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Rohit was moving towards the hill to have a sight scene from its top. At a certain point in his way, he found that the angle of depression to top of the hill was45^{\circ}. After travelling 30m towards the hill, he found that the angle of depression to the top of the hill now became 60^{\circ}. Find the height of the hill.

  • Option 1)

    69.80m

  • Option 2)

    70.98m

  • Option 3)

    68.47m

  • Option 4)

    None

 

Answers (1)

best_answer

 

Angle of Depression -

If an object is below the horizontal line from the eye, we have to lower our head to view the object.

- wherein

angle of depression

 

 from fig

\Delta AMC

\tan 45^{\circ} = \frac{h}{30+BC}

\therefore BC=h-30 \:\:\:\: - (i)

Now in \Delta NBC

\tan 60^{\circ} = \frac{h}{bc}

\sqrt{3} = \frac{h}{h-30}

\therefore h\sqrt{3}-30\sqrt{3} = h

\therefore h=\frac{30\sqrt{3}}{\sqrt{3}-1} = 15 (\sqrt{3}+1) \sqrt{3} = 15 (3+\sqrt{3})

                                                                    =15\times 4.732

                                                                   = 70.98

 


Option 1)

69.80m

This solution is incorrect.

Option 2)

70.98m

This solution is correct.

Option 3)

68.47m

This solution is incorrect.

Option 4)

None

This solution is incorrect.

Posted by

prateek

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