From the top of an upright pole $24\sqrt{3}$ feet high, the angle of elevation of the top of an upright tower was $60^\circ$. If the foot of the pole was $60$ feet away from the foot of the tower, what is the height (in feet) of the tower?

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**Solution:**

1. Let the height of the pole be $h_p = 24\sqrt{3}$ feet.
2. Let the height of the tower be $h_t$ feet.
3. The horizontal distance between the pole and the tower is $d = 60$ feet.

From the top of the pole, the angle of elevation to the top of the tower is $60^\circ$. This forms a right triangle where:
- The vertical side is the difference in height between the tower and the pole: $h_t - h_p$.
- The base is the distance between the pole and the tower: $60$ feet.

Using the tangent function:

$$	an 60^\circ = \frac{h_t - h_p}{d}$$

We know $	an 60^\circ = \sqrt{3}$, so:

$$\sqrt{3} = \frac{h_t - 24\sqrt{3}}{60}$$

Multiply both sides by 60:

$$60 	imes \sqrt{3} = h_t - 24\sqrt{3}$$

Add $24\sqrt{3}$ to both sides:

$$h_t = 60\sqrt{3} + 24\sqrt{3} = (60 + 24)\sqrt{3} = 84\sqrt{3}$$

**Answer:** The height of the tower is $84\sqrt{3}$ feet.

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