# A bird is sitting on the top of a vertical   pole 20 m high and its elevation from a point O on the ground is 450.  It flies off horizontally straight away from the point O.After one second, the elevation of the bird from O is reduced to 300. Then the speed (in m/s) of the bird is : Option 1) Option 2) Option 3) Option 4)

As we learnt in

Height and Distances -

The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.

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$In \: \Delta PQQ \:, tan45^{\circ}=\frac{PQ}{OQ}$

$\Rightarrow OQ=20m$

$In \: \Delta {P}'{Q}'O, \: tan30^{\circ}=\frac{2O}{OQ'}\Rightarrow OQ'=20\sqrt{3}$

$Thus \: QQ'=20\sqrt{3}-20=20(\sqrt{3}-1)$

$Thus \: speed=20\left ( \sqrt{3}-1 \right ) \: \: =20\left ( \sqrt{3}-1 \right ) \: m/sec$

Option 1)

This option is incorrect

Option 2)

This option is correct

Option 3)

This option is incorrect

Option 4)

This option is incorrect

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