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)

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.

-

 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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