A boat moving at 7.8 km/hr relative to the water is crossing a river 3.5 km wide in which the current is flowing at 2.8 km/hr. At what angle upstream should the boat head to reach a point on the other shore directly opposite the starting point?

Respuesta :

Answer:[tex]\theta =21.03^{\circ}[/tex]

Explanation:

Given

velocity of boat with respect to river [tex]=7.8 km/hr[/tex]

velocity of river is [tex]2.8 km/hr[/tex]

River is 3.5 km wide

Suppose boat leaves at an angle of [tex]\theta [/tex]  with vertical such that it reaches exactly opposite end of initial Point

Therefore

[tex]7.8\sin \theta [/tex]component will balance the river Flowso that sin component helps to cross the river

[tex]7.8\sin \theta =2.8[/tex]

[tex]\sin \theta =0.3589[/tex]

[tex]\theta =sin^{-1}(0.3589)[/tex]

[tex]\theta =21.03^{\circ}[/tex]

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