Amazing question. However, since drawings aren’t always to scale, you should say at the beginning that points D & E are tangent to the circle. Without that information, the solution just assumes it.
@ghhdcdvv5069 Жыл бұрын
تمرين جيد جميل. رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويحميكم ويرعاكم جميعا. تحياتنا لكم من غزة فلسطين .
@waiphyoemg16683 жыл бұрын
I use Pythagorean theorem. Let CD=x. Draw BC. Angle ACB is right angle because this angle subtends on the diameter. In rt triangle ACB, BC^2 = AB^2 - AC^2=9^2 - (5+x)^2 = 81 - 25 - 10x -x^2 =56 -10x - x^2 In rt triangle DCB, BC^2 = DB^2 - CD^2 = 6^2 - x^2 = 36 - x^2 therefore, 56 - 10x - x^2 =36 - x^2 So, 10x=20 then x=2 CD=2
@philipkudrna56433 жыл бұрын
Nice problem! I had figured out that the diameter of the semi-circle must be nine. Knowing Al-Akashi’s Law of Cosines would have very much helped me in calculating angle alpha, which I had arrived at correct 38,94 degrees, but only with help of a calculator using tangent and the height of the inner smaller triangle. Very cumbersome and far less elegant. From there I also arrived at x=2 using Thales, but also only with a calculator. I like that it is possible in a more elegant way without the help of a calculator! Unfortunately I hadn‘t realized that you could also have solved it using Pythagoras and the Triangles ACB and DCB easily without a calculator!
@procash19683 жыл бұрын
Superb Sir. Wish someone like you was my Maths teacher at school. If so, then I may have done my graduation & post graduation in Maths & become a maths teacher myself
@MrPaulc2222 ай бұрын
2-tangents give D=9, so R = 9/2 Thales gives ABC and therefore, BCD, as right triangles. CD is x. CB is y. (x+5)^2 + y^2 = 9^2 (1) y^2 + x^2 = 6^2 (2) x^2 + 10x + 25 + y^2 = 81 (1) y^2 + x^2 = 36 (2) x^2 = 81 - y^2 - 10x - 25 (1) x^2 = 36 - y^2 (2) 56 - y^2 - 10x = 36 - y^2 Simplify: 20 = 10x x = 2 Wow! Astonishingly, I have used a simpler, cleaner method than your good self. As this is such a rarity, I might frame it :) .
@thirumalaithathachary31113 жыл бұрын
We can do it without involving trigonometry. Let CD=x. Join points C and B.. ' Triangle ACB and triangle DCB are right triangles. Therefore CB^2=36-x^2 ( Pythogorean Theorem ). Also, AC^2+BC^2=AB^2. Diameter of the circle=6+3=9. ( obvious ). Therefore (5+x)^2 + (36-x^2) = 81. Solve this equation to get x=2
@bkj5563 жыл бұрын
This the level which I wanted. Thanks....
@Sultan-nx9jn Жыл бұрын
It can be solved without trigonometry as well. If we take a length of CD as x, then CB will be equal to 9^2-(5+x)^2 or 6^2 - x^2. Therefore, 9^2-(5+x)^2=6^2-x^2 => 81-36=5*(5+2*x) => 5+2*x=9 => x=2 or CD=2.
@mustafizrahman28223 жыл бұрын
What a tremendous question it is! But I have failed to solve it. Thanks a lot for posting. I have saved it.
@PreMath3 жыл бұрын
No worries. Keep it up dear 😃
@HassanLakiss3 жыл бұрын
Thank you for a nice question and as usual a good explanation
@wafikhwijeh40063 жыл бұрын
It's exciting problem and very interesting to me when i was watching your kindy explaning... Then i ask you sir to present most of your solution problems if that is possible with most best Regardes fir you sir.
@johnbrennan33723 жыл бұрын
Very good method------ always interesting.
@PreMath3 жыл бұрын
Thank you John! Cheers!👍 Keep it up dear 😃
@lindafromcalifornia11553 жыл бұрын
That was so cool to watch you work that out.
