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The length of the shadows of a vertical pole of height $h$ , thrown by the sun's rays at three different moments are $h$ , $2 h$ and $3 h$ . The sum of the angles of elevation of the rays at these three moments is equal to

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\frac{π}{2}

\frac{π}{3}

\frac{π}{4}

\frac{π}{6}

answer is A.

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Detailed Solution

It is given that the length of the shadows of the vertical pole of height

h

which is thrown by the sun's rays at three different moments are

h

2 h

and

3 h

.
According to the given data the figure has been drawn below.
Question Image

Let

OA

be the vertical pole of height

h

and

O P_{1} = h, O P_{2} = 2 h

and

O P_{3} = 3 h

be the lengths of shadow.
In

AO P_{1}

, we have

tan θ = P e r p e n d i c u l a r b a s e

\tan θ_{1} = \frac{OA}{O P_{1}}

\Rightarrow \frac{h}{h} = 1

\Rightarrow θ_{1} = \frac{π}{4}

ΔAO P_{2}

, we have

tan θ = P e r p e n d i c u l a r b a s e

\tan θ_{2} = \frac{OA}{O P_{2}}

\Rightarrow \frac{h}{2 h} = \frac{1}{2}

\Rightarrow

θ_{2} = \tan^{- 1} \frac{1}{2}

Similarly, in

ΔAO P_{3}

, we have ,

tan θ = P e r p e n d i c u l a r b a s e

\tan θ_{3} = \frac{1}{3}

θ_{3} = \tan^{- 1} (\frac{1}{3})

Therefore, sum of the angles of elevation of the eyes is,

\Rightarrow θ_{1} + θ_{2} + θ_{3}

\Rightarrow \frac{π}{4} + \tan^{- 1} (\frac{1}{2}) + \tan^{- 1} (\frac{1}{3})

(\tan 45 ° = 1 = \frac{π}{4})

\Rightarrow \frac{π}{4} + \tan^{- 1} (\frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{3} \times \frac{1}{2}})

[∵ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - xy})]

\Rightarrow \frac{π}{4} + \tan^{- 1} (\frac{\frac{5}{6}}{\frac{5}{6}})

\Rightarrow \frac{π}{4} + \tan^{- 1} (1)

\Rightarrow \frac{π}{4} + \frac{π}{4}

(\tan 45 ° = 1 = \frac{π}{4})

\Rightarrow \frac{π}{2}

So the sum of the angles of elevation of the rays is

\frac{π}{2}

.
So the correct option is 1.

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