From the top of an upright pole $17.75 ext{ m}$ high, the angle of elevation of the top of an upright tower was $60^\circ$. If the tower was $57.75 ext{ m}$ tall, how far away (in meters) from the foot of the pole was the foot of the tower?

Question

From the top of an upright pole $17.75 	ext{ m}$ high, the angle of elevation of the top of an upright tower was $60^\circ$. If the tower was $57.75 	ext{ m}$ tall, how far away (in meters) from the foot of the pole was the foot of the tower?

Accepted Answer

**Solution:**

1. Let the distance between the foot of the pole and the foot of the tower be $x$ meters.

2. The height difference between the tower and the pole is:
   
   $$AC = 57.75 - 17.75 = 40 	ext{ m}$$

3. From the top of the pole, the angle of elevation to the top of the tower is $60^\circ$. Using the right triangle formed,

$$	an 60^\circ = \frac{AC}{x}$$

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

$$\sqrt{3} = \frac{40}{x} \implies x = \frac{40}{\sqrt{3}}$$

5. Rationalize the denominator:

$$x = \frac{40}{\sqrt{3}} 	imes \frac{\sqrt{3}}{\sqrt{3}} = \frac{40\sqrt{3}}{3} 	ext{ m}$$

**Answer:** $\boxed{\frac{40\sqrt{3}}{3}}$ meters

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