thing ob my phone
if you make a perpendicular line from B to AC, and call the point D, angle BDC is a 90* angle. in ABC, you know angle A=30* and angle B=105* so angle C=45*. this means triangle BCD is a special 1:1:sqrt2 triangle, the hypotenus (BC) is sqrt2 so the 2 other sides (BD,CD) are 1. since angle D is a perpendicular, triabgle ABD has angle A =30* and angle D=90*, so angle B=60*. this means we have another special triangle with 1:sqrt3:2=BD:AD:AB. we know that BD is 1 so we can substitute: 1:sqrt3:2=1:AD:AB so it is a 1x multiplication which means AD=sqrt3 (AB is irrelevant). recall we want to know AC, that is also equal to AD+CD=sqrt3+1 and that’s the awnser
edit: there is probably an easier way with like the cos rule but i dont know that (is it a^2=b^2+c^2-2ab•cos(α)?)
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Super duper bumper easy question!!
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Draw the perpendicular BH to AC. We now have 2 triangles: BHC and BHA. The triangle BHC is a right isosceles in which the hyposetenuse BC =sqrt of 2 so the sides BH=CH = 1.
The triangle BHA is half of a equilateral triangle in which AH is the height and AB is the side, BH half of the side, so AB= 2 BH= 2, and AH= 1/2 ABX sqrt of 3, thus AH= 1/2X2X sqrt of 3= sqrt of 3.
Thus AC=AH+CH= sqrt of 3+1.
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Though a simple sum, very good explanation.
Answer 1+ sqrt 3 =2.732
Since the triangle angles are 30, 45, and 105, with one side = sqrt 2, two triangles,30-60-90 and 45, 90 45
can be formed by drawing a perpendicular line from AC to angle 105. let's label the point on AC as P,
then PA =sqrt 3 and PC =1. Therefore AC = 1 +sqrt 3 = 2.732.
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I have done within limit
Thanks
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We make a perpendicular line BM on AC, where M is on AC. If A is 30 and B is 105 then C is 45 so it means triangle MBC is isosceles right triangle so BM will be equal to sqrt of 2 then we calculate AM which will be 2 times BM as the 30 angle is there so AM will be 2 times sqrt of 2 then use Pythagorean theorem in triangle MBC to get MC then do the sum and u will get AC
Edit: I was almost right as the hypotenuse of BMC will be BC
Got it in one, with the law of sines (which is good exercise because I never learned it in school).
Nice exercise! 👍
(BTW, I'm glad you agree the use of standard triangles makes life easier 😉)
ps If you want the last one exactly, the sine of 105 is (1+sqrt(3))/(2*sqrt(2).
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sin105=sin(180-75)=sin75=(sqrt2+sqrt6)/4
Use sine rule,
sin30/sqrt2=sin75/AC
AC=(sqrt2*(sqrt2+sqrt6)/4)/(1/2)=2*sqrt2(sqrt2+sqrt6)/4
=(sqrt2+sqrt6)/sqrt2=(1+sqrt3)
AC=(1+sqr3) units and that is our answer.
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Well done Shadrana - but it would have been even nicer if you had included the line below. Jai Bhaarat Maa.
sin(75°) = sin(45°+30°) = sin(45°).cos(30°) + cos(45°).sin(30°) = (√2/2).(√3/2 + 1/2 ) = (√2).(1+√3)/4 = (√2+√6)/4.
.
Draw a line from B perpendicular to AC intersection at D, ABD is 30 degree and 60 degree right triangle and BDC are 45-45 degree right triangle, BD=CD=1, AD=1.732, so AC=2.732
Draw a perpendicular from B to AC to point D. So BD = CD = 1. Tan 30 = 1/AD. So AD = 1.73 So AC = 2.73
Nice video....grettings from zagreb Croatia
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Amazing keep it up
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√3+1ですね
BH⊥ACとなる点Hを作ればわかりやすい
優れたアプローチ
ありがとう！乾杯！
揺れ続ける😀
Нello dear,
It's not essential to draw an altitude from
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حل اخر
نفرض M مركز الدايره الماره برؤوس المثلثABC
MA =MB=MC =جزر٢
قياس الزاويه AMC= ١٥٠
وبالتالي يمكن حساب طولAc منالمثلث AMC
Also it seems a method to calculate/find exact value for sin(105°)=(1+sqrt(3))/(2*sqrt(2))~0.966...
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Draw the height bd,then bd= dc=1. Then tan (a) to find ad= square of 3
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Great video :)
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very well explained, thanks for sharing, happy holidays
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Before watching: AC is 1 + sqrt(3). Reason: Create a point D on AC, so that it creates half of a square (meaning BD=DC) Since BC is the diagonale of this square and is sqrt(2), we know that the sidelinght of this square must be 1. The remaining figure ABD is a 30/60/90 triangle. As the short side is 1 (side of square), we know that AB must be 2 and AD must be sqrt(3). Thus AC = 1 + sqrt(3). (under one minute).
After watching: exactly what I did! (The law of sines method is less pretty and a bit messy, as you need a calculator to do it...)
Super Philip
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Actually, students should be taught to think of sqrt2 as the diagonal of a unit square and sqrt3 as the longer diagonal of a rhombus created by putting together two unit equilateral triangles. Then they no longer need to memorize the basic trig values of 45deg, 60deg, 30 deg, and, if they have a little more ability to visualise, even that of 15degs.
