A river is flowing at a speed of 3 m/s doe east. A boat, whose speed in still water is 6 m/s, sis steered in the direction due to the north. Find the angle made by the resultant velocity of summer w.r.t river flow.

30^{\circ}

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37^{\circ}

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53^{\circ}

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{63.4}^{\circ}

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Solution

The correct option is D ${63.4}^{\circ}$
Speed of river, $V_{r} = 3 m / s$
Speed of boat, $V_{b} = 6 m / s$
$\begin{matrix} ∴ tan θ = \frac{V_{b}}{V_{R}} = \frac{6}{3} tan θ = 2 θ = {tan}^{- 1} (2) = 63.4 \end{matrix}$

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A river is flowing at a speed of 3 m/s doe east. A boat, whose speed in still water is 6 m/s, sis steered in the direction due to the north. Find the angle made by the resultant velocity of summer w.r.t river flow.

The correct option is D 63.4∘Speed of river, Vr=3m/sSpeed of boat, Vb=6m/s∴tanθ=VbVR=63tanθ=2θ=tan−1(2)=63.4

The correct option is D ${63.4}^{\circ}$
Speed of river, $V_{r} = 3 m / s$
Speed of boat, $V_{b} = 6 m / s$
$\begin{matrix} ∴ tan θ = \frac{V_{b}}{V_{R}} = \frac{6}{3} tan θ = 2 θ = {tan}^{- 1} (2) = 63.4 \end{matrix}$