Câu trả lời:
a) Ta có: cos\(\widehat{A}\)= \(\dfrac{AC^2+AB^2-BC^2}{2.AC.AB}\)=\(\dfrac{4,5^2+6^2-7,5^2}{2.4,5.6}\)=0
=>\(\widehat{A} \)= 90o
=> △ABC vuông tại A
b) * cos\(\widehat{B}\)= \(\dfrac{BC^2+AB^2-AC^2}{2.BC.AB}\)=\(\dfrac{7,5^2+6^2-4,5^2}{2.7,5.6}\)=\(\dfrac{4}{5}\)
=> \(\widehat{B}\)= 36o52'11,63"
* \(\widehat{C}\)= 180o - (\(\widehat{A} + \widehat{B}\)) = 180o - (90 + 36o52'11,63") = 53o7'48,37"
* Ta có: ah = bc => h = \(\dfrac{b.c}{a}\)= \(\dfrac{AC.AB}{BC}\)= \(\dfrac{4,5.6}{7,5}\)= 3,6 (cm)