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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

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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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