Question

Answer 1

\(\widehat{ABC}=90^o+15^o30'=105,5^o\)

\(\widehat{CAB}=90^o-45^o=45^o\)

\(\widehat{ACB}=180^o-105,5^o-45^o=29,5^o\)

Ta có:

\(\dfrac{AC}{sin105,5}=\dfrac{AB}{sin29,5}\)

\(\Rightarrow AC=\dfrac{120.sin105,5}{sin29,5}=234,8\left(m\right)\)

Ta có:

\(sin45=\dfrac{CH}{AC}\)

\(\Rightarrow CH=234,8.sin45=166\left(m\right)\)