Question

The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 10.4degrees. If skyscraper B is 1464 feet​ tall, how far apart are the two​ skyscrapers? Assume the bases of the two buildings are at the same elevation.
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Part 1
The distance from skyscraper A to skyscraper B is
 
​(Round to two decimal places as​ needed.)

Answer 1

To find the distance between the two skyscrapers, we can use the tangent of the angle of inclination. The angle of inclination θ is 10.4 degrees, and the height of skyscraper B (the opposite side of the triangle) is 1464 feet. We want to find the distance (the adjacent side of the triangle), which we can denote as \( d \).

The tangent of an angle in a right triangle is defined as:

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

In this case, we have:

\[
\tan(10.4^\circ) = \frac{1464}{d}
\]

Rearranging to solve for \( d \):

\[
d = \frac{1464}{\tan(10.4^\circ)}
\]

Now, let's calculate \( \tan(10.4^\circ) \) using a calculator or trigonometric table, and then calculate \( d \):

\[
\tan(10.4^\circ) \approx 0.1833 \quad (\text{using a calculator})
\]

Now substituting this value into the equation for \( d \):

\[
d \approx \frac{1464}{0.1833} \approx 7996.25 \text{ feet}
\]

Thus, rounding to two decimal places, the distance from skyscraper A to skyscraper B is approximately:

\[
\boxed{7996.25 \text{ feet}}
\]