\(\Delta ABC\) có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\text{ ( Tổng 3 góc tam giac ) }\)
\(\Rightarrow\widehat{A}+\widehat{C}=180^o-\widehat{B}=180^o-55^o=125^o\)
Ta có: \(3\widehat{A}=2\widehat{B}\Rightarrow\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}\)
\(\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{A}+\widehat{B}}{2+3}=\dfrac{125}{5}=25\) ( Áp dụng tính chất dãy tỉ số bằng nhau )
\(\dfrac{\widehat{A}}{2}=25\Rightarrow\widehat{A}=25.2=50^o\)
\(\dfrac{\widehat{B}}{3}=25\Rightarrow\widehat{B}=25.3=75^o\)
Vì \(\Delta ABC=\Delta PQR\left(gt\right)\)
\(\Rightarrow\widehat{A}=\widehat{P}=55^o\)
\(\Rightarrow\widehat{B}=\widehat{Q}=50^o\)
\(\Rightarrow\widehat{C}=\widehat{R}=75^o\)
Vậy \(\widehat{P}=55^o\\ \widehat{Q}=50^o\\ \widehat{R}=75^o\)
