Question

Tính số x trong các hình sau:

Answer 1

a, (+) Ta có: \(\widehat{K}=90^o;\widehat{B}=40^o;\widehat{H}=90^o\)
Xét \(\Delta IKB\) có:
\(\widehat{KIB}+\widehat{K}+\widehat{B}=180^o\) ( tổng 3 góc trong 1 tam giác)
\(\widehat{KIB}+90^o+40^o=180^o\)
\(\widehat{KIB}=180^o-40^o-90^o\)
\(\widehat{KIB}=50^o\)
(+) Ta có: \(\widehat{KIB}=\widehat{HIA}=50^o\) ( 2 góc đối đỉnh)
(+) Xét \(\Delta HIA\) có:
\(\widehat{H}+\widehat{A}+\widehat{HIA}=180^o\) ( tổng 3 góc trong 1 tam giác)
\(90^o+\widehat{A}+50^o=180^o\)
\(\widehat{A}=180^o-50^o-90^o\)
\(\widehat{A}=40^o\)
hay \(\widehat{x}=40^o\)

Answer 2

b,Ta có: \(\widehat{D}=90^o;\widehat{C}=25^o;\widehat{E}=90^o\)
(+) Xét \(\Delta DOC\) có:
\(\widehat{D}+\widehat{DOC}+\widehat{C}=180^o\) ( tổng 3 góc trong 1 tam giác)
\(90^o+\widehat{DOC}+25^o=180^o\)
\(\widehat{DOC}=180^o-25^o-90^o\)
\(\widehat{DOC}=65^o\)
(+) Ta có: \(\widehat{DOC}=\widehat{EOB}=65^o\) ( 2 góc đối đỉnh)
(+) Xét \(\Delta EOB\) có:
\(\widehat{E}+\widehat{EOB}+\widehat{B}=180^o\) ( tổng 3 góc trong 1 tam giác)
\(90^o+65^o+\widehat{B}=180^o\)
\(\widehat{B}=180^o-90^o-65^o\)
\(\widehat{B}=25^o\)
hay \(\widehat{x}=25^o\)