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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Heights and Distances

NTA Abhyas 2020 · Mathematics · Heights and Distances

The length of the shadows of a vertical pole of height h , thrown...

The length of the shadows of a vertical pole of height hh, thrown by the sun's rays at three different moments are h,2hh, 2h and 3h3h. The sum of the angles of elevation of the rays at these three moments is equal toThe length of the shadows of a vertical pole of height h , thrown by the sun's rays at three different moments are h,

Options

  1. Aπ2\dfrac{\pi}{2}
  2. Bπ3\dfrac{\pi}{3}
  3. Cπ4\dfrac{\pi}{4}
  4. Dπ6\dfrac{\pi}{6}

Correct answer

A. π2\dfrac{\pi}{2}

Step-by-step solution

Let OAOA be the vertical pole of height hh and OP1,OP2,OP3O{P}_{1}, O{P}_{2}, O{P}_{3} be the lengths of shadow.

In ∆AOP1∆ AO{P}_{1} we have,

tanθ1=OAOP1=hh=1tan {\theta}_{1} = \dfrac{OA}{O{P}_{1}} = \dfrac{h}{h} = 1

⟹θ1=π4⟹{\theta}_{1} = \dfrac{\pi}{4}

In ∆AOP2∆ AO{P}_{2} we have,

tanθ2=OAOP2=22h=12tan {\theta}_{2} = \dfrac{OA}{O{P}_{2}} = \dfrac{2}{2h} = \dfrac{1}{2}

⟹θ2=tan⁡−112⟹{\theta}_{2} = \tan^{-1} \dfrac{1}{2}

Similarly, tanθ3=OAOP3=h3h=13tan {\theta}_{3} = \dfrac{OA}{O{P}_{3}} = \dfrac{h}{3h} = \dfrac{1}{3}

θ3=tan⁡−113{\theta}_{3} = \tan^{-1} \dfrac{1}{3}

∴∴ Sum of the angles

=θ1+θ2+θ3{= \theta}_{1} + {\theta}_{2} + {\theta}_{3}

=π4+tan⁡−112+tan⁡−113= \dfrac{\pi}{4} + \tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{3}

The length of the shadows of a vertical pole of height h , thrown by the sun's rays at three different moments are h,

=π4+tan⁡−1(12+131−12×13)<br/><br/>=π4+= \dfrac{\pi}{4} + \tan^{-1} \left(\dfrac{\dfrac{1}{2} + \dfrac{1}{3}}{1 - \dfrac{1}{2} \times \dfrac{1}{3}}\right)<br /><br />=\dfrac{\pi}{4} +\tan^{-1}11

=π4+π4=π2= \dfrac{\pi}{4} + \dfrac{\pi}{4} = \dfrac{\pi}{2}
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