Question

Answer 1

Câu 17:

a: ΔABC vuông tại A

=>\(\hat{B}+\hat{C}=90^0\)

mà \(\hat{B}-\hat{C}=20^0\)

nên \(\hat{B}=\frac{90^0+20^0}{2}=55^0;\hat{C}=55^0-20^0=35^0\)

Câu 16:

a: a//b

a⊥ AB

Do đó: b⊥AB

b: Ta có: \(\hat{BDC}+\hat{CDb}=180^0\) (hai góc kề bù)

=>\(\hat{BDC}=180^0-75^0=105^0\)

a//b

=>\(\hat{C_1}=\hat{CDb}\) (hai góc đồng vị)

=>\(\hat{C_1}=75^0\)

Ta có: \(\hat{C_1}+\hat{C_2}=180^0\) (hai góc kề bù)

=>\(\hat{C_2}=180^0-75^0=105^0\)

Ta có: BD//AC
=>\(\hat{ACD}+\hat{CDB}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ACD}=180^0-105^0=75^0\)