Find the Area of the Quarter Circle | Step-by-Step Tutorial

[@procash1968]

Superb as usual PreMath Teacher is really good What I like the most is his way of taking time & patiently explaining so that even a Maths Hater will understand & start likeing Maths Respects to you Maths Guru ji. Keep up this good work

[@PreMath]

So nice of you Pracash dear. You are awesome. Thank you! Cheers! Stay blessed😀 Love and prayers from the USA!

[@СергейГришин-е4и]

Решение гораздо проще, а данные о равенстве отрезков АВ и ВС излишне.

[@jayquirk2297]

Nice problem and great solution, as always. Here's an alternative that uses less trigonometry. Mirror the original quarter circle to create a circle with the triangle replicated four times. The outer edges of the four triangles will create a square inscribed within the circle. Each side of the square will sqrt(3) + 1 in length. (the a and b sides of the triangle) From there, the radius is the center of the square to the outside diagonal, r= side(length)/2 * sqrt(2) = (sqrt(3)+1)*sqrt(2)/2 . Then use standard area of a circle divide by four and the equation simplifies to pi*(2+sqrt(3)/4 and you can use a calculator to get the rest of the way.

[@PreMath]

Very well done. Thank you Jay! Cheers! Keep rocking😀

[@mahatmapodge]

That's brilliant, I did it the same way as the video, but for triangle ADB, but yours is a much more elegant solution, the diameter of the circle is the hypotenuse of an equilateral right triangle with side length (AD)+(DC) that's elegantly simple, I like it

[@youngboydan4153]

Very cool! Thanks for sharing!!!

[@waheisel]

I used the cosine sum formula. I like your way, more elegant.

[@josemath6828]

Facil por trigonometría

[@buxeessingh2571]

You can also get the radius BD at the end by Ptolemy's theorem. Quadrilateral ABCD is cyclic because its opposite angles are supplementary.

[@PreMath]

Good idea Thank you Buxees! Cheers! Keep rocking😀

[@josemath6828]

Correcto amigo.

[@timeonly1401]

Beautiful and elegant way to get the radius! I only learned about Ptolemy's theorem, cyclic quadrilaterals, their properties, etc., because of this post (We covered quadrilaterals in high school geometry, but I don't think we covered the cyclic stuff).Thx! Those ancient geometers and mathematicians continue to amaze and astound me!! The longer I live, the smarter these guys get! ;-)

[@ProfessorDBehrman]

There is a slight error: cos 105 degrees = -0.25882, NOT -0.256. The final answer should be: 2.9311

[@phungpham1725]

Thank you so much for your nice explanation. Honestly, I forgot all about trigonometry, I don't know the Ptolemy's theorem either! So I find a way to solve the problem using pure geometry: We know that AC=2, BC= sqrt of 2, CD=1, now we need to calculate the radius BD: Just draw the circle of which the diameter is AC, so this circle goes pass point A and D-----> the angle BDC=angleBAC= 45 degrees( having the same arc BC), and the angle DBC=angle CAB=30 degrees ( having the same arc CD). Calculating BD: consider the triangle BCD: drop the height CH to BD, now we have 2 special right triangles: BHC is a 30-90-60 triangle so BH = 1/2 (sqrt of 2)x sqrt of 3 CHD is a right isosceles triangle so DH = 1/2(sqrt of 2) Thus BD = 1/2 (sqrt of 2)x sqrt of 3 + 1/2 (sqrt of 2) = (sqrt of 3+1)/sqrt of 2 sq BD= 2+ sqrt of 3 The area of the quarter circle= 1/4 x pi x (2+sqrt of 3) =2.931 sq units

[@ОльгаСоломашенко-ь6ы]

This will be more accurate. Cos(105)=cos(45+60)=cos(45)cos(60°)-sin(45)sin(60)=√2/2*1/2-√2/2*√3/2=(√2-√6)/4. r2=2+1-2*√2*(√2-√6)/4=3-(4-4√3)/4=3-1+√3=2+√3=2+1.732=3.732.

[@PreMath]

Wow. Great job Olga Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@заводмихельсона]

Вы могли это решить не выходя за рамки 8 класса.Кроме того,что катет против угла 30 градусов вдвое меньше гипотенузы и теоремы Пифагора здесь ничего и не надо.Зачем вы им подсказали,как найти косинус суммы.Теперь они посвятят этому следующий ролик.И да,есть решение этой задачи старое,как яйцо динозавра,для произвольного угла.Глобальное дополнительное построение,и ничего считать не придется,от слова совсем.

