Calculam aria triunghiului folosind formula lui Heron.
[tex]p= \frac{AB+BC+AC}{2}= \frac{9+12+7}{2}= \frac{28}{2} =14.[/tex]
[tex] A_{ABC}= \sqrt{p(p-AB)(p-BC)(p-AC)}= \\ = \sqrt{14*(14-9)(14-12)(14-7)}= \\ = \sqrt{14*5*2*7}= \\ =14\sqrt{5} ~ (cm ^{2}) [/tex]
[tex] A_{ABC} = \frac{BC*AB*sinB}{2} =\ \textgreater \ sinB= \frac{2* A_{ABC} }{BC*AB}= \frac{28 \sqrt{5} }{12*9}= \frac{7 \sqrt{5} }{27} .[/tex]
[tex]sin ^{2} B+ cos^{2}B=1=\ \textgreater \ cosB= \sqrt{1- sin^{2}B } = \sqrt{1- \frac{245}{729} }= \sqrt{ \frac{484}{729} }= \frac{24}{27} . [/tex]
[tex]tgB= \frac{sinB}{cosB}= \frac{ \frac{7 \sqrt{5} }{27} }{ \frac{24}{27} } = \frac{ 7\sqrt{5} }{24} [/tex]
[tex]ctgB= \frac{1}{tgB} = \frac{24}{ 7\sqrt{5} } = \frac{ 24\sqrt{5} }{35}. [/tex]
