Toán Lớp 9: Alo
Cho tam giác ABC (no vuông) có BC = 8cm, gócC=43, gócB=32. Tính AC
Best toán ơi, cứu cháu
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: Alo
Cho tam giác ABC (no vuông) có BC = 8cm, gócC=43, gócB=32. Tính AC
Best toán ơi, cứu cháu
Comments ( 1 )
Trong\,\Delta ABC:\\
\widehat A + \widehat B + \widehat C = {180^0}\\
\Leftrightarrow \widehat A = {180^0} – {43^0} – {32^0} = {105^0}
\end{array}$
\dfrac{{BC}}{{\sin \widehat A}} = \dfrac{{AC}}{{\sin \widehat B}} = \dfrac{{AB}}{{\sin \widehat C}}\\
\Leftrightarrow \dfrac{8}{{\sin {{105}^0}}} = \dfrac{{AC}}{{\sin {{32}^0}}}\\
\Leftrightarrow AC = \dfrac{{8.\sin {{32}^0}}}{{\sin {{105}^0}}} = 4,55\left( {cm} \right)\\
Vậy\,AC = 4,55cm
\end{array}$