[@silvaadamyan9098]

cos(105?) = cos(90? + 15?) = - sin(15?) ~ - 0,2588 - this way is much simpler

[@josemath6828]

Muy bien.

[@ШурикФумо]

The shame and the degradation of the science, i guess. My teacher would smack me upside the head for the approximate answer).

[@JoshiSings_1345]

A good teacher can inspire hope, ignite the imagination, and instill a love of learning. you are such a great teacher sir

[@PreMath]

Wow, thank you Sai So nice of you dear Keep rocking😀 Love and prayers from the USA!

[@josemath6828]

Mu ybien.

[@youngboydan4153]

I love your videos! Thank you. Here was my process: Dropped an altitude (let’s call it DE) from D to line BC creating a right triangle BDE with BD as the hypotenuse. Then to get the legs of that triangle to do pythag thrm, I used the little right triangle DCE I created with DC as hypotenuse. Since angle DCE is 75 degrees, DE= sin75 and CE= cos75 Then did Pythagorean Theorem to get hypotenuse of triangle BDE, but I didn’t square root it since we will square it for the radius squared anyway ((Sin75)^2 + ((sqrt2)+cos75)^2) * pi/4 = approx 2.93 Maybe not the easiest way, but it’s how my brain worked it out ;)

[@PreMath]

Very well done. Great effort. Thank you Young! Cheers! Keep rocking😀

[@josemath6828]

Correcto amigo.

[@geraldillo]

I love these kind of puzzles , thanks for posting! I did it differently; Because triangle ACD is a right angle triangle, angle ACD = (180-90-30)º= 60º I drew a line from point D, downward and parallel to line AB I call the intersection with the horizontal radius point E angle ECD = 180º-(angle BCA +angle ACD)º= 180-(45+60)º= 75º Because line CD=1; line CE = cosine 75º line DE = sine 75º cosine 75º= (√6-√2)/4 sine 75º= (√6+√2)/4 the hight of triangle BDE= sine 75º the base of triangle BDE= (line BC+line CE)=(√2+cos 75º)= √2+(√6-√2)/4= (√6+3√2)/4 line BD= radius Now you can use the pythagorean theorem to solve for r; (line BC+line ce)^2+(line DE)^2=r^2 (√2+cos 75º)^2+(sin 75º)^2=r^2 ((√6+3√2)/4)^2+((√6+√2)/4)^2= r^2 Doing the math gives; r^2=2+√3 Area of the quarter circle= pi*r^2/4= pi(2+√3)/4

[@PreMath]

Great job Geraldillo Thank you! Cheers! Keep rocking😀

[@francois8422]

BD = radius sought (of the quarter circle) O= midpoint of AC = circumference center for A, B, C, D OA=OB=OD=OC=1 Angle< BOD=2*

[@philipkudrna5643]

Yes!! I did it exactly the same way. Since I didn’t know the properties of the 30-60-90 Triangles by heart, I had to use the law of sines to arrive at CD=1. AB=BC=sqrt(2) was easy and the I even remembered by heart the law of cosines formula and did the correct calculation! I am really proud of myself, because that was not an easy one and required rather advanced (in my view) knowledge of trigonometry. An interesting problem. Keep them coming!

[@PreMath]

Super Philip Thank you! Cheers! Keep rocking😀

[@josemath6828]

Muy bien.

[@eleall5295]

Here is more desirable solution On step3, I found 180° - 75° = 105° I used some trigonometric theorem that if theta is acute angle, cos(180° - theta) = - cos(theta) And I drew two expressions On triangle ABD, use law of cosines r² = 5 - 2*sqrt(6)*cos75° On triangle BCD, use law of cosines r² = 3 - 2*sqrt(2)*cos105° = 3 + 2*sqrt(2)*cos75° We can solve the problem by multiplying sqrt(3) and both sides of 2nd equation, add each side of 1st and 2nd equation Therefore cos75° is gone Only r² remains (sqrt(3)+1)*r² = 5+3*sqrt(3) The reason why this solution is more desirable is this solution have a great flexibility to problem with non-common angles Even if problem is presented with non-common angles, which have trouble of using Angle Sum Identities, my solution is already generalized to non-common angles Thanks for reading

[@marioalb9726]

Searching the inscribed square, its side is: s = 2 cos 30° + 2 sin 30° s = √3 + 1 = 2,732 cm Diameter of circle is equal to diagonal of inscribed square d²= 2s² d = 3,8637 cm Area = ¼ . ¼πd² Area = 2,931 cm² Much easier than this complicated video !!

