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Answer 1

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Answer 2

$\large{\underline{\underline{\mathfrak{\red{\sf{Answer-}}}}}}$

The height of the tower is 20√3 m.

$\large{\underline{\underline{\mathfrak{\red{\sf{Explanation-}}}}}}$

Let the height of tower be AB.

• The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it is 60°

$\implies$ DC = 40 m.

Also, DB = DC + BC

In ∆ABC,

TanØ = $\sf{\dfrac{P}{B}}$

$\implies$ tan60° = $\sf{\dfrac{AB}{BC}}$

We know that, Tan60° = √3.

$\implies$ √3 = $\sf{\dfrac{AB}{BC}}$

$\implies$ √3BC = AB

$\implies$ BC = $\sf{\dfrac{AB}{\sqrt3}}$ ⠀⠀⠀⠀⠀⠀⠀⠀—eq (1)

Now,

In ∆ABD,

TanØ = $\sf{\dfrac{P}{B}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\implies$ tan30° = $\sf{\dfrac{AB}{BD}}$

We know that, $\sf{\dfrac{1}{\sqrt3}}$ = $\sf{\dfrac{AB}{BD}}$

$\implies$ BD = √3AB

$\implies$ DC + BC = √3AB

$\implies$ BC = √3AB - DC

$\implies$ BC = √3AB - 40⠀⠀⠀⠀⠀⠀—eq (2).

_______________

From eq (1) and (2),

$\sf{\dfrac{AB}{\sqrt3}}$ = √3AB - 40

$\implies$ AB = √3(√3AB) - 40√3

$\implies$ AB = 3AB - 40√3

$\implies$ AB - 3AB = -40√3

$\implies$ $\cancel{-}$ 2AB = $\cancel{-}$ 40√3

$\implies$ AB = $\sf\cancel{\dfrac{40\sqrt3}{2}}$

$\implies$ AB = $\sf{20\sqrt3}$ m

Hence, the height of the tower is 20√3 m.

________________________

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