Giải:
a) Xét \(\Delta AHC\left(\widehat{AHC}=90^o\right)\) có: \(\widehat{HAC}+\widehat{C}=90^o\)
\(\Rightarrow\widehat{HAC}=50^o\) ( do \(\widehat{C}=40^o\) )
\(\Rightarrow\widehat{HAD}=\widehat{CAD}=25^o\)
Xét \(\Delta AHD\left(\widehat{AHD}=90^o\right)\) có: \(\widehat{HAD}+\widehat{ADH}=90^o\)
\(\Rightarrow\widehat{ADH}=65^o\) ( do \(\widehat{HAD}=25^o\) )
b) Xét \(\Delta AHK\left(\widehat{AKH}=90^o\right)\) có: \(\widehat{AHK}+\widehat{HAK}=90^o\)
\(\Rightarrow\widehat{AHK}=40^o\) ( do \(\widehat{HAK}=50^o\) )
\(\Rightarrow\widehat{HAB}=40^o\)
Xét \(\Delta ABH\left(\widehat{AHB}=90^o\right)\) có: \(\widehat{ABH}+\widehat{HAB}=90^o\)
\(\Rightarrow\widehat{ABH}=50^o\) ( do \(\widehat{AHB}=40^o\) )
hay \(\widehat{ABC}=50^o\)
Vậy a) \(\widehat{ADH}=65^o\)
b) \(\widehat{ABC}=50^o\)
Tự vẽ hình nha.
a.\(\Delta AHC\left(\widehat{AHC}=90^o\right)\)có:
\(\widehat{HAC}+\widehat{HCA}=90^o\)
\(\Rightarrow\widehat{HAC}+40^o=90^o\)
\(\Rightarrow\widehat{HAC}=50^o\)
Có: AD là phân giác \(\widehat{HAC}\) nên \(\widehat{HAD}=\widehat{DAC}=\frac{\widehat{HAC}}{2}=\frac{50^o}{2}=25^o\)
Xét \(\Delta AHD\left(\widehat{AHD}=90^o\right)\)có:
\(\widehat{HAD}+\widehat{HDA}=90^o\)
\(\Rightarrow25^o+\widehat{HDA}=90^o\)
\(\Rightarrow\widehat{HDA}=65^o\)
b. Xét \(\Delta AHK\left(\widehat{AKH}=90^o\right)\)có:
\(\widehat{HAK}+\widehat{AHK}=90^o\)
\(\Rightarrow50^o+\widehat{AHK}=90^o\)
\(\Rightarrow\widehat{AHK}=40^o\)
\(\Rightarrow\widehat{AHK}=\widehat{HAB}=40^o\)
Xét \(\Delta ABH\left(\widehat{AHB}=90^o\right)\)có:
\(\widehat{ABH}+\widehat{BAH}=90^o\)
\(\Rightarrow\widehat{ABH}+40^o=90^o\)
\(\Rightarrow\widehat{ABH}=50^o\)
hay \(\widehat{ABC}=50^o\)