[@AnonimityAssured]

I made it 2.931 by applying the cosine rule to the upper triangle rather than the lower one. An alternative method was to reproduce the right triangle four times, to get a square with side-lengths √3 + 1, using the handy 30° triangle. That made r² = ((√3+1)√2/2)² = √3 + 2, so that the area of the quarter circle = (√3 + 2)π/4 ≈ 2.931.

[@PreMath]

Super job Julian Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@josemath6828]

Correcto .

[@kennethstevenson976]

I got the same answer, the same way.

[@dougustine]

I really like how you started with a strategy, I am almost 50 and I love these videos.

[@dougustine]

I thought you would avoid using the cosine law. If you are to jump into sin cosine etc, it would have been faster to look at sin(45)*H to get BC which is Root2/2*2 or root2, Cos to get AD. It seems simpler in my mind to stick to one system. But again I LOVE your channel and watch almost everyone that pops up. Sometimes I go back and try to solve the same equation using different methods. Keep up the problems

[@Ivan-Matematyk]

From Ptolemy's theorem follows 2*r=sqrt(3)*sqrt(2)+1*sqrt(2). Thus r^2=(sqrt(3)+1)^2/2.

[@PreMath]

Great! Thank you! Cheers! Keep rocking😀

[@drpkmath12345]

Very nice problem my friend! Very thorough step-by-step tutorial for sure. Keep up the good work my friend!

[@PreMath]

So nice of you my friend Thank you! Cheers! Keep rocking😀

[@sameerqureshi-kh7cc]

New formula of math.....soft spoken person + mathemagician = teacher of premath 😊👍🌹

[@PreMath]

Wow, thank you Sameer So nice of you dear Keep rocking😀

[@robertberg1609]

Nice problem and solution as always. Turns out that cos(105) equals -sin(15) which can be derived by using sin(45-30). Long story short. You get cos(105) = ((sqrt(2) - sqrt(6))/4. Put that in the expression for r

[@MAREKROESEL]

Elegant version: Let S be a centre of AC. BS = 1, SD = 1, => BSD is isosceles,

[@davidp4427]

Interesting, you can rotate the diagram through the other three quadrants which results in an inscribed square with side equal to the two sides of the original triangle (1 + square root of 3). The diagonal of the inscribed square is the diameter of the circle. Calculating the diameter and area, you get the area of the quadrant as 0.93 * pi.

[@3057luis]

You can also use AD and DC straight lines to get two equations in the form y=ax+b. Solving this system equation will give the x,y coordinates of point D, and than r^2=x^2+y^2.

[@tiborkote]

Was using the same first 3 steps, but then drawed a vertical line from D point downward to BC extension, calling it E point at the end. Created a small right triangle (CDE) where can can calculate the horizontal (CE) and vertical (ED) sides with Cos (75) and sin (75). This way skipped to use of Law of Cosines, maybe using up part of its definition. Then radius can calculate from BED right triangle as we know the BE and ED sides.

[@marioalb9726]

Searching right triangle with hypotenuse BD = r One cathetus is a = 2 sin 30°. sin 75° a = 0.9659 cm Similarity of triangles: 2 / 1 = r / 0,9659 r = 1,9318 cm Area = ¼πr² Area = 2,931 cm² ( Solved √ ) Much easier than this complicated video

[@EnnioPiovesan]

A,B,C,D are four points in a circle with center in the midpoint of AC (triangles ABC and ADC are both rectangles). By Ptolemy's theorem AC*BD=AD*BC+AB*DC so: 2r=√3√2 + √2 --> r=(√2/2)(1+√3). Area=(1/4)π (1/2)(1+√3)² = (1/8)π(4+2√3)=(π/4)(2+√3)

[@illyriumus2938]

Once i learned the cos formula from you, now I solve everything with cos 😂. I am really getting good at this thanks to you!!

[@PreMath]

So nice of you dear. You are awesome. Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@josemath6828]

Correcto.

