tag:blogger.com,1999:blog-6933544261975483399.post4640714858441566346..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 211: 60 Degree Triangle, AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-71771729318623456992023-03-04T23:58:17.457-08:002023-03-04T23:58:17.457-08:00With normal notation, AB=c, AC=b and BC=a
By cosin...With normal notation, AB=c, AC=b and BC=a<br />By cosine law, b^2=a^2+c^2-2ac(cos60)<br />b^2=a^2+c^2-ac<br />------------------------------------------<br />Sc=c^2(sin60)/2<br />Sa=a^2(sin60)/2<br />S=acsin(60)/2<br />-------------------------------------------<br />(1)<br />Sb=b^2(sin60)/2<br />=(a^2+c^2-ac)*sin60/2<br />=Sa+Sc-S<br />-------------------------------------------<br />(2)<br />It is easy to see S=sqrt(SaSc)<br />By replacing the S in the equation obtained in (1)<br />Sb=Sa+Sc-sqrt(SaSc)Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-14987607047724971782020-09-06T03:12:07.647-07:002020-09-06T03:12:07.647-07:00The use of Al-Kashi's theorem or cosine rule a...The use of Al-Kashi's theorem or cosine rule and trigonometry can be avoided by sticking to strictly euclidean concepts.<br />As <B=60°, if we let G be the foot of the altitude from A to BC, ABG is half an equilateral triangle and we have BG=AB/2=c/2.<br />Applying Euclid's propositions 13 of Book 2, in ABC we have b^2=a^2+c^2-2.BG.BC or b^2=a^2+c^2-a.c (1).<br />Any triangle with <B=60° will have an area of sqrt(3).a.c/4 (derived from Pythagore).<br />Using this formula for S, Sa, Sb and Sc and (1) yields the equations S=sqrt(Sa.Sc), Sb=Sa+Sc-S and Sb=Sa+Sc-sqrt(Sa.Sc) QEDGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28426464065903366712009-03-27T21:10:00.000-07:002009-03-27T21:10:00.000-07:00We've S(a)+S(c)-S(b)= V3a^2/4 + V3c^2/4 -V3b^2/4 w...We've S(a)+S(c)-S(b)= V3a^2/4 + V3c^2/4 -V3b^2/4 where V=square root. Hence, LHS=(V3/4)(a^2+c^2-b^2)--(1). By cosine rule in Tr. ABC,b^2=a^2+c^2-2ac*cos(60) or b^2=a^2+c^2 -2ac*(1/2) or ac=a^2+c^2-b^2. Using (1), S(a)+S(c)-S(b)= V3ac/4<BR/>Now S=asin(60)*c/2 = V3ac/4. Thus, S(a)+S(c)-S(b)= S or S(b)= S(a)+S(c)-S <BR/>We've S(a)=V3a^2/4 and S(c)=V3c^2/4.<BR/>Hence, S(a)*S(c)=3a^2*c^2/16 or (S(a)*S(c))^(1/2)=V3ac/4 which is = S as shown earlier. Hence,<BR/>S(b)= S(a) + S(c)-(S(a)*S(c))^(1/2)<BR/>Ajit:ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com