@ovalteen44043 жыл бұрын
I figured out AC by placing point o at the center of the semicircle and drawing an isosceles triangle with both radii, splitting it into 2 right triangles, using the cosine to find the adjacent bisector of AC, and doubling it. So 4.5/(7/9)*2=7. Of course I've never been known to do things the easy way.
@predator17023 жыл бұрын
Thank you teacher. 👏👏👍
@PreMath3 жыл бұрын
You are very welcome Thank you 👍 Keep it up dear 😃
@ferroggon3 жыл бұрын
I applied the Pythagorean theorem to the DCB and ACB triangles.
@ohiovic12363 жыл бұрын
same
@markp72623 жыл бұрын
I also did that. As long as you knew Thales, it is a much simpler solve without needing to use Law of Cosines. 9^2 - (x+5)^2 = 6^2 - x^2, solve for x.
@ohiovic12363 жыл бұрын
@@markp7262 Agree .. Bravoo
@Mathematician61243 жыл бұрын
I drew a perpendicular from D to AB at point P to derive the cos theta. Let AP be equal to x and PB be equal to (9-x).This gave 25-x2=36-81+18x-x2, finally x =35/9 and cos theta equals AP /AD =7/9.
@وجديالنصرالله3 жыл бұрын
اسهل مما ظننت..شكرا
@harikatragadda3 жыл бұрын
ACB and DCB are both right triangles with a common side CB. (5+CD)²-9²=6²-CD² CD=2
@gandharvagrover83963 жыл бұрын
How is DCB right triangle? AD doesn't include the diameter for 90° angle between radius and tangent. Or you see it other way... Just asking...
@tosuchino64653 жыл бұрын
@@gandharvagrover8396 In a semicircle, the angle between the lines drawn from the edges of the diameter to ANY point on its arc is always 90 degrees.
@rolandhutapea63902 жыл бұрын
Yes, I did the same simple way...no need using cosine law
@xyz92502 жыл бұрын
Great idea. But it should be 9^2 - (5+CD)^2 ,
@mahalakshmiganapathy64553 жыл бұрын
Simply superb
@theoyanto Жыл бұрын
Great example, didn't get, had to watch the video, and didn't think to use cosine law, and you did say in your note that you would be using it, so my fault, should pay more attention, thanks again 👍🏻
@242math3 жыл бұрын
love this question, very well explained, thanks for sharing
@luigipirandello59193 жыл бұрын
Wonderful, Sir.
@xyz92502 жыл бұрын
I didn’t apply the law of cosine, but dropped a line from D perpendicular to AB, call it DG, as it’s shared side of two right angle triangles, AD^2 - AG^2 = DB^2 - BG^2, as BG= 9 - AG, can get AG= 35/9, as AC/AB = AG/AD , (5+ DC) / 9 = ( 35/9)/ 5, and DC= 2
@pedroloures33103 жыл бұрын
Nice problem! I tried using chords theorems but it didn't work very well. After some time I also came up with this solution. Thanks for the video.
@مجديمحمود-ذ5ذ2 жыл бұрын
فيثاغورث
@kunalchakraborty97352 жыл бұрын
I have solved this by pythagorous theorem 9^2 - ( 5 + CD) ^ 2 = 6 ^ 2 - CD^2 By solving this equation CD can be found. @cd=2
AB=3+6=9. Angle ACB =90 degrees. Combine Pythagoras in triangle ABC and triangle BCD. Resulting in DC=2.