[@michaelkouzmin281]

I solved it using analitical geometry assuming B=[0;0] and finding equations for lines containing [AD] and [DC], thus coordinates of D [(sqrt(6)+3sqrt(2))/4; (sqrt(2)+sqrt(6))/4]. Then r^2=x^2+y^2; r^2=sqrt(3)+2; Asec= pi(sqrt(3)+2)/4= 2.9311 sq.units

[@PreMath]

Very well done. Thank you Michael! Cheers! Keep rocking😀

[@LogicQuest]

Love your work..keep posting these questions..

[@PreMath]

So nice of you dear Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@ExpressStaveNotation]

I realised that all the points ABCD themselves lie on a circle centered at mid AC (because of the right angles at B and D). This circle is radius 1. Call the center O. OCD is equilateral so angle COD is 60. Angle BOC is 90. So BOD is 150. Then I used the cos rule on triangle BOD. r^2 = 1^2 + 1^2 - 2 (1)(1)cos150. = 2 - 2(-r3/2). = 2+r3. So Area = (2+r3)pi/4.

[@alainpeugny1146]

Yes! And this triangle BOD is isosceles as BO = OD = 1. As said, BOD angle equals 150° and so the two side angles, OBD and ODB, are both equal to 15°. The radius can be easily calculated : r = OB.cos15 + OD.cos15 = 2.cos15 = 1.9318516. r^2 = 3.73205 (and not 3.724 as estimated on the video… 😎)

[@luigipirandello5919]

Amazing solution. Thank you Sir. You are a inspired teacher. Everything you explain become easy to understand. I feel myself motivated.

[@PreMath]

So nice of you Luis dear. You are awesome. Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@josemath6828]

Muy bien.

[@emmanuelcollet224]

I used a copy and rotation by 90 degres of the 2 triangles. I have a cord with a length = 1 + sqrt(3). I define x as the rayon minus the length of the isoceles triangle, x = r - sqrt(2) use the cord theorem with the diameter and the cord 1 * sqrt(3) = x * RestOfDiameter 1 * sqrt(3) = (r - sqrt(2)) * (r + sqrt(2)) so r^2 = 2 + sqrt(3) and so area of the quarter circle = pi / 4 * r^2

[@montynorth3009]

Also works using triangle ABD. Same use of cosine formula. Sides, the square roots of 2 & 3 and the included angle of 75 degrees.

[@PreMath]

Excellent Monty Thank you! Cheers! Keep rocking😀

[@johnryder1713]

Nothing like a good Triangle problem, Thanks a lot

[@PreMath]

Most welcome! Thank you John! Cheers! Keep rocking😀

[@pranavamali05]

Wow nice question, the quality of questions u bring makes me view and like ur vedio everyday

[@PreMath]

Glad to hear that So nice of you Pranav Thank you! Cheers! Keep rocking😀

[@Deluthan77]

Another trig-free method: (1) As before, triangle ACD is 30-60-90, from which we get that angle ACD=60 degrees and CD has length 1. (2) Now, add a point P on line AC such that BP is the perpendicular bisector of AC. Also draw in the line DP. (3) It is easy to show that BPC is a 45-45-90 triangle while DPC is an equilateral triangle and all segments connected to P are length 1. (4) Hence, BPD is a 150-15-15 isosceles triangle. (5) Hence, angle DBC is 30 degrees and CDB is 45 degrees. (6) Finally, calculate the radius BD by partitioning the triangle BCD into 30-60-90 and 45-45-90 sub-triangles. This leads to a radius squared of r^2=2+sqrt(3)

[@predator1702]

Amazing solution 👍😊, thank you teacher 🙏.

[@PreMath]

You're welcome 😊 You are the best Thank you! Cheers! Keep rocking😀

[@gaopengjiang4294]

Connet BD and set two lines from D to be perpendicular with BA and BC at points E and F, respectively. Then, the triangle ADE and triangle DCF are similar triangles. Based on the ratio rules between sides and the Pythagorean theorem, it could be found that BD (the radius of the circle) is twice the length of DF. And DF could also be solved by the ratio rules between the sides of triangle ADE and triangle DCF. It is found to be (sqrt(6)+sqrt(2))/4. Thus, the radius is (sqrt(6)+sqrt(2))/2, and the area of the quarter circle is pi*(2+sqrt(3))/4.

