The angle of elevation of the top of a tower from a certain point is 60°. If the observer moves 10 m away from the tower, the angle of elevation of the top of the tower decreases by 15°. The height of the tower is:

Let h = height of the tower, x = initial distance from the tower.

Initial angle of elevation = 60°

After moving 10 m away, angle of elevation = 60° - 15° = 45°

New distance from tower = x + 10

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At the first position:

$\tan(60°) = \frac{h}{x}$

$\sqrt{3} = \frac{h}{x}$

$h = x\sqrt{3}$ ... ①

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At the second position:

$\tan(45°) = \frac{h}{x + 10}$

$1 = \frac{h}{x + 10}$

$h = x + 10$ ... ②

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From equations ① and ②:

$x\sqrt{3} = x + 10$

$x\sqrt{3} - x = 10$

$x(\sqrt{3} - 1) = 10$

$x = \frac{10}{\sqrt{3} - 1}$

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Rationalizing the denominator:

$x = \frac{10}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$x = \frac{10(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2}$

$x = \frac{10(\sqrt{3} + 1)}{3 - 1}$

$x = \frac{10(\sqrt{3} + 1)}{2}$

$x = 5(\sqrt{3} + 1)$

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From equation ②:

$h = x + 10$

$h = 5(\sqrt{3} + 1) + 10$

$h = 5\sqrt{3} + 5 + 10$

$h = 5\sqrt{3} + 15$

$h = 5(1.732) + 15$

$h = 8.66 + 15$

$h = 23.66$ m

Therefore, the height of the tower is approximately 23.65 m.