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)

 

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.

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