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A tower T_{1} of height 60 m is located exactly opposite to a tower T_{2} of height 80 m on a straight road. From the top of T_{1} if the angle of depression of the foot of T_{2} is twice the angle of elevation of the top of T_{2} then the width (in m) of the road between the feet of the towers T_{1} and T_{2} is :

  • Option 1)

    10\sqrt{2}

  • Option 2)

    10\sqrt{3}

  • Option 3)

    20\sqrt{3}

  • Option 4)

    20\sqrt{2}

 

Answers (2)

best_answer

As we learned,

 

Angle of Elevation -

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

- wherein

angle of elevation

 

 

\tan \theta =\frac{BC}{AC}=\frac{20}{\omega }\tan 2\theta =\frac{2\left ( \frac{20}{\omega } \right )}{1-\frac{400}{\omega ^{2}}}=\frac{80}{\omega }

\Rightarrow \: \omega ^{2}=800

\Rightarrow \: \omega =20\sqrt{2}

\tan \theta =\frac{20}{x},\: \tan 2\theta =\frac{60}{d}

\Rightarrow \: \frac{60}{d}=\frac{40ld}{\left ( 1-400ld^{2} \right )}

\Rightarrow \: d=20\sqrt{3}


Option 1)

10\sqrt{2}

Option 2)

10\sqrt{3}

Option 3)

20\sqrt{3}

Option 4)

20\sqrt{2}

Posted by

Himanshu

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