Question

Answer 1

Có \(AC=400m;\widehat{DAB}=57^0;\widehat{DAC}=43^0\)

\(\widehat{ABD}=90^0-\widehat{BAD}=90^0-57^0=33^0\)

\(\widehat{BAC}=\widehat{BAD}-\widehat{CAD}=57^0-43^0=14^0\)

Có \(\dfrac{BC}{sin\widehat{BAC}}=\dfrac{AC}{sin\widehat{ABC}}\)\(\Leftrightarrow\dfrac{BC}{sin14^0}=\dfrac{400}{sin33^0}\)\(\Leftrightarrow BC\approx177,68\left(m\right)\)