Question

Answer 1

Xét tam giác BCD: \(BD=\dfrac{CD}{\sin CBD}=\dfrac{10}{\sin3^050'}\approx150\left(m\right)\)

Đổi \(400m=0,4km;150m=0,15km\)

\(t_{AB}=\dfrac{0,4}{4}=0,1\left(h\right)\\ t_{BD}=\dfrac{0,15}{3}=0,05\left(h\right)\\ \Rightarrow t_{tổng}=0,1+0,05=0,15\left(h\right)=9\left(phút\right)\)