The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's altitude is $30$ the when it was $45$ . Prove that the height of tower is $x (\sqrt{3} + 1)$ metres.

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Solution

Let $h$ be the height of the tower.

In $Δ A B C$ ,

$\frac{A B}{B C} = tan 45^{\circ}$

$\frac{h}{y} = 1$

$h = y$

In $Δ A B D$ ,

$\frac{A B}{B D} = tan 30^{\circ}$

$\frac{h}{2 x + y} = \frac{1}{\sqrt{3}}$

$2 x + y = \sqrt{3} h$

$2 x + h = \sqrt{3} h$

$h = \frac{2 x}{\sqrt{3} - 1}$

$h = (\sqrt{3} + 1) x m e t e r s$

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The length of the shadow of a tower standing on level plane is found to be 2x metres longer when the sun' s altitude is $30^{\circ}$ than when it was $45^{\circ}$ . Prove that the height of tower is $x (\sqrt{3} + 1)$ metres.

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Q. The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30°than when it was 45°. The height of the tower in metres is

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(\sqrt{3} + 1) x

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The length of the shadow of a tower standing on a level plane is found to be 2x metres longer when the sun's altitude is $30^{\circ}$ then when it is $45^{\circ}$ . The height of the tower = ___ m.

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The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's altitude is 30 the when it was 45. Prove that the height of tower is x(√3+ 1) metres.

Let h be the height of the tower. In ΔABC, ABBC=tan45∘ hy=1 h=y In ΔABD, ABBD=tan30∘ h2x+y=1√3 2x+y=√3h 2x+h=√3h h=2x√3−1 h=(√3+1)xmeters

The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's altitude is $30$ the when it was $45$ . Prove that the height of tower is $x (\sqrt{3} + 1)$ metres.