If the angles of elevation of the top of a  tower from three collinear points A, B and C, on a line leading to the foot of the

tower, are 300, 450 and 600 respectively,then the ratio, AB : BC, is :

  • Option 1)

    \sqrt{3}:1

  • Option 2)

    \sqrt{3}:\sqrt{2}

  • Option 3)

    1:\sqrt{3}

  • Option 4)

    2:3

 

Answers (1)

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.

-

In  \Delta PQA, \tan 30^{o}=\frac{PQ}{AQ}

\Rightarrow AQ=\sqrt3h                                ...................... ( 1 )

In \Delta PQB, \tan 45^{o}=\frac{PQ}{BQ}

\Rightarrow BQ=h                                         .................... ( 2 )

In \Delta PQC, \tan 60^{o}=\frac{PQ}{CQ}

\Rightarrow CQ=\frac{h}{\sqrt 3}                                    ..................... ( 3 )

Now,  \frac{AB}{BC}=\frac{AQ-BQ}{BQ-CQ}=\frac{\sqrt3-1}{1-\frac{1}{\sqrt 3}}

=\sqrt 3\:\: \frac{(\sqrt 3-1)}{\sqrt 3-1}

=\sqrt 3

AB:BC=\sqrt 3:1


Option 1)

\sqrt{3}:1

This option is correct.

Option 2)

\sqrt{3}:\sqrt{2}

This option is incorrect.

Option 3)

1:\sqrt{3}

This option is incorrect.

Option 4)

2:3

This option is incorrect.

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