Question

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Answer 1

Step-by-step explanation:

here is the solution .the required value is 50 m

Answer 2

$\large\underline{ \underline{ \sf \maltese{ \: Correct \: Question:- }}}$

From the top of a vertical building of 50√3 m height on a level ground, the angle of depression of an object on the same ground is observed to be 60°. Find the distance of the object from the foot of the building.

$\large\underline{ \underline{ \sf \maltese{ \: To\:Find:- }}}$

Distance of the object from the foot of the building.

$\large\underline{ \underline{ \sf \maltese{ \: Concept\:used:- }}}$

Heights and Distances.

$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

We know that,

$\sf{{\boxed{\sf{\red{tan\theta ={\dfrac{opposite}{adjacent}}}}}}}$

$\sf{{\implies}{\dfrac{AB}{BC}}={tan 60^{\circ}}}$

$\sf{{\implies}{\dfrac{AB}{BC}}=\sqrt 3}$

$\sf{{\implies}{\dfrac{50\sqrt3}{BC}}=\sqrt 3}$

$\sf{{\implies}{BC=}{\dfrac{50\sqrt3}{\sqrt3}}}$

$\sf{{\implies}{BC=50m}}$

Hence, the distance of the object from foot of building is 50m.