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How to solve this problem- A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 300. After walking for 10 minutes from

A man is walking towards a vertical pillar in a straight path, at a uniform speed.  At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 300.  After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 600. Then the time taken (in minutes) by him, from B to reach the pillar, is :

  • Option 1)

    6

  • Option 2)

    10

  • Option 3)

    20

  • Option 4)

    5

 
Answers (1)
295 Views

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.

-

 Let PQ represent pillar. Say height = h

In \Delta PAQ ,  \tan 30^{o}=\frac{PQ}{AQ}

\Rightarrow AQ=\sqrt 3h

In \Delta PQB,    \tan 60^{o}=\frac{PQ}{BQ}\:\:\Rightarrow BQ=\frac{h}{\sqrt 3}

Now, the person is walking at a uniform speed, hence time taken will be proportional to distance travelled.

Thus, 

\frac{T_{BQ}}{T_{AB}}=\frac{BQ}{AB}=\frac{\frac{h}{\sqrt 3}}{\sqrt 3h-\frac{h}{\sqrt 3}}

=\frac{\frac{1}{\sqrt 3}}{\sqrt 3-\frac{1}{\sqrt 3}}=\frac{\frac{1}{\sqrt 3}}{\frac{2}{\sqrt 3}}=\frac{1}{2}

\Rightarrow T_{BQ}=\frac{T_{AB}}{2}=\frac{10 \min}{2}=5\min

Hence, time required is 5 minutes.

 


Option 1)

6

This option is incorrect.

Option 2)

10

This option is incorrect.

Option 3)

20

This option is incorrect.

Option 4)

5

This option is correct.

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