A bird is flying up at angle ${sin}^{- 1} (\frac{3}{5})$ with the horizontal. A fish in a pond looks at that bird when it is vertically above the fish. The angle at which the bird appears to fly (to be fish)is $(μ_{ω} = \frac{4}{3})$

${sin}^{- 1} (\frac{3}{5})$

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${sin}^{- 1} (\frac{4}{5})$

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$45^{\circ}$

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${sin}^{- 1} (\frac{3}{16})$

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Solution

The correct option is B
${sin}^{- 1} (\frac{4}{5})$

given that $\frac{d x}{d t} = v$
In the figure
$\frac{3 h}{x + y} = \frac{h}{y}$
$∴ 3 y = x + y$
or $y = \frac{x}{2}$
$∴ \frac{d y}{d t} = \frac{1}{2} (\frac{d x}{d t})$
or speed of shadow $= \frac{v}{2}$

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A bird is flying up at angle sin^-1(3/5) with the horizontal. A fish in a pond looks at that bird when it is vertically above the fish. The angle at which the bird appears to fly (to be fish)is (μ_ω=4/3)

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A bird is flying up at angle sin−1(35) with the horizontal. A fish in a pond looks at that bird when it is vertically above the fish. The angle at which the bird appears to fly (to be fish)is (μω=43)

The correct option is B sin−1(45) given that dxdt=v In the figure 3hx+y=hy ∴3y=x+y or y=x2 ∴dydt=12(dxdt) or speed of shadow =v2

A bird is flying up at angle ${sin}^{- 1} (\frac{3}{5})$ with the horizontal. A fish in a pond looks at that bird when it is vertically above the fish. The angle at which the bird appears to fly (to be fish)is $(μ_{ω} = \frac{4}{3})$

The correct option is B
${sin}^{- 1} (\frac{4}{5})$

given that $\frac{d x}{d t} = v$
In the figure
$\frac{3 h}{x + y} = \frac{h}{y}$
$∴ 3 y = x + y$
or $y = \frac{x}{2}$
$∴ \frac{d y}{d t} = \frac{1}{2} (\frac{d x}{d t})$
or speed of shadow $= \frac{v}{2}$