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#1
20 àÁÉÒ¹ 2011, 20:46
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µÃÕ⡳ ¤ÃѺ
¡Ó˹´ãËé ÊÒÁàËÅÕèÂÁ ABC ÁÕ $\frac{sin A}{sin B} =\sqrt{2}$ áÅÐ $\frac{tan A}{tan B} =\sqrt{3}$
¨§ËÒ¤èҢͧ A,B ¤×ͼÁ·ÓÍÍ¡ÁÒáÅéÇäÁèä´éÍèФÃѺ ¢ÍÇÔ¸Õ·ÓẺ¡ÒèѴÃÙ»¹Ð¤ÃѺẺ¹Õé $cos^2 B = x$ $sin^2 B = 1 - x$ $tan^2 B =\frac{1-x}{x}$ $sin^2 A=2-2x$ $tan^2 A = \frac{3-3x}{x}$ $Êèǹ cos^2 A = \frac{2x}{3}$ ¼ÁàËÅ×Íá¤èËÒ x ÍèФÃѺáµè¶éÒ᷹ŧä»ã¹ÊÁ¡Òà $\frac{sin A}{sin B} =\sqrt{2}$ $\frac{tan A}{tan B} =\sqrt{3}$ Áѹ¨Ðä´é¤ÓµÍºáºº¹ÕéÍèФÃѺ 2=2 à¾×è͹ºÍ¡ÇèÒãËéËÒ¨Ò¡ $sin^2 A +cos^2 A = 1$ áÅéǨÐä´é áµè¼ÁÂѧ§ÍÂÙè·ÓäÁµéͧËÒ x ¨Ò¡ $sin^2 A +cos^2 A = 1$ ¤ÃѺáµèÃÙé¤ÓµÍºáÅéÇ ÍÂÒ¡¨ÐËÒ§§ 
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#2
20 àÁÉÒ¹ 2011, 22:04
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$sin^2A=2sin^2B$
$cos^2A=\frac{2}{3}cos^2B $ $2sin^2B+\frac{2}{3}cos^2B=1$ $2(1-cos^2B)+\frac{2}{3}cos^2B=1$ $6-4cos^2B=3$ $3-4c0s^2B=0$ $(\sqrt{3}-2cosB )(\sqrt{3}+2cosB)=0$ ⨷Âì¡Ó˹´ãËé$B$ à»ç¹ÁØÁã¹ÊÒÁàËÅÕèÂÁ ´Ñ§¹Ñ鹤èÒ $cosB$ äÁèà¡Ô¹ 180 ͧÈÒ à»ç¹ÁØÁáËÅÁ¡çä´é ÁØÁ»éÒ¹¡çä´é $cosB=\pm \frac{\sqrt{3}}{2} $ $sinB=\frac{1}{2} $ ´Ñ§¹Ñé¹ÁØÁ $B$ à»ç¹ä´é·Ñé§ $30$ ËÃ×Í $150$ ͧÈÒ áµèàÃҵѴÁØÁ $150$ ͧÈÒÍÍ¡à¾ÃÒкǡ¡Ñº¤èÒÁØÁ$A$ ·ÕèËÒä´éáÅéÇ à¡Ô¹$180$ ͧÈÒ $sinA=\frac{1}{\sqrt{2} } $ $cosA=\pm \frac{1}{\sqrt{2} }$ ´Ñ§¹Ñé¹ÁØÁ $A$ à»ç¹ä´é·Ñé§ $45$ ËÃ×Í $135$ ͧÈÒ áµèàÃҵѴÁØÁ $135$ ͧÈÒÍÍ¡à¾ÃÒкǡ¡Ñº¤èÒÁØÁ$B$ ·ÕèËÒä´éáÅéÇ à¡Ô¹$180$ ͧÈÒ ´Ñ§¹Ñ鹤èҢͧÁØÁ$A,B$ ·ÕèÊÍ´¤Åéͧ¡Ñºâ¨·Âì¤×Í $45,30$
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20 àÁÉÒ¹ 2011 22:06 : ¢éͤÇÒÁ¹Õé¶Ù¡á¡éä¢áÅéÇ 2 ¤ÃÑé§, ¤ÃÑé§ÅèÒÊØ´â´Â¤Ø³ ¡ÔµµÔ
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#4
21 àÁÉÒ¹ 2011, 00:45
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$mn = tan^2 A - sin^2 A$
$ = sin^2 A(\frac{1}{cos^2 A} - 1 )$ $ = sin^2 A(\frac{1-cos^2 A}{cos^2 A})$ $ = sin^2 A$ $tan^2 A$ $m + n = 2 tan A$ $m - n = 2 sin A$ $m^2 - n^2 = (m+n)(m-n)$ $= (2 tan A) (2 sin A)$ $= 4 sin A$ $tan A$ $ = 4 \sqrt{mn}$
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