Školjka

%V0 Primjenom sinusovog poučka dobivamo $ (\dfrac{\sin{\beta}}{\sin{\gamma}} + \dfrac{\sin{\gamma}}{sin{\beta}})\cos{\alpha} + (\dfrac{\sin{\gamma}}{\sin{\alpha}} + \dfrac{\sin{\alpha}}{sin{\gamma}})\cos{\beta} + (\dfrac{\sin{\alpha}}{\sin{\beta}} + \dfrac{\sin{\beta}}{sin{\alpha}})\cos{\gamma} = 3 \newline \Leftrightarrow \sin^2{\beta}\sin{\alpha}cos{\alpha} + \sin^2{\beta}\sin{\gamma}cos{\gamma} + \sin^2{\alpha}\sin{\beta}cos{\beta} + \sin^2{\alpha}\sin{\gamma}cos{\gamma} + \sin^2{\gamma}\sin{\alpha}cos{\alpha} + \sin^2{\gamma}\sin{\beta}cos{\beta} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}(\sin{\alpha}\cos{\beta} + \sin{\beta}\cos{\alpha} ) + \sin{\alpha}\sin{\gamma}(\sin{\alpha}\cos{\gamma} + \sin{\gamma}\cos{\alpha} ) + \sin{\gamma}\sin{\beta}(\sin{\gamma}\cos{\beta} + \sin{\beta}\cos{\gamma} ) = 3\sin{\alpha}\sin{\beta}\sin{\gamma} \newline \Leftrightarrow \sin{\alpha}\sin{\beta}\sin{(\alpha + \beta)} + \sin{\gamma}\sin{\beta}\sin{(\gamma + \beta)} + \sin{\alpha}\sin{\gamma}\sin{(\alpha + \gamma)} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} $ $ \sin{(\alpha + \beta)}=\sin({\pi - \gamma)} = \sin{\gamma} $ Uvrštavanjem dobivamo $ 3\sin{\alpha}\sin{\beta}\sin{\gamma} = 3\sin{\alpha}\sin{\beta}\sin{\gamma} $