Question

Bài 9,10,12,15

Answer 1

Bài 15: AB//CD

=>\(\hat{A}+\hat{D}=180^0\) (hai góc trong cùng phía)

=>\(\hat{D}=180^0-120^0=60^0\)

ABCD là hình thang cân

=>\(\hat{A}=\hat{B}\)

=>\(\hat{B}=120^0\)

ABCD là hình thang cân

=>\(\hat{C}=\hat{D}\)

=>\(\hat{C}=60^0\)

Bài 12:

AD//BC

=>\(\hat{DAB}+\hat{ABC}=180^0\)

mà \(\hat{DAB}-\hat{ABC}=20^0\)

nên \(\hat{DAB}=\frac{180^0+20^0}{2}=100^0;\hat{ABC}=100^0-20^0=80^0\)

Ta có: \(\hat{BAD}+\hat{BCD}=150^0\)

=>\(\hat{BCD}=150^0-100^0=50^0\)

AD//BC

=>\(\hat{DCB}+\hat{ADC}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ADC}=180^0-50^0=130^0\)

Bài 10:

AB//CD

=>\(\hat{BAD}+\hat{ADC}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ADC}=180^0-100^0=80^0\)

AB//CD

=>\(\hat{ABC}+\hat{BCD}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ABC}=180^0-50^0=130^0\)

BÀi 9:

\(\hat{B}+\hat{D}=100^0;\hat{B}+\hat{C}=200^0\)

=>\(\hat{B}+\hat{C}-\hat{B}-\hat{D}=200^0-100^0\)

=>\(\hat{C}-\hat{D}=100^0\)

mà \(\hat{C}+\hat{D}=120^0\)

nên \(\hat{C}=\frac{100^0+120^0}{2}=110^0;\hat{D}=120^0-110^0=10^0\)

\(\hat{B}+\hat{D}=100^0\)

=>\(\hat{B}=100^0-10^0=90^0\)

Xét tứ giác ABCD có \(\hat{A}+\hat{B}+\hat{C}+\hat{D}=360^0\)

=>\(\hat{A}=360^0-90^0-110^0-10^0=270^0-120^0=150^0\)