Question

Tứ giác ABCD có góc A = 110°, góc D = 70°. Tính số đo góc B, C biết tỉ số giữa 2 góc là 7 phần 3.

Answer 1

`\hat{B}/\hat{C}=7/3`

`=>\hatB=7/3hatC`

Ta có:`hatA+hatB+hatC+hatD=360^o`

`=>hatB+hatC=360^o-hatA-hatD=360^o-180^o=180^o`

`=>7/3hatC+hatC=180^o`

`=>10/3hatC=180^o`

`=>hatC=54^o`

`=>hatB=7/3hatC=126^o`

Answer 2

Ta có: \(\angle\left(A\right)+\angle\left(B\right)+\angle\left(C\right)+\angle\left(D\right)=360^0\)

\(=>\angle\left(B\right)+\angle\left(C\right)=360-110-70=180^0\left(1\right)\)

lại có: \(\dfrac{\angle\left(B\right)}{\angle\left(C\right)}=\dfrac{7}{3}=>\angle\left(B\right)=\dfrac{7\angle\left(C\right)}{3}\left(2\right)\)

thế(2) vào(1)\(=>\angle\left(C\right)+\dfrac{7\angle\left(C\right)}{3}=180=>\angle\left(C\right)=54^0\)

\(=>\angle\left(B\right)=180-54=126^o\)