Determina i lati del triangolo AB nella figura.
$AC= 30sen(35°)^{-1}≅52,3~cm$;
$BC= 30sen(70°)^{-1}≅31,9~cm$;
$AB= 52,3cos(35°)+31,9cos(70°)=42,842+10,910 ≅ 53,8~cm$.
TAN(35°) = 30/x-----> x = 42.84 cm
TAN(70°) = 30/y------> y = 10.92 cm
AB = x + y-----> AB = 42.84 + 10.92-----> AB = 53.76 cm
AC = x/COS(35°) =42.84/COS(35°)------> AC = 52.30 cm
BC = y/COS(70°) =10.92/COS(70°)------> BC = 31.93 cm
AC = 30/sen 35° = 52,30 cm
BC = 30/sen 70° = 31,93 cm
AB = CH/tan 35°+CH/tan 70° = 30/(1/tan 35°+1/tan70°) = 53,76 cm