tag:blogger.com,1999:blog-2787161678592839240.post4792110507526296876..comments2024-03-28T15:06:26.752+05:30Comments on www.mathsblog.in : Maths Blog for School Teachers & Students: ഉത്തരങ്ങള് മൂന്നുതരമായി.....ഇനിയും...?വി.കെ. നിസാര്http://www.blogger.com/profile/14303804236214732024noreply@blogger.comBlogger24125tag:blogger.com,1999:blog-2787161678592839240.post-76720329670819253962011-06-04T19:44:27.468+05:302011-06-04T19:44:27.468+05:30ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ...ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ആങ്കിള് Q 45 ഡിഗ്രീയാണ്.<br />ആങ്കിള് BCQ = 45 ഡിഗ്രി<br />ആങ്കിള് DCQ = 45 ഡിഗ്രി<br />AP = PQ<br />അതുകൊണ്ട് ആങ്കിള് DCA = DCQ ബൈ 2 =22.5 Degree<br />അതുകൊണ്ട് ആങ്കിള് DAC = 67.5<br />Angle CAB = 22.5 Degree<br />Angle PCB =67.5 Degree<br />Angle B = 90 Degree<br />So Angle CPB = 22.5<br />So Angle A + P = Qmanojhttps://www.blogger.com/profile/14581923669862547293noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-150228053899705272011-06-04T19:44:21.503+05:302011-06-04T19:44:21.503+05:30ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ...ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ആങ്കിള് Q 45 ഡിഗ്രീയാണ്.<br />ആങ്കിള് BCQ = 45 ഡിഗ്രി<br />ആങ്കിള് DCQ = 45 ഡിഗ്രി<br />AP = PQ<br />അതുകൊണ്ട് ആങ്കിള് DCA = DCQ ബൈ 2 =22.5 Degree<br />അതുകൊണ്ട് ആങ്കിള് DAC = 67.5<br />Angle CAB = 22.5 Degree<br />Angle PCB =67.5 Degree<br />Angle B = 90 Degree<br />So Angle CPB = 22.5<br />So Angle A + P = Qmanojhttps://www.blogger.com/profile/14581923669862547293noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-25118981147227663882011-06-04T19:43:46.592+05:302011-06-04T19:43:46.592+05:30ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ...ആങ്കിള് CQB ഒരു മട്ടത്രികോണം ആയതുകൊണ്ട് ആങ്കില് ആങ്കിള് Q 45 ഡിഗ്രീയാണ്.<br />ആങ്കിള് BCQ = 45 ഡിഗ്രി<br />ആങ്കിള് DCQ = 45 ഡിഗ്രി<br />AP = PQ<br />അതുകൊണ്ട് ആങ്കിള് DCA = DCQ ബൈ 2 =22.5 Degree<br />അതുകൊണ്ട് ആങ്കിള് DAC = 67.5<br />Angle CAB = 22.5 Degree<br />Angle PCB =67.5 Degree<br />Angle B = 90 Degree<br />So Angle CPB = 22.5<br />So Angle A + P = Qmanojhttps://www.blogger.com/profile/14581923669862547293noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-45892818898072280382009-10-19T16:52:50.106+05:302009-10-19T16:52:50.106+05:30Good Question.
Needs some kind of Mathematical Ene...Good Question.<br />Needs some kind of Mathematical Energy.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-36717636334555393212009-10-18T07:49:07.695+05:302009-10-18T07:49:07.695+05:30dear johnsir, go through my doubt of 6 th comment ...dear johnsir, go through my doubt of 6 th comment of oct 12 post.if any possibility pl comment.anonymousgeetha can also participate.VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-36140361864803244502009-10-18T07:48:45.725+05:302009-10-18T07:48:45.725+05:30Answer to the question using CoOrdinate geometry
L...Answer to the question using CoOrdinate geometry<br />Let A(0,0) P(a,0) Q(2a,0) B(3a,0) C(3a,a)<br />Slop of AB =m1 = 0 ( take it as x axis)<br />Slop of PC = m2 = tan Y = (a-0)/(3a -a) = 1/2<br />slop of QC = Tan Z = ( a-0)/ 3a-2a) =1<br />slop of AC = tan X = ( a-0)/(3a -0) = 1/3<br />Tan ( X+Y) = (tanX +tanY)/ 1- tanX *TanY<br /> = ( 1/3 +1/2 )/1- 1/3*1/2 = 1<br /><br />Tan z =1 <br />Tan ( X +Y) = Tan Z<br />since X , Y , Z are in the first quadrant, X+Y = Z<br />( Higher secondary students can use this method)JOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-33217036429040785832009-10-18T07:43:48.313+05:302009-10-18T07:43:48.313+05:301105 is a product 0f 5,13 &17(three prime nos....1105 is a product 0f 5,13 &17(three prime nos.)write 1105 as the sum 0f two different sqs .in 4 ways.VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-2560977957882731332009-10-18T07:27:56.048+05:302009-10-18T07:27:56.048+05:30(2n-1 )*(2n-1)+(3n+5)(3n+5) =(3n+1)(3n+1)+(2n+5)(2...(2n-1 )*(2n-1)+(3n+5)(3n+5) =(3n+1)(3n+1)+(2n+5)(2n+5) = R , second order ramanujan numberJOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-90860454793992612422009-10-18T07:18:35.044+05:302009-10-18T07:18:35.044+05:30Ramanujam number can be written as the sum of the...Ramanujam number can be written as the sum of the square of numbers in 6 different ways .try anonymous.VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-877545744705891082009-10-18T07:08:05.672+05:302009-10-18T07:08:05.672+05:3010sq+11sq+12 sq=13sq+14sq=365.
