a) Xét tam giác AHB ta có:
\(tanB=\dfrac{AH}{BH}\)
\(\Rightarrow AH=tanB\cdot BH\) (1)
Xét tam giác AHC ta có:
\(tanC=\dfrac{AH}{HC}\)
\(\Rightarrow AH=tanC\cdot HC\) (2)
Từ (1) và (2)
\(\Rightarrow tanC\cdot HC=tanB\cdot BH\)
\(\Rightarrow tan30^o\cdot\left(BC-HB\right)=tan42^o\cdot BH\)
\(\Rightarrow tan30^o\cdot15-tan30^o\cdot BH=tan42^o\cdot BH\)
\(\Rightarrow BH\approx6\left(cm\right)\)
\(\Rightarrow AH=tan42^o\cdot6\approx5,4\left(cm\right)\)
b) \(\widehat{A}=180^o-42^o-30^o=108^o\)
Theo định lý sin ta có:
\(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}\)
\(\Rightarrow AC=\dfrac{BC\cdot sinB}{sinA}=\dfrac{15\cdot sin42^o}{sin108^o}\approx10,6\left(cm\right)\)
