Question

Giải giúp mình với🥺😖

Answer 1

a. \(BC^2=289=64+225=AC^2+AB^2\) nên ABC vuông A

b. \(P_{ABC}=AB+BC+CA=40\left(cm\right)\)

\(\sin B=\dfrac{AC}{BC}=\dfrac{15}{17}\approx62^0\\ \Rightarrow\widehat{B}\approx62^0\\ \Rightarrow\widehat{C}\approx90^0-62^0=28^0\)

c. Áp dụng HTL: \(\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=\dfrac{64}{17}\left(cm\right)\\CH=\dfrac{AC^2}{BC}=\dfrac{225}{17}\left(cm\right)\\AH=\dfrac{AB\cdot AC}{BC}=\dfrac{120}{17}\left(cm\right)\end{matrix}\right.\)