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dowód
: 23 lut 2007, o 14:48
autor: PanDragon
W trójkącie ABC dane są: \(\displaystyle{ |\sphericalangle BAC| = }\) oraz \(\displaystyle{ |\sphericalangle ABC| = \beta}\). Wykaż, że: \(\displaystyle{ \frac{|BC| + |AC|}{|AB|} = \frac{cos\frac{\alpha - \beta}{2}}{cos\frac{\alpha + \beta}{2}}}\)
Z góry dziekuje za pomoc
dowód
: 23 lut 2007, o 20:43
autor: Lady Tilly
\(\displaystyle{ |AB|=c}\)
\(\displaystyle{ |BC|=a}\)
\(\displaystyle{ |AC|=b}\)
\(\displaystyle{ \frac{b}{sin\beta}=\frac{a}{sin\alpha}}\)
\(\displaystyle{ b{\cdot}sin\alpha=a{\cdot}sin\beta}\)
\(\displaystyle{ b=\frac{a{\cdot}sin\beta}{sin\alpha}}\)
\(\displaystyle{ \frac{c}{sin(180^{o}-(\alpha+\beta))}=\frac{a}{sin\alpha}}\)
\(\displaystyle{ c=\frac{a{\cdot}sin(180^{o}-(\alpha+\beta))}{sin\alpha}}\)
\(\displaystyle{ \frac{a+b}{c}=\frac{sin\alpha+sin\beta}{sin(180^{o}-(\alpha+\beta))}}\)