Question

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Answer 1

Bài 5:

ΔABC=ΔPQR

=>\(\widehat{B}=\widehat{Q}\)

=>\(\widehat{B}=55^0\)

\(3\cdot\widehat{A}=2\cdot\widehat{C}\)

=>\(\widehat{A}=\dfrac{2}{3}\cdot\widehat{C}\)

Xét ΔABC có \(\widehat{A}+\widehat{C}+\widehat{B}=180^0\)

=>\(\dfrac{2}{3}\cdot\widehat{C}+\widehat{C}=180^0-55^0=125^0\)

=>\(\dfrac{5}{3}\cdot\widehat{C}=125^0\)

=>\(\widehat{C}=125^0:\dfrac{5}{3}=75^0\)

\(\widehat{A}=\dfrac{2}{3}\cdot75^0=50^0\)

ΔABC=ΔPQR

=>\(\widehat{P}=\widehat{A}=50^0;\widehat{R}=\widehat{C}=75^0\)

Bài 6:

Xét ΔDEF có \(\widehat{D}+\widehat{E}+\widehat{F}=180^0\)

mà \(\widehat{D}:\widehat{E}:\widehat{F}=2:3:5\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{\widehat{D}}{2}=\dfrac{\widehat{E}}{3}=\dfrac{\widehat{F}}{5}=\dfrac{\widehat{D}+\widehat{E}+\widehat{F}}{2+3+5}=\dfrac{180^0}{10}=18^0\)

=>\(\left\{{}\begin{matrix}\widehat{D}=2\cdot18^0=36^0\\\widehat{E}=3\cdot18^0=54^0\\\widehat{F}=5\cdot18^0=90^0\end{matrix}\right.\)

ΔABC=ΔDEF

=>\(\widehat{A}=\widehat{D}=36^0;\widehat{B}=\widehat{E}=54^0;\widehat{C}=\widehat{F}=90^0\)