Question

Tính góc x

lm hộ mình, mình đang cần gấp.

Answer 1

IK//EF

=>\(\widehat{IKF}+\widehat{OFE}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{OFE}+140^0=180^0\)

=>\(\widehat{OFE}=40^0\)

\(\widehat{IEF}+\widehat{E_1}=180^0\)(hai góc kề bù)

=>\(\widehat{IEF}+130^0=180^0\)

=>\(\widehat{IEF}=50^0\)

Xét ΔOEF có \(\widehat{EOF}+\widehat{FEO}+\widehat{EFO}=180^0\)

=>\(x+50^0+40^0=180^0\)

=>\(x=90^0\)

Answer 2

Lời giải:

Bổ sung điều kiện: $IK\parallel EF$.

Vì $IK\parallel EF$ nên:

$\widehat{OIK}=\widehat{OEF}$ (2 góc đồng vị)

$=180^0-130^0=50^0$

$\widehat{OKI}=180^0-\widehat{IKF}=180^0-140^0=40^0$

Xét tam giác $OIK$ thì:

$x=180^0-(\widehat{OIK}+\widehat{OKI})=180^0-(50^0+40^0)=90^0$