From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60°. The pole subtends an angle 30° at the top of the tower. Then the height of the tower is:

Let the pole be at point A (base) with height 20 m (top at point P) and the tower be at point B (base) with height $$h$$ (top at point T). From the base of the pole (A), the angle of elevation to the top of the tower (T) is 60°. Let the horizontal distance between the pole and tower be $$d$$. Then $$\tan 60° = \frac{h}{d} \Rightarrow h = d\sqrt{3}$$.

The pole subtends an angle of 30° at the top of the tower T. From T, looking at the pole which extends from A (ground level) to P (height 20 m), the angle subtended is $$\angle ATP = 30°$$. Let $$\alpha$$ be the angle of depression from T to A and $$\beta$$ be the angle of depression from T to P, so that $$\tan \alpha = \frac{h}{d}$$ and $$\tan \beta = \frac{h - 20}{d}$$. The angle subtended by the pole at T is $$\alpha - \beta = 30°$$.

By the tangent subtraction formula, $$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \tan 30° = \frac{1}{\sqrt{3}}$$, which gives $$\frac{\frac{h}{d} - \frac{h-20}{d}}{1 + \frac{h}{d}\cdot\frac{h-20}{d}} = \frac{1}{\sqrt{3}}$$, then $$\frac{\frac{20}{d}}{1 + \frac{h(h-20)}{d^2}} = \frac{1}{\sqrt{3}}$$ and hence $$\frac{20d}{d^2 + h(h-20)} = \frac{1}{\sqrt{3}}$$.

Substituting $$h = d\sqrt{3}$$ into the denominator yields $$d^2 + h(h-20) = d^2 + d\sqrt{3}(d\sqrt{3} - 20) = d^2 + 3d^2 - 20d\sqrt{3} = 4d^2 - 20d\sqrt{3}$$, so that $$\frac{20d}{4d^2 - 20d\sqrt{3}} = \frac{1}{\sqrt{3}}$$. Simplifying gives $$\frac{20}{4d - 20\sqrt{3}} = \frac{1}{\sqrt{3}}$$, hence $$20\sqrt{3} = 4d - 20\sqrt{3}$$, $$4d = 40\sqrt{3}$$ and $$d = 10\sqrt{3}$$.

Finally, $$h = d\sqrt{3} = 10\sqrt{3} \times \sqrt{3} = 30$$. Therefore, the height of the tower is 30 meters.