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 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 $T_{1}$ is:

10 \sqrt{2}

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10 \sqrt{3}

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20 \sqrt{3}

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20 \sqrt{2}

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Solution

The correct option is B $20 \sqrt{3}$
We know, $t a n (Θ) = \frac{O p p o s i t e}{A d j a c e n t}$

In $△ A B C, tan x = \frac{20}{A C}$

In $△ A C D, tan 2 x = \frac{60}{A C}$

$\frac{2 tan x}{1 - {tan}^{2} x} = \frac{60}{A C}$ ............. Using, $t a n 2 A = \frac{2 t a n A}{1 - {t a n}^{2} A}$

$\frac{2 \times \frac{20}{A C}}{1 - \frac{400}{A C^{2}}} = \frac{60}{A C}$

$\frac{40 A C}{A C^{2} - 400} = \frac{60}{A C}$

$40 A C^{2} = 60 A C^{2} - 24000$

$20 A C^{2} = 24000$

$A C^{2} = 1200$

$A C = 20 \sqrt{3} m$

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A block, released from rest from the top of a smooth inclined plane of angle of inclination $θ_{1}$ , reaches the bottom in time $t_{1}$ . The same block, released from rest from the top of another smooth inclined plane of angle of inclination $θ_{2}$ , reaches the bottom in time $t_{2}$ . If the two inclined planes have the same height, the relation between $t_{1}$ and $t_{2}$ is

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A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. From the top of T1, if the angle of depression of foot of T2 is twice the angle of elevation of the top of T2 , then the width (in m) of the road between the feet of the towers T1 and T1 is:

The correct option is B 20√3We know, tan(Θ)=OppositeAdjacentIn △ABC,tanx=20ACIn △ACD,tan2x=60AC2tanx1−tan2x=60AC ............. Using, tan2A=2tanA1−tan2A2×20AC1−400AC2=60AC40ACAC2−400=60AC40AC2=60AC2−2400020AC2=24000AC2=1200AC=20√3m