A tower T _ 1 of height 60m is located exactly opposite to a tower...

A tower

{T}_{1}

of height

60m

is located exactly opposite to a tower

{T}_{2}

of height

80m

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}

A tower T _ 1 of height 60m is located exactly opposite to a tower T _ 2 of height 80m on a straight road. From the top

Options

A $20\sqrt{2}$
B $10\sqrt{2}$
C $10\sqrt{3}$
D $20\sqrt{3}$

Correct answer

D. $20\sqrt{3}$

Step-by-step solution

Let, the width of the road the road is

d.

If the angle of the elevation is

\theta,

then

tan\theta = \dfrac{20}{d},

here

20

is the diﬀerence between the heights of

{T}_{1}

and

{T}_{2}.

Given that, the angle of depression is twice of the angle of elevation.

\tan 2\theta = \dfrac{60}{d}

We know,

\tan 2\theta = \dfrac{2tan \theta}{1 - \tan^{2} \theta}

∴ \dfrac{60}{d} = \dfrac{40 / d}{1 - \left(400 / d^{2}\right)}

⇒ \dfrac{400}{d^{2}} = \dfrac{1}{3}

d = 20\sqrt{3}

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