ithine mashanmar th...10sq+11sq+12 sq=13sq+14sq=365.<br />ithine mashanmar thanne veno?<br />6 sq+8sq+11sq+12sq= 365.<br />iniyum venamenkil ariyikkuka.VIJAYAN N Mhttps://www.blogger.com/profile/01196709026427033192noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-89789344435917323982009-10-17T22:37:28.898+05:302009-10-17T22:37:28.898+05:30അല്ലയോ കണക്ക് മാഷുമ്മാരേ, 365 നെ സംഖ്യകളുടെ വര്ഗങ്...അല്ലയോ കണക്ക് മാഷുമ്മാരേ, 365 നെ സംഖ്യകളുടെ വര്ഗങ്ങളുടെ തുകയായി എത്ര രീതിയില് എഴുതാന് കഴിയും? എത്ര രീതി ഇവിടെ വരുമെന്ന് നോക്കട്ടെ.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-19588000311290546072009-10-17T10:56:33.977+05:302009-10-17T10:56:33.977+05:30I ivite higher secondary students to solve this pr...I ivite higher secondary students to solve this problem using co ordinate geometry. Take the coordinates of P,Q ,B and P by taking a as the origin. Using the slopes of AC,PC ,QC and AB we can find the angles in the general form<br /><br /><br />Second year students can solve it using Vectot equations.This will be a good assignmet for youJOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-41265603083704313892009-10-16T23:34:07.711+05:302009-10-16T23:34:07.711+05:30Excellent.This is the real dicussion on a problem....Excellent.This is the real dicussion on a problem.Now we got 3 different solutions.Let us think ,share and appreciateJOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-58394180699179927842009-10-16T23:06:18.596+05:302009-10-16T23:06:18.596+05:30we can proove angleA=anglePCQ.
letAP=PQ=QB=a
QC=ro...we can proove angleA=anglePCQ.<br />letAP=PQ=QB=a<br />QC=root 2a,PC=root5a,AC=root10a.<br />Consider triangle's PQC,CQA..<br />PQ/CQ =a/root2a=1/root2<br />QC/QA=root2 a/2a=1/roo2angle A is common<br />so two triangle's are simillar .<br />angle A=anglePCQ<br />angle P+anglePCQ=angleCQB.<br />angle A+angle P=angle Q<br /><br />so X+Y=Z<br /><br />thomas.kulathuvayalsajan paulhttps://www.blogger.com/profile/11067627975897595499noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-75575989110939291102009-10-16T21:16:04.999+05:302009-10-16T21:16:04.999+05:30ശരിയുത്തരം നല്കിയ വിജയന് സാറിനും മുരളീധരന് സാറിന...ശരിയുത്തരം നല്കിയ വിജയന് സാറിനും മുരളീധരന് സാറിനും അഭിനന്ദനങ്ങള്. ജോണ് സാര് പറഞ്ഞ പോലെ ഇനിയും പുതിയ രീതികളെന്തെങ്കിലുമുണ്ടെങ്കില് വരട്ടെ. അതെല്ലാം നമുക്ക് ഗുണകരമാകുകയേയുള്ളു. കമന്റ് ബോക്സിലെ ഈ വിജ്ഞാനചര്ച്ചകള് അധ്യാപകരെന്ന നിലയില് നമുക്കെത്ര ഉപകാരപ്രദമാണ്. ഇനിയുമുണ്ടാകും വേറെന്തെങ്കിലും മാര്ഗങ്ങള്... അല്പം നീണ്ടതാണെങ്കിലും ഉടനെ തന്നെ മറ്റൊരു രീതി ഉടനെ ബ്ലോഗ് ടീമിന്റെ പേരില് പ്രതീക്ഷിച്ചോളൂ.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-72645677881349119432009-10-16T19:55:13.391+05:302009-10-16T19:55:13.391+05:30Yes ..... I am expecting this from Muralidharan ...Yes ..... I am expecting this from Muralidharan sir.The proof is completly based on first principles.I hope all mathematics teachers read the post will appreciate the two methods in the commend. Vijan sirs method of solving the problem is suitable in the 10 th standard while doing trigonometry.Surely I will give this in the class as a problem in the application of trigonometric ratiosJOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-59993137379852183742009-10-16T19:12:17.362+05:302009-10-16T19:12:17.362+05:30My name is MURALEEDHARANMy name is MURALEEDHARANMURALEEDHARAN.C.Rhttps://www.blogger.com/profile/12387715646323771032noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-29782182768902628332009-10-16T18:53:21.277+05:302009-10-16T18:53:21.277+05:30Let S,R be the points in DC such that PS=SR=RC
