We are given a scenario where the angles of elevation to the bottom and top of a transmission tower are 45° and 60° respectively, with the tower mounted on a 20 m high building. We need to determine the height of the tower.

Let's break the problem down step by step:

Step 1: Assign Variables

• Let $h_t$ represent the height of the transmission tower.
• The height of the building is given as 20 m.
• Let $d$ be the horizontal distance from the point on the ground to the base of the building.

Step 2: Use Trigonometry

We'll use tangent functions because we have right-angled triangles formed by the ground, the building/tower, and the line of sight.

For the bottom of the transmission tower (building only):

Since the angle of elevation to the bottom of the tower is 45°, we apply: $\tan(45^\circ) = \frac{\text{Height of the building}}{d}$ $\tan(45^\circ) = \frac{20}{d}$ Since $\tan(45^\circ) = 1$, we have: $d = 20 \, \text{m}$

For the top of the transmission tower:

The angle of elevation to the top of the transmission tower is 60°, and the total height (building + tower) is $20 + h_t$. Using the tangent formula again: $\tan(60^\circ) = \frac{20 + h_t}{d}$ $\sqrt{3} = \frac{20 + h_t}{20}$ Multiplying both sides by 20: $20\sqrt{3} = 20 + h_t$ Now solve for $h_t$: $h_t = 20\sqrt{3} - 20$ $h_t \approx 34.64 - 20 = 14.64 \, \text{m}$

Final Answer:

The height of the transmission tower is approximately $14.64 \, \text{m}$. The correct option is A) 14.64 m.

Do you need more details or have any further questions? Here are some related questions you might find helpful:

1. How do we calculate horizontal distance when different angles of elevation are given?
2. What is the role of the tangent function in right-angled triangles?
3. How can we use trigonometric identities for solving elevation problems?
4. What happens if the angle of elevation changes?
5. How can this method apply to more complex shapes?

Tip: Always ensure you identify the reference point and correct height/distance when solving angle of elevation problems.