Nice methods, thank you teacher 🙏.
You are welcome dear
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Wonderful Sir, as usual amazing
I did Sum of two angles for AC
sin(45+60) = sin45cos60+sin60cos45
The triangle could be drawn better to reflect the correct degrees of the angles. They are off even to the naked eyes.
Both method are Good!
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Think of triangle (15°, 15°, 150°)
AC/√2=√2(2+√3)/AC
AC^2=√2√2(2+√3)
AC=√(4+2√3)=1+√3
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Sine rule! Voila! I can sleep in peace
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من المغرب شكرا على المجهودات
يمكن استعمال علاقة sin في المثلثABC.
Using Law of sines, it gonna be √2/sin(30)=AC/sin(105) and solving gave AC=√3+1
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sin rule will apply
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Sir, Pakistan has send flowers for you 🌹🌹🌹🌹
So nice of you Sameer
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?=1+squrt(3), because sin(105)=sin(75)=(1/4)*(squrt(2)+squrt(6))
I solved with sinuses' formula and it took 3 mins
AC/sin(105)=√2/sin(30)
Sin(105) = cos(15)
Cos(15) = (√3+1)/(2√2)
Sin(30)= 1/2
AC = 2√2 (√3+1)/(2√2)
AC = √3+1
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sin(30°)=1/2
@@seroujghazarian6343 Thank you
Corrected the slip.
I commit such slips. I will be alert in future before registering my response
AC=√3+1≈2,732≈2,73
Right triangle BDC
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So nice of you Mahalakshmi
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Right triangle ABD
Because you have given same time problem
x + 105° + 30° = 180°
x + 135° = 180°
x = 45°
#RightTriangle
Why can’t we use sine law and cosine law
AC = AD + DC
AC = sqrt (3) + 1
AC = 1 + sqrt (3)
I got the first solution just by looking at it.
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Geometrical proof is easily comprehensible by secondary level students,I think..
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The screen is black. I heard the voice though.
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√3+1 = 2.732 easily.
Super
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I've done the first method,I could not remember the sine method.
No worries
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Usaria Lei do Seno
Sir as you know I am also a youtuber. I am using Camtacia studio for recording as well as editing my videos before uploading. But I like this software you and Mr. presh talwalkar are using. Can I have this software name or downloading link plzzz....
Please send an email at premathchannel@mail.com
Thanks for asking😀
#lawofsine #sine #Trigonometry
Sum of angles = 180°
sin(105°)=sin(90°+15°)=cos(15°)=cos(45°-30°)=cos(45°)cos(30°)+sin(45°)sin(30°)=(sqrt(2)/2)(sqrt(3))/2)+(sqrt(2)/2)(1/2)=(sqrt(6)+sqrt(2))/4
---------------------
sin(30°)/sqrt(2)=sin(105°)/AC
sqrt(2)/4=(sqrt(2)(sqrt(3)+1))/(4AC)
(sqrt(3)+1)/AC=1
AC=sqrt(3)+1
#triangle
Id say sine(105)sqrt(8)
Sine Law.
#sumofangles #anglesum #sum
Первый метод зачетный! Второй-фигня!
Еще мне нравится первый способ! Однако я хочу поделиться обоими методами и позволить моей аудитории решать.
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Ты классный 😀
@@PreMath 😉👍
AC=2sqrt2sin75
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Hi sir 👍❤️🙏🙏🙏🌆🌆🎄🎄⛄🎉🎉⛄🎄🌆
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#306090specialtriangle #306090triangle
Just laughing at how this video solves both in over 6 minutes. 😉
2nd Comment and 5th like
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Just reminding y'all that this is for grade 4
AB*AB)+(BC*BC)=AC and
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have u reached the comment i had dropped for previous one episode?
Yes, I just saw your comment now and replied!
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Is it 11 class
Ponto H | AC | Que seja um ponto sobre a verdade! Se Abaixarmos uma Perpendicular do Ponto B com (105) º ao Ponto H, Dividindo o Ponto B com (105) º em Dois Ângulos Diferentes, Um com (45) º e o Outro (30) º (60) º Teremos um Triângulo direito com) º (90) º! BHC também é | BH | ^ 2 + | HC | ^ 2 = | BC | ^ 2 de | BH | = | HC | = x; => 2 * x ^ 2 = 2; => x ^ 2 = 1; x = 1; Será! Portanto, x = 1; x = | HC |; | HC | = 1; É possível! | BH | = 1; Oposto (30) º No Triângulo Den (30) º (60) º (90) º | BH | = 1; O oposto (60) º será | AH | = √3! E a resposta à sua pergunta | AC | Seu comprimento é | AC | = | AH | + | HC |; Será a soma de seus comprimentos! Isto também | AC | = √3 + 1 ; Vai estar em forma!
Ans : (_/3 + 1)/_/2. *(Solved in few seconds mentally)
Except it's sqrt(3)+1, not divided by sqrt(2).
And I'm calling BS on your having solved it in your head in seconds.
😂
@@northeastchronicles9311 What's your problem???
Honestly this question seems to me easiest question!! There is nothing called tuf I found in this problem
Well done keep going we really need your number or any ather place can we share to you what we con fuse
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