Question

Ariver 5 Km wide is flowing at the rate of 2 Km/H. The
minimum time taken by a boat to cross the river with a
speed 5 Km/
Hin still water) is approximately​

Answer 1

Answer:

Tan theta = 4/3

theta = tan⁻¹4/3= 53.1⁰

magnitude of resultant velocity

OB'= √ OA² + OB²

= √3² + 4²

=5 km/h

Distance of swimmer= OB' x time

= 5 x 15/60

=1.25 km

Answer 2

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