Question

Answer 1

Bài 6: \(\hat{C}-\hat{B}=36^0\)

=>\(\hat{C}=\hat{B}+36^0\)

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

=>\(\hat{B}+2\cdot\hat{B}+\hat{B}+36^0=180^0\)

=>\(4\cdot\hat{B}=144^0\)

=>\(\hat{B}=36^0\)

=>\(\hat{A}=2\cdot36^0=72^0;\hat{C}=36^0+36^0=72^0\)

Bài 7:

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

=>\(\hat{A}+\hat{C}=180^0-70^0=110^0\)

=>\(180^0-3\cdot\hat{C}+\hat{C}=110^0\)

=>\(2\cdot\hat{C}=180^0-110^0=70^0\)

=>\(\hat{C}=35^0\)

=>\(\hat{A}=180^0-3\cdot35^0=75^0\)

b: ED//BC

=>\(\hat{AED}=\hat{ACB}\) (hai góc đồng vị) và \(\hat{DEB}=\hat{EBC}\) (hai góc so le trong)

mà \(\hat{ACB}=\hat{EBC}\left(=\frac12\cdot\hat{ABC}\right)\)

nên \(\hat{AED}=\hat{BED}\)

=>ED là phân giác của góc AEB