a, Ta có: \(\widehat{BAC}=90^o\)
Mà \(\widehat{HAC}=61^o\)
\(\Rightarrow\widehat{BAH}=\widehat{BAC}-\widehat{HAC}=90^o-61^o=29^o.\)
Xét △BAH vuông tại H có:
\(\widehat{BAH}+\widehat{BHA}+\widehat{ABH}=180^o\) ( Tổng 3 góc trong △ )
Mà \(\widehat{BAH}=29^o\)
\(\widehat{BHA=90^o}\)
\(\Rightarrow\widehat{ABH}=180^o-\widehat{BAH}-\widehat{BHA}=180^o-29^o-90^o=61^o.\)
b, Xét △KEG vuông tại K có:
\(\widehat{GEK}+\widehat{EKG}+\widehat{EGK}=180^o\left(t3g\Delta\right)\)
Mà \(\widehat{EGK}=44^o\)
\(\widehat{EKG}=90^o\)
\(\Rightarrow\widehat{GEK}=180^o-90^o-44^o=46^o\)
Ta có: \(\widehat{FEG}-90^o\)
Mà \(\widehat{GEK}=46^o\)
\(\Rightarrow\widehat{FEK}=90^o-46^o=44^o.\)
a: x=góc HAC=61 độ
b: x=góc G=44 độ
