Ta có
\(\dfrac{BH}{3}=\dfrac{CH}{5}=\dfrac{bH+CH}{3+5}=\dfrac{BC}{8}=\dfrac{16}{8}=2\\ =>\left\{{}\begin{matrix}BH=2.3=6\left(cm\right)\\CH=2.5=10\left(cm\right)\end{matrix}\right.\)
Xét tam giác ABC vuông tại A có
\(AH^2=BH.CH\\ =>AH=\sqrt{6.10}=2\sqrt{15}\)
Xét tam giác AHC vuông tại H có
\(tanC=\dfrac{AH}{HC}=\dfrac{2\sqrt{15}}{10}\\ =>\widehat{C}=37^o46'\)
Xét tam giác AHB vuông tại H có
\(tanB=\dfrac{AH}{BH}=\dfrac{2\sqrt{15}}{6}\\ =>\widehat{B}=52^o14'\)
Ta có: \(\dfrac{BH}{HC}=\dfrac{3}{5}\) \(\Rightarrow5BH-3HC=0\) (2)
\(BC=BH+CH=16\) (1)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}BH=6\\CH=10\end{matrix}\right.\)
Xét tam giác ABC vuông A, đcao AH:
\(AH=\sqrt{BH.CH}=\sqrt{6.10}=2\sqrt{15}\left(cm\right)\)
\(AB=\sqrt{AH^2+BH^2}=\sqrt{\left(2\sqrt{15}\right)^2+6^2}=4\sqrt{6}\left(cm\right)\)
\(\cos B=\dfrac{AB}{BC}=\dfrac{4\sqrt{6}}{16}=\dfrac{\sqrt{6}}{4}\)
\(\Rightarrow\widehat{B}\simeq52,23^o\)
\(\Rightarrow\widehat{C}=180^o-52,23^o-90^o=37,77^o\)