pls answer fast.... I'll mark u ask brainliest for ever irrilavent answers will be reported pls fast guys . Created by psupriya789. Math

pls answer fast.... I'll mark u ask brainliest for ever irrilavent answers will ...

suchi2863

Step-by-step explanation:

here is the solution .the required value is 50 m

psupriya789 I wont to make as brainliest but so, problem is accuring so, it ll take time

psupriya789 thanks a lot suchi.. :)

psupriya789 ur so, brilint

psupriya789 Hlo reply na ??

suchi2863 welcome Supriya

psupriya789 :)

psupriya789 suchi i marked u as brainliest :)

TheDukeAYUSH

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

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.

psupriya789 thanks a lot dost

psupriya789 thank you so much dost

TheDukeAYUSH Most Welcome :)

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