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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)
$20\sqrt{2}$

Option 2)
$20\left ( \sqrt{3} -1\right )$

Option 3)
$40\left ( \sqrt{2} -1\right )$

Option 4)
$40\left ( \sqrt{3} -\sqrt{2}\right )$

Answers (1)

best_answer

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)

$20\sqrt{2}$

This option is incorrect

Option 2)

$20\left ( \sqrt{3} -1\right )$

This option is correct

Option 3)

$40\left ( \sqrt{2} -1\right )$

This option is incorrect

Option 4)

$40\left ( \sqrt{3} -\sqrt{2}\right )$

This option is incorrect

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