[@1luffiz]

Ptolemy's theorem: (1×√2+√3×√2)/2=R. Square=πR^2/4=((√6+√2)/2)^2×π/4

[@prime3011]

Very interesting. But I don't like to use scientific calculater (be cause in Ukrainian schools we never used calcilators). And here we uses it for cos(105). So I tried not to use it and found a method. The Middle of AC - let it will be G. It is the center of circle around right triangle ACD. So AG=CG=GD=1. Let's calculate the BG. BG. It is Mediane of our triangle ABC and a bissectrise of this triangle and the hight of this triangle. So ffrom right triangle AGB we may calculate that BG=1. So, we see that it is a part of our circle around triangle ACD and BG, BD are rediuses of this circle. And so angle BGD is 2 angles BAD = 2*(45+30)=2*75=150 degrees. And cos(150)=-cos(30). So we use the theorem of cosines in triangle BGD: BD^2=1+1-2*1*1(-sqrt(3)/2)=2+sqrt(3). And then can calculate the Area=pi*(2+sqrt(3))/4. And than if use pi=3.14 and sqrt(3)=1.732 we may calculate and receive the same answer.

[@banazamalsadiwala7516]

Thank you so much for this video.

[@PreMath]

You are so welcome! Good to hear from you Banaz So nice of you dear Cheers! Keep rocking😀

[@josemath6828]

Correcto.

[@mcorruptofficial6579]

Hello friend, thanks for the example. It would be better, i guess, if you made reasoning this problem in backward way. That is, first how to find the quarter-circle area? Ok, we have to get the radius. Second, to find the radius r equals BD we have to consider the line BD as a leg of the triangle BCD. Third, we should account all the legs and angles inside of the triangle BCD And so on... - i think your got this idea - backward way 👍 So we do in electronic-circuit inventing/engineering. Kind Regards ✊

[@michaelstahl1515]

I choosed the same way but I choosed triangle ABD for solution. The ressult was as well. I`m happy to see asolution with cosinus theorem .

[@geoffreyparfitt7003]

If you complete the problem using trig values as surds, 3.724 is replaced by (2 + root 3).

[@PreMath]

Yes, we could take that route as well. Thank you Geoffrey! Cheers! Keep rocking😀

[@crispinmcsticks]

Nice video! Couldn’t you get BC and AB via the Pythagorean Theorem? If it is given that BC = AB, and let AB = BC = x, you get X^2 + X^2 = 2^2 2X^2=4 X^2=(4/2) X^2=2 X=sqrt(2) Or is this a false assumption?

[@PreMath]

You are right, we can use Pythagorean theorem as well. 45-45-90 theorem is same as Pythagorean theorem. Thank you Crispin! Cheers! Keep rocking😀

[@josemath6828]

Buen planteamiento.

[@242math]

got it, thanks for the challenge bro

[@PreMath]

So nice of you my friend Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@ClaudioBrogliato]

Got a similar result with a different process. Once I had the quadrilateral sides and angles I noticed that the opposite angles both sum to 180°. Hence the quadrilateral is a ciclic one and you can apply the nice formula m n = ( a times c) + (b times d) where m and n are the diagonals and a b c d the sides.

[@vcvartak7111]

You don't need a cosine rule. Sum of the product of two opposite sides of cyclic quadrilateral is product of the diagonals. Hence r*2=sqrt(3)*sqrt(2) + 1*sqrt(2)

[@mightyconquers]

This really help me in class

[@PreMath]

Great Mighty Thank you! Cheers! Keep rocking😀

[@apollogamer5949]

Thanks 😊... Btw How do you do??

[@PreMath]

You are welcome Apollo! I'm doing fine😀 Thank you! Cheers! Keep rocking😀

[@millipro1435]

good 👍❤️♥️ now it's our turn to ask you x - x ^-1 = 1 finds the value of x knowing that the answer is in gold

[@PreMath]

x - x ^-1 = 1 => x^2 - x-1=0 => Use quadratic formula, we get: x=(1+- sqrt5)/2 two solution! Cheers😀

[@millipro1435]

@@PreMath yeep and my reference was to show the golden ratio 😉❤️

[@nirupamasingh2948]

You go to the minutest detail vvvvnice way of solving

[@washingtoncostasilva625]

I used the Ptolomeu theorem in the quadrilateral and found a better approximation of the radius.

