Thấy : \(\widehat{PBQ}=\widehat{BQA}-\widehat{BPQ}=55^o-15^o=40^o\)
Ta có : \(\dfrac{PQ}{sin40^o}=\dfrac{BQ}{sin15^o}\Rightarrow BQ=\dfrac{100}{sin40^o}.sin15^o\approx40,27\left(m\right)\)
\(\dfrac{BQ}{sin90^o}=\dfrac{AB}{sin55^o}\Rightarrow AB\approx40,27.sin55^o\approx33\left(m\right)\)
Đúng 1
Bình luận (0)
