Question

'You are standing 100 ft away from a veyr tall builidnd. Your line of sight is 6 ft from the ground. In f the angle of elevation (measured by a protractor) to the top of the building is 80 degrees, how tall is the building?"

Answer 1

To find the height of the building, we can use some basic trigonometry. We will set up a right triangle where:

- The distance from the observer to the base of the building is the adjacent side (100 ft).
- The height of the building above the observer's eye level is the opposite side.
- The angle of elevation to the top of the building is 80 degrees.

Let \( h \) be the height of the building above the line of sight (which is 6 ft from the ground). We can use the tangent of the angle of elevation:

\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]

In this case,

\[
\tan(80^\circ) = \frac{h_{\text{above eye level}}}{100}
\]

We solve for \( h_{\text{above eye level}} \):

\[
h_{\text{above eye level}} = 100 \cdot \tan(80^\circ)
\]

Now, we calculate \( \tan(80^\circ) \) using a calculator:

\[
\tan(80^\circ) \approx 5.6713
\]

Then, substituting this value:

\[
h_{\text{above eye level}} \approx 100 \cdot 5.6713 \approx 567.13 \text{ ft}
\]

Now, to find the total height of the building, we must add the height of the observer's line of sight (6 ft) to this value:

\[
\text{Total height of the building} = h_{\text{above eye level}} + 6 \approx 567.13 + 6 \approx 573.13 \text{ ft}
\]

Therefore, the tall building is approximately **573.13 feet** tall.