Question

The angle of inclination from the base of skyscraper A to the top of skyscraper B is approximately 14.2degrees. If skyscraper B is 1451 feet​ tall, how far apart are the two​ skyscrapers? Assume the bases of the two buildings are at the same elevation.

Answer 1

Answer:

5,734.3 feet (nearest tenth)

Step-by-step explanation:

The problem can be modeled as a right triangle where:

• The triangle's height (vertical leg) is the height of skyscraper B.
• The triangle's base (horizontal leg) is the distance between the bases of the two skyscrapers.
• The angle between the triangle's base and hypotenuse is the angle of inclination from the base of skyscraper A to the top of skyscraper B.

To find the distance between the two skyscrapers, we can use the tangent trigonometric ratio.

$\boxed{\begin{array}{l}\underline{\textsf{Tangent trigonometric ratio}}\\\\\sf \tan(\theta)=\dfrac{O}{A}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ \theta is the angle.}\\\phantom{ww}\bullet\;\textsf{ O is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{ A is the side adjacent the angle.}\end{array}}$

In this case:

• θ = 14.2°
• O = 1451 ft
• A = distance between the two buildings

Substitute these values into the ratio and solve for A:

$\tan 14.2^{\circ}=\dfrac{