Question

Answer 1

a, \(AB=\sqrt{BC^2-AC^2}=\sqrt{399}\left(cm\right)\left(pytago\right)\)

\(\sin B=\dfrac{AC}{BC}=\dfrac{25}{32}\approx\sin51^022'\Leftrightarrow\widehat{B}\approx51^022'\\ \widehat{C}=90^0-\widehat{B}\approx90^0-51^022'=38^038'\)

b, \(BC\cdot\sin B\cdot\sin C=BC\cdot\dfrac{AB}{BC}\cdot\dfrac{AC}{BC}=\dfrac{AB\cdot AC}{BC}\left(1\right)\)

Áp dụng HTL: \(AH\cdot BC=AC\cdot AB\Leftrightarrow AH=\dfrac{AC\cdot AB}{BC}\left(2\right)\)

Từ (1)(2) ta được đpcm

\(c,\cos^2B+\tan^2B\cdot\cos^2B=\cos^2B+\dfrac{\sin^2B}{\cos^2B}\cdot\cos^2B=\cos^2B+\sin^2B=1\)