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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?

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Answer:
\theta =21.03^(\circ)

Step-by-step explanation:

Given

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

velocity of river is
2.8 km/hr

River is 3.5 km wide

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

Therefore


7.8\sin \thetacomponent will balance the river Flowso that sin component helps to cross the river


7.8\sin \theta =2.8


\sin \theta =0.3589


\theta =sin^(-1)(0.3589)


\theta =21.03^(\circ)

A boat moving at 7.8 km/hr relative to the water is crossing a river 3.5 km wide in-example-1
User Marla
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