Let a vertical tower AB have its end A on the level ground.  Let C be the mid-point of AB and P be a point on the ground such that AP=2AB.  If \angle BPC= \beta, then tan β is equal to :

 

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{2}{9}

  • Option 3)

    \frac{4}{9}

  • Option 4)

    \frac{6}{7}

 

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.

-

 Let, AC=CB=h, AB=2h and AP=4h

Now, \:\tan \angle BPA=\frac{2h}{4h}=\frac{1}{2}

\tan \angle CPA=\frac{h}{4h}=\frac{1}{4}

\tan \beta=\tan (\angle BPA- \angle CPA)=\frac{tan(\angle BPA)-\tan(\angle CPA)}{1+\tan(\angle BPA)\:\tan(\angle CPA)}

=\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{8}}=\frac{\frac{1}{4}}{\frac{9}{8}}=\frac{2}{9}

\tan \beta=\frac{2}{9}

 


Option 1)

\frac{1}{4}

This option is incorrect.

Option 2)

\frac{2}{9}

This option is correct.

Option 3)

\frac{4}{9}

This option is incorrect.

Option 4)

\frac{6}{7}

This option is incorrect.

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