Dro...Let S,R be the points in DC such that PS=SR=RC<br />Drow PS & QR<br />Also drow QC & QS<br />Complete the square whose adj. sides are QC & QS<br />Let E be the 4th corner point of the square<br />Consider the 2 rht angled triangles PBC & AEC ,their perpendicular sides are in the ratio 1:2<br />It follows that these triangles are similar (a:b=x:y implies that a:x=b:y)<br />Hence their corresponding angles are equal<br />ie angle CPB (y) = angle CAE<br />ie x+y = angle CAB +y=45=z<br />Hence the resultMURALEEDHARAN.C.Rhttps://www.blogger.com/profile/12387715646323771032noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-77913022336928183782009-10-16T12:34:24.740+05:302009-10-16T12:34:24.740+05:30There is no "pidivashi " for a mathemati...There is no "pidivashi " for a mathematics teacher.I just invite others to solve the problem in another way.That is what is called divergent thinking .I think you are a mathematics teacher of high school classes. Solving problem is not an end. The dynamism of geometric concepts shoud be realised by the learner by our instructions and guidance.That is why we tell the children to attempt an assignment in different ways.How to apply basic theorems in rati, proportion and similarity will give us another experience to the learner.Dear anonymus,I am not a complicated mathematics instructorJOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-6401675995554714962009-10-16T11:12:12.350+05:302009-10-16T11:12:12.350+05:30Majority of the Std VIII IT Practical Questions ar...Majority of the Std VIII IT Practical Questions are from Second Term. What is the aim of this examination?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-12779846493534305702009-10-16T10:44:38.174+05:302009-10-16T10:44:38.174+05:30ഏത് രീതി ഉപയോഗിച്ചായാലും ഉത്തരം കിട്ടിയല്ലോ ജോണ് ...ഏത് രീതി ഉപയോഗിച്ചായാലും ഉത്തരം കിട്ടിയല്ലോ ജോണ് മാഷേ, കണക്ക് ടീച്ചേഴ്സ് ഇത്ര പിടിവാശി പിടിക്കണോ?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-76145956216539555272009-10-16T07:34:42.883+05:302009-10-16T07:34:42.883+05:30The proof given by Vijayan sir is based on trigono...The proof given by Vijayan sir is based on trigonometry.Try to prove using pure geometric concepts.I think geometric method will give us the real enjoyment of problem solving.JOHN P Anoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-18769700296749324962009-10-16T07:26:29.625+05:302009-10-16T07:26:29.625+05:30let AD=AP=PQ=QB=k
THEN
then angle Z=45
tanX=k/3k=...let AD=AP=PQ=QB=k<br />THEN <br />then angle Z=45<br />tanX=k/3k= 0.3333<br />X=18 DEGREE<br />TAN Y =k/2k= 0.5<br />Y= 27 DEGREE<br />therefore X+Y =18+27=45<br />THAT IS X+Y=Zvijayanhttps://www.blogger.com/profile/02345885159669791726noreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-29788734388027206802009-10-16T06:13:45.769+05:302009-10-16T06:13:45.769+05:30Good Question!
Let me think...
GeethaGood Question!<br />Let me think...<br /><br />GeethaAnonymousnoreply@blogger.com