Đáp án:
\[\widehat A = 80^\circ ;\,\,\,\widehat B = 60^\circ ;\,\,\,\,\widehat C = 40^\circ \]
Giải thích các bước giải:
\(2\widehat B = 3\widehat C \Leftrightarrow \widehat B = \dfrac{3}{2}\widehat C\)
Tổng 2 góc trong một tam giác bằng \(180^\circ \) nên ta có:
\(\begin{array}{l}
*)\\
\widehat {BAD} + \widehat B + \widehat {ADB} = 180^\circ \\
\Leftrightarrow \widehat {BAD} + \dfrac{3}{2}\widehat C + 80^\circ = 180^\circ \\
\Leftrightarrow \widehat {BAD} + \dfrac{3}{2}\widehat C = 100^\circ \\
\Leftrightarrow \widehat {BAD} = 100^\circ - \dfrac{3}{2}\widehat C\\
*)\\
\widehat {BAC} + \widehat B + \widehat C = 180^\circ \\
\Leftrightarrow 2\widehat {BAD} + \dfrac{3}{2}\widehat C + \widehat C = 180^\circ \\
\Leftrightarrow 2\widehat {BAD} + \dfrac{5}{2}\widehat C = 180^\circ \\
\Leftrightarrow 2.\left( {100^\circ - \dfrac{3}{2}\widehat C} \right) + \dfrac{5}{2}\widehat C = 180^\circ \\
\Leftrightarrow \dfrac{1}{2}\widehat C = 20^\circ \\
\Leftrightarrow \widehat C = 40^\circ \Rightarrow \widehat B = \dfrac{3}{2}\widehat C = 60^\circ \\
\widehat {BAC} = 180^\circ - \widehat B - \widehat C = 80^\circ
\end{array}\)
Vậy tam giác ABC có \(\widehat A = 80^\circ ;\,\,\,\widehat B = 60^\circ ;\,\,\,\,\widehat C = 40^\circ \)
