#31
|
||||
|
||||
ÍéÒ§ÍÔ§:
«Öè§Áѹ¨Ðà»ç¹ ¨Ø´àÃÔèÁµé¹¢Í§ µÃÕ⡳ÁÔµÔ áÅÐ ¤Ô´ Ptolemy's Theorem (à¡ÕèÂǡѺǧ¡ÅÁ)´éǤÃѺ ¤¹ä·ÂÁÑ¡àÃÕ¡ª×èÍÇèÒ »·ÍàÅÁÕ ÍФÃѺ
__________________
10 àÁÉÒ¹ 2010 20:27 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¤usÑ¡¤³Ôm |
#32
|
||||
|
||||
ãªèàŤÃѺ ààÅéÇãªéËÒ sin 1 ä´é»èÒǤÃѺ
__________________
¤ÇÒÁàÊÕ觷Õè¹èÒ¡ÅÑÇ·ÕèÊØ´äÁèãªè¡ÒáéÒÇ仢éҧ˹éÒààµè¤×Í¡ÒÃËÂØ´ÍÂÙè¡Ñº·Õè |
#33
|
||||
|
||||
ãªéÇԸբͧ Ptolemy ËÒ¤ÍÃì´áÅéÇ ãªéÊٵà $sin (x/2) = (cord x)/120$¤ÃѺ
__________________
|
#34
|
||||
|
||||
¤Ó¶ÒÁàà»Å¡æ¤ÃѺ
1.ãËé xààÅÐy à»ç¹¨Ó¹Ç¹àµçÁºÇ¡ààÅéÇ x=(13^2)-(12^2)ààÅÐy^2=(x^2)-((20+35-16-19)^2) ¨§ËÒ¤èҢͧ [(y-99)(y-97)(y-95).............(y-3)(y-1)]/[(y-99)(y-97)(y-95).............(y-3)(y-1)] 2.Çѹ˹Ö觢³Ð·Õè´.ª.¡ ä¡è à´Ô¹ä»ÃéÒ¹¢¹Á»Ñ§à¢ÒÊǹ·Ò§¡Ñºà´ç¡ 4¤¹ààµèÅФ¹¾Ò¹éͧÁÒÍÕ¡ 4¤¹ ¹éͧ¢Í§à´ç¡ààµèÅФ¹ÍØéÁààÁǤ¹ÅÐ 4µÑÇ ¶ÒÁÇèÒÁÕ¤¹ààÅÐààÁǨӹǹà·èÒä÷Õèä»ÃéÒ¹¢¹Á»Ñ§ ààµè§àͧ¤ÃѺ
__________________
¤ÇÒÁàÊÕ觷Õè¹èÒ¡ÅÑÇ·ÕèÊØ´äÁèãªè¡ÒáéÒÇ仢éҧ˹éÒààµè¤×Í¡ÒÃËÂØ´ÍÂÙè¡Ñº·Õè 11 àÁÉÒ¹ 2010 15:47 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¹Ñ¡Ê׺ÍѨ©ÃÔÂÐ |
#35
|
||||
|
||||
1.ãËé xààÅÐy à»ç¹¨Ó¹Ç¹àµçÁºÇ¡ààÅéÇ x=(13^2)-(12^2)ààÅÐy^2=(x^2)-((20+35-16-19)^2)
¨§ËÒ¤èҢͧ [(y-99)(y-97)(y-95).............(y-3)(y-1)]/[(y-99)(y-97)(y-95).............(y-3)(y-1)] $x=(13^2)-(12^2) = 5^2 = 25$ ààÅÐ $y^2=(x^2)-((20+35-16-19)^2)=25^2-20^2=15^2$ ¨Ðä´é¤èÒ y = +15 ËÃ×Í (y-15) = 0, ´Ñ§¹Ñ鹤èҢͧ $\dfrac{(y-99)(y-97)(y-95).............(y-3)(y-1)}{(y-99)(y-97)(y-95).............(y-3)(y-1)}$ ¨Ö§äÁè¹ÔÂÒÁ¤ÃѺ |
#36
|
||||
|
||||
ÍéÒ§ÍÔ§:
à¾ÃÒФÓÇèÒ "Êǹ·Ò§¡Ñº" äÁèä´é¡Ó˹´ÇèҨеéͧÍÍ¡ÁÒ¨Ò¡ÃéÒ¹¢¹Á»Ñ§àÊÁÍ令ÃѺ |
#37
|
||||
|
||||
¶Ù¡µéͧ¤ÃѺ ¤Ø³Puriwatt
__________________
¤ÇÒÁàÊÕ觷Õè¹èÒ¡ÅÑÇ·ÕèÊØ´äÁèãªè¡ÒáéÒÇ仢éҧ˹éÒààµè¤×Í¡ÒÃËÂØ´ÍÂÙè¡Ñº·Õè |
#38
|
||||
|
||||
·èÒ·Ò§¨Ðà§Õº¨Ñ§àÅ _kup
|
#39
|
||||
|
||||
Post ¡Ñ¹Ë¹èͤÃѺ
__________________
¤ÇÒÁàÊÕ觷Õè¹èÒ¡ÅÑÇ·ÕèÊØ´äÁèãªè¡ÒáéÒÇ仢éҧ˹éÒààµè¤×Í¡ÒÃËÂØ´ÍÂÙè¡Ñº·Õè |
#40
|
||||
|
||||
0.017452¤ÃѺ¼Á
__________________
(- -'') |
#41
|
||||
|
||||
Êâµ¹àΨ¹ì ÍÂÙè·Õèä˹ (¹Õèä§â¨·ÂìËÔ¹)
__________________
Fight Fight Fight to ......... |
#42
|
|||
|
|||
$Cosec 10^o - \sqrt{3}Sec 10^o=?$
|
#43
|
||||
|
||||
á¡ÐÃÐËÇèÒ§à¤Ã×èͧËÁÒÂà·èҡѺàͧà¹éÍ
$\displaystyle \csc 10^\circ - \sqrt{3}\sec 10^\circ = \frac{4}{\sin 20^\circ}(\frac12\cos 10^\circ-\frac{\sqrt3}{2}\sin 10^\circ)=4\frac{\sin 20^\circ}{\sin 20^\circ}=4$
__________________
¤¹ä·ÂÃèÇÁã¨ÍÂèÒãªéÀÒÉÒÇÔºÑµÔ ½Ö¡¾ÔÁ¾ìÊÑÅѡɳìÊÑ¡¹Ô´ ªÕÇÔµ(¤¹µÍºáÅФ¹¶ÒÁ)¨Ð§èÒ¢Öé¹àÂÍÐ (¨ÃÔ§æ¹Ð) Stay Hungry. Stay Foolish. 19 ¡Ã¡®Ò¤Á 2010 19:56 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 1 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ nongtum |
#44
|
|||
|
|||
ÊØ´ÂÍ´àŤÃѺ
|
|
|