[@andrewkoziel7470]

The was a bracket missing in the previous comment. The more precise solution is: Area = PI *(2+SQRT(3))/4 = 2.93114585... This could be obtained from the drawing by comparing similar triangles.

[@williamangelogonzales148]

If E is midpoint of AC, you can use cosine law where in the side length are 1 and 1 and the angle in between is 150 deg

[@moth36]

Thank you very much for this nice video

[@sumithpeiris8440]

In cyclic quadrilateral ABCD, use Ptolemy's Theorem 2 X r = V2 X (V3 + 1) r = V2 / 2 + V6 / 2 Sumith Peiris Moratuwa Sri Lanka

[@bentels5340]

Did it almost the same way. But I don't see why you are being inexact about it -- the answer is (3 + 2sqrt(2)sin(15))*pi/4 .

[@PreMath]

Good job Ben Keep rocking😀

[@WaiWai-qv4wv]

very thanks

[@PreMath]

Most welcome Wai Wai So nice of you dear Thank you! Cheers! Keep rocking😀

[@thed5818]

Good solution. (waiting for harder questions)

[@PreMath]

Thank you! Cheers! Keep watching😀 Greetings from the USA!

[@thed5818]

@@PreMath Thanks, greetings from Turkey! 🙂

[@mahalakshmiganapathy6455]

Thank you sir very good problem

[@PreMath]

So nice of you Mahalakshmi Thank you dear! Cheers! Keep rocking😀 Greetings from the USA!

[@servenserov]

1. Complete to full citcle; 2.Square (1+√3) × (1+√3); 3. Diagonal of a square = diameter of a circle = =√((1+√3)²+(1+√3)²); ................

[@sedatguzel4480]

cos 105 değerini bilmeyenler için : B ve D açılarının toplamı 180 olduğu için ABCD dörtgeni teğetler dörtgenidir. Aynı yayı gören çevre açılar birbirine eşittir. Buna göre , DBC açısı 30 ve BDC açısı 45 derecedir. C açısından BD kenarına dikme indirip 45-45-90 ve 30-60-90 üçgeninden de sonucu bulabilirler.

[@Okkk517]

The radius square is equal to (√2+cos(75°))^2 + sin(75°)^2 =√3+2.

[@williamwingo4740]

If it were nit-picking time, someone might say that the numerical value of -0.256 for cos 105 degrees is still only approximate. A purist would use the sum-of-angles formula and say that 105 degrees = 60 degrees plus 45 degrees; then from the sum-of-two-angles formula cos 105 = cos 60 cos 45 - sin 60 sin 45; then from the 30-60 and 45-45 right triangle identities, omitting the surgical details, we have: cos 105 = (1/2) (1 / sqrt(2)) - (sqrt(3) / 2) (1 / sqrt(2)) = 1 / (2 sqrt(2)) - sqrt(3) / (2 sqrt(2)) = (1 - sqrt(3)) / (2 sqrt(2)) which is an exact solution; but approximately -0.2588 by my calculator; which gives the same figure for cos 105 directly. A Google search also confirms -0.2588 [1]. I'm not sure where your figure of -0.256 came from--but hey, both figures are approximate. This gives rise to some slight differences between us in the rest of the calculation. Normally, I would be reluctant to contradict and/or argue with the professor; but then I got a lot of practice at that thirty years ago when I was a graduate student in medical school. (OTOH, see confession below [2]). Anyway, getting back on topic and using this exact figure for cos 105, we have r squared = 2 + 1 - (2 sqrt(2) ((1 - sqrt(3)) / (2 sqrt(2)) = 2 + 1 - (1 - sqrt(3)) = 2 + 1 - 1 + sqrt(3) = 2 + sqrt(3); a surprisingly simple result. The calculator gives approximately 3.7321. Then the area of the quadrant is (pi r squared / 4) = (pi) (2 + sqrt(3)) / 4 and again the calculator gives approximately 2.9311. Thank you, ladies and gentlemen; I'll be here all week.😎 1. www.google.com/search?client=avast-a-1&q=cosine+105+degrees&oq=cosine+105+degrees&aqs=avast..69i64.7j0j7&ie=UTF-8 2. And if it were confession time, I might admit that I had a hard time with the calculator--until I realized it was set to radians.