@dipinds64463 жыл бұрын
Nice👍
@PreMath3 жыл бұрын
Thank you Dipin! Cheers! 👍 Keep it up dear 😃
@tahasami5973 жыл бұрын
Thank for premath
@billylowgroundvytaszukas57972 жыл бұрын
beautiful
@kiranbarnwal85033 жыл бұрын
CD=2 by laws of sine and cosine Nice question
@PreMath3 жыл бұрын
Excellent Kiran 👍 Keep it up dear 😃
@phanimaheswara74923 жыл бұрын
Thank you sir
@PreMath3 жыл бұрын
So nice of you Phani Thank you 👍 Keep it up dear 😃
@pranavamali053 жыл бұрын
Awesome
@mauriciosahady Жыл бұрын
Teorema de Pitágoras duas vezes. That's enough! Triangulos ACB and ADB
@recepgocmen68483 жыл бұрын
very good
@rodolforathakrishnan81283 жыл бұрын
cute and cool! 💎💪🏽😉
@PreMath3 жыл бұрын
Thank you Rodolfo 👍 Keep it up dear 😃
@kamaljain52283 жыл бұрын
Since you did not really use \alpha, but just cos(\alpha), it is possible to solve the problem without trigonometry.
@ExpressStaveNotation3 жыл бұрын
I wasn't content to just find CD = 2. So I worked out the radius of the green circle. (6r2/5)
@Mathematician61243 жыл бұрын
You can make it harder. Find the length of the chord or find out the radius. Would you like to do this? It is a good sum. Please respond, would you like to make it harder?
Yes. So did I. No need for trig to solve this proboem.
@niltonsilveira41992 жыл бұрын
CB=y: CD^2 + y^2=36 and (5+CD)^2+y^2=81 ====> CD=2
@qobilruzmatov483 жыл бұрын
Nice good
@PreMath3 жыл бұрын
Thank you 👍 Keep it up dear 😃
@VolkGreg3 жыл бұрын
AB = AF + FB = AE + DB = 3 + 6 = 9 cos a = (AB^2 + AD^2 - DB^2) / (2 AB AD) = (9^2 + 5^2 - 6^2) / (2*9*5) = 70/90 = 7/9 AC = AD + DC = AB cos a DC = AB cos a - AD = 9 (7/9) - 5 = 2
@theophonchana50253 жыл бұрын
Angle ACB = 90°
@theophonchana50253 жыл бұрын
cos (alpha) = (5 + CD) ÷ 9
@adgf1x2 жыл бұрын
CD=7-5=2 unit
@micke_mango2 жыл бұрын
I'm sorry, but I think this is sloppy math! At the very least, we're presented far to little information at the beginning of the problem. * The only tangent point we know of is F. D and E is never mentioned to be tangent points so the lines that presumably (?) connect to these points aren't necessarily tangents. Which means that we cannot use the two-tangent theorem without adding assumptions. * We also don't know if AC is a straight line, i. e. there could be an angle in D, between AD and DC (≠ 180°). That means we cannot use Thales' theorem without proving (or assuming) that point D is a point on the straight line AC. Your intentions with this problem are understandable, but unfortunately not mathematically obvious - the necessary geometric details are understated. A good practice is to describe how the geometry details are constructed in the beginning of the problem - that will pinpoint necessary assumptions
@مجديمحمود-ذ5ذ2 жыл бұрын
بمكن الحل بتطبيق فيثاغورث اسهل جدا جدا
@theophonchana50253 жыл бұрын
AD = 5 AE = 3 BD = 6 CD = ?
@theophonchana50253 жыл бұрын
AB = 9
@Nishant-us2il3 жыл бұрын
Watch at 1.5 x , it's slow other wise
@theophonchana50253 жыл бұрын
CD + 5 = 7 CD = 2
@PreMath3 жыл бұрын
Keep it up Theo dear 😃
@theophonchana50253 жыл бұрын
Diameter = 9
@theophonchana50253 жыл бұрын
BD = BF BF = 6
@theophonchana50253 жыл бұрын
AE = AF AF = 3
@sumitsagar19933 жыл бұрын
2
@TomKaren942 жыл бұрын
Good one. But, again, too many steps from 5+CD=7 to CD=2.Too detailed explanation. I jump ahead through these videos out of frustration.
@Teamstudy45953 жыл бұрын
2nd comment
@PreMath3 жыл бұрын
Thank you Jayant 👍 Keep it up dear 😃
@Teamstudy45953 жыл бұрын
Really tricky but. After seeing. Comments I realized x =2 (@ns)