Question

An observer is 175 meters from the base of a building that has a communication tower built on top. The observer measures the angle of elevation to the top of the building and the top of the communication tower to be 55.01° and 57.53° respectively. Calculate the height of the communication tower

Answer 1

The height of the communication tower is approximately
`\(8.47 \, \text{m}\)`.

To calculate the height of the communication tower, we can use trigonometry. Let's denote the height of the building as
`\(h_{\text{building}}\)` and the height of the communication tower as
`\(h_{\text{tower}}\)`.

First, let's consider the angle of elevation to the top of the building. Using the tangent function:

`\[ \tan(\text{angle of elevation to building}) = \frac{h_{\text{building}}}{\text{distance to building}} \]`

`\[ \tan(55.01°) = \frac{h_{\text{building}}}{175 \, \text{m}} \]`

Now, solve for
`\(h_{\text{building}}\)`:

`\[ h_{\text{building}} = 175 \, \text{m} * \tan(55.01°) \]`

Next, consider the angle of elevation to the top of the communication tower:

`\[ \tan(\text{angle of elevation to tower}) = \frac{h_{\text{building}} + h_{\text{tower}}}{\text{distance to tower}} \]`

`\[ \tan(57.53°) = \frac{h_{\text{building}} + h_{\text{tower}}}{175 \, \text{m}} \]`

Now, solve for
`\(h_{\text{tower}}\)`:

`\[ h_{\text{tower}} = 175 \, \text{m} * \tan(57.53°) - h_{\text{building}} \]`

Now, substitute the value of
`\(h_{\text{building}}\)` into this equation to find
`\(h_{\text{tower}}\)`:

`\[ h_{\text{tower}} = 175 \, \text{m} * \tan(57.53°) - (175 \, \text{m} * \tan(55.01°)) \]`

Let's calculate the expression to find the height of the communication tower:

`\[ h_{\text{tower}} = 175 \, \text{m} * \tan(57.53°) - (175 \, \text{m} * \tan(55.01°)) \]`

Using a calculator:

`\[ h_{\text{tower}} = 175 \, \text{m} * 1.4724 - (175 \, \text{m} * 1.4284) \]\[ h_{\text{tower}} \approx 257.94 \, \text{m} - 249.47 \, \text{m} \]\[ h_{\text{tower}} \approx 8.47 \, \text{m} \]`

Therefore, the height of the communication tower is approximately
`\(8.47 \, \text{m}\)`.