[@homerotreto9391]

Not familiar with Celsius?🤔🤭

[@rishabhjain728]

By using Ptolemy theorem 2r=√2(√3+1) r=(√3+1)/√2 Area=π(2+√3)/4 =2.9311sq.unit (approx)

[@orkuntanverdi3005]

How do we know the value of cos105 while we r on exam?

[@andrewkoziel7470]

The more precise solution is: Area = PI *(2+SQRT(3)/4 = 2.93114585... This could be obtained from the drawing by comparing similar triangles.

[@subramaniankrishnaswami7196]

cos(105) = -0.259 not -0.253. Hence the answer is 2.93 not 2.92

[@rgrod6801]

I'm surprised nobody caught the error written on the upper left side @2:13 stating AB = AC as a given. It is not! I know it's just a typo error that happens to anybody once in a while. Otherwise it's still a good presentation.

[@yosepupithani5441]

cosine ratio is heart of the problem

[@PreMath]

Excellent Yosepu Thank you! Cheers! Keep rocking😀

[@yosepupithani5441]

@@PreMath thank you very much for video sir

[@billcame6991]

I solved it by figuring out that a square inside the full circle would have an edge of (1 + sqrt(3)) and figuring out that the radius would be an edge of an equilateral triangle from a 1/4 of that square with 45/45/90 degrees.

[@jdd2918]

Great vid

[@PreMath]

So nice of you Jedd Thank you! Cheers! Keep rocking😀 Greetings from the USA!

[@ΚΑΝΑΒΟΣΜΑΝΩΛΗΣ]

I didn't try...I hate calculations when the method is known...Thanks again

[@Teamstudy4595]

Ans : Pie(1 + _/3)/4

[@judydinallo3490]

great problem for homework

[@PreMath]

So nice of you Judy Thank you! Cheers! Keep rocking😀

[@josemath6828]

Muy interesante.

[@misterenter-iz7rz]

No compound angle formula, an isosceles triangle insides, 1,1 between angle 150, r^2=2+sqrt(3), then the answer is pi/4(2+sqrt(3)).😊

[@misterenter-iz7rz]

premath is progressing in large step.🎉🎉🎉

[@jussiano2761]

Olá tudo bem! O raio r^2= 2 + 3 - 2Cos(30+45).(6^1/2)

[@PreMath]

Obrigado pelo bom feedback. Você é demais. Cuide-se querido😀

[@brianfogarty8838]

Can anyone tell me why the instructor was able to rationlize a=2/Sq rt of 2 to a=square rt of 2. I appreciate it

[@homerotreto9391]

By multiplying for one (sqrt2/sqrt2)

[@kazkaz1003]

I used Ptolemy's theorem because I dont know the value of cos105°

[@PreMath]

Great jpb Thank you! Cheers! Keep rocking😀

[@sandanadurair5862]

Thank you. Today I learnt Ptolemy theorm. I could see ABCD as cyclic quadrilateral.. AB=BC= √2 AD = √3 CD= 1 AC = 2 BD = r As per Ptolemy theorem AB.CD=BC.AD =AC.BD √2.1+√2.√3 = 2.r r = (1+√3)/√2 r^2 = 3.732 Area = 3.732*π/4 = 0.933π sq.units Good practice for me. Thank you once again

[@Sergey-Primak]

AD+CD=sqrt(3)+1 R=[AD+CD]*cos 45 = [AD+CD]/sqrt(2) S = Pi*R^2/4 = Pi*[sqrt(3)+1]^2/8 = 2,93 unfortunately I cannot attach the image AD+CD - side of square inscribed in the circle

[@josemath6828]

Muy bien.

[@susennath6035]

Good

[@PreMath]

Thank you Susen! Cheers! Keep rocking😀

[@richardli681]

Cos 105 isn’t exactly accurate, I found a way that includes root 3 as the answer(roughly 1.732), which is more accurate than cos. It didn’t involve trig such as cos

[@dianeweiss4562]

Much better if we could find an expression for cos(105).

[@barryday9107]

cos(105) = cos(45+60) = cos(45).cos(60) - sin(45).sin(60) = sqrt(2)/4-sqrt(6)/4. Done. I think there was some misrounding in the calculation.

[@dianeweiss4562]

Use 1/2 angle formula for double angle of 210.

