দশম অধ্যায়ঃ ত্রিভুজের ধর্মাবলী
[tex] {a \over {\sin A}} = {b \over {\sin B}} = {c \over {\sin C}} = 2R [/tex]
[tex] \cos A = {{{b^2} + {c^2} - {a^2}} \over {2bc}}[/tex]
[tex] \cos B = {{{c^2} + {a^2} - {b^2}} \over {2ca}} [/tex]
[tex] \cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}[/tex]
[tex] a = b\cos C + c\cos B[/tex]
[tex] b = c\cos A + a\cos C[/tex]
[tex] c = a\cos B + b\cos A[/tex]
[tex] \sin {A \over 2} = \sqrt {{{(s - b)(s - c)} \over {bc}}}[/tex]
[tex] \sin {B \over 2} = \sqrt {{{(s - c)(s - a)} \over {ca}}}[/tex]
[tex] \sin {C \over 2} = \sqrt {{{(s - a)(s - b)} \over {ab}}}[/tex]
[tex] \cos {A \over 2} = \sqrt {{{s(s - a)} \over {bc}}}[/tex]
[tex] \cos {B \over 2} = \sqrt {{{s(s - b)} \over {ca}}}[/tex]
[tex] \cos {C \over 2} = \sqrt {{{s(s - c)} \over {ab}}}[/tex]
[tex]\tan {A \over 2} = \sqrt {{{(s - b)(s - c)} \over {s(s - a)}}} = {{(s - b)(s - c)} \over \Delta }[/tex]
[tex] \tan {B \over 2} = \sqrt {{{(s - c)(s - a)} \over {s(s - b)}}} = {{(s - c)(s - a)} \over \Delta }[/tex]
[tex] \tan {C \over 2} = \sqrt {{{(s - a)(s - b)} \over {s(s - c)}}} = {{(s - a)(s - b)} \over \Delta }[/tex]
- 107 views