[@skakdosmer]

The “right angle” in the drawing is obviously less than 90˚. This makes me a little suspicious: Is it even possible to make the drawing correctly?

[@devkundan539]

Why b=1?

[@vidyadharjoshi5714]

Step 2 - AB = BC ( AB is not equal to AC but AB = sqrt2 )

[@__RamadhanSetyo]

Better show how to find w.out calculator. I mean, Show how to get the value of cos105°

[@alexniklas8777]

Решил аналогично

[@mahinnazu5455]

Sir can I give your some maths my Facebook page???

[@PreMath]

Please send email to: [e-mail] Cheers

[@kimphatnguyen9401]

Easy

[@PreMath]

Awesome Kim! Cheers! Keep rocking😀

[@owolabiquadri5405]

Good day premaths. I need clarification on this: if the figure is a quarter circle is the radius not longer than AB and BC as indicated in the diagram. I will like to know why we used AB as the radius. Thank you.

[@luborjakubec9212]

2.93

[@Teamstudy4595]

Sorry Sir I Solved it differently!

[@PreMath]

No worries! Keep rocking😀

[@simonedaems1141]

COS 105° = (- 0,259).

[@olivier9125]

on calculator cos de 105°= -0.2588190451 and not -0.256 so who is right ?

[@PreMath]

Dear Oliver, we are taking approximate values to avoid big string. Keep rocking😀

[@olivier9125]

@@PreMath approximate value would rather -0.259 and not -0.256

[@msafasharhan]

Please remove the black writing you do it in this video and the previouse video thank you

[@misterenter-iz7rz]

Attempt your old puzzle.😂, I guess you would refuse to adopt compoind angle formulas, even cosine rule? sqrt(2)×sqrt(2), a right- angled triangle, 2, 1, sqrt(3), a right- angled triangle, angle between 45+60=105, cos 105=cos 45 cos 60-sin 45 sin 60=sqrt(2)/2 1/2-sqrt(2)/2 sqrt(3)/2=1/4 sqrt(2)(1-sqrt(3)), r^2=2+1+(sqrt(3)-1)=2+sqrt(3)=1/2(4+2sqrt(3))=1/2(1+sqrt(3))^2, r=1/sqrt(2) (1+sqrt(3)), 1/2(sqrt(2)+sqrt(6)), therefore the area is 1/4 pi 1/4(8+2sqrt(12))=pi/4 1/2(4+2sqrt(3))=pi/4(2+sqrt(3)).😊

[@mustafizrahman2822]

Failed to solve it😥

[@PreMath]

No worries! You gave your honest shot. Keep rocking😀 Greetings from the USA!

[@sandanadurair5862]

Dont worry Rahman. I was looking for your solution!!!!!! I appreciate your humbleness at this age.

[@sandanadurair5862]

Hi Rahman I am 4.5 times elder to you. I used to watch your quick responses!!!!!!!!

[@mustafizrahman2822]

@@sandanadurair5862 Wow! Thanks.

[@christopherbalme123]

tu aurais pu être plus précis r² = (√2)² + (1)² - 2√2 x 1 x cos (105°) r² = 3 - 2√2 x cos (105°) calculons cos (105°) précisément cos (105°) = cos (45°) x cos (60°) - sin (45°) x sin (60°) cos (105°) = ((√2)/2 )x 1/2 - ((√2)/2) x ((√3)/2 ) cos (105°) = √2/4 - √6/4 remplaçons cos (105°) r² = 3 - 2√2 (√2/4 - √6/4) r² = 3 - (2 x √2 x √2 )/4 + (2 x √2 x √2 x √3)/4 r² = 3 - (2 x 2)/4 + (2 x 2 x √3)/4 r² = 3 - 1 + √3 r² = 2 + √3 calculons l'aire aire = 1/4 x π x r² aire = 1/4 x π( √3 + 2 ) aire = 1/4 x ((√3)π + 2π ) aire = ((√3)π + 2π )/4 ≈ 2,93114585 ≈ 2,93 et pas 2,92

[@заводмихельсона]

8 минут посвятить говнозадаче в два действия.Либо одним дополнительным микропостроениемДа еще и не получить точный ответ.Это что,африканские студенты на жизнь зарабатывают?

[@АнтонГробовщиков]

задачка решается проще без тригонометрии)) r^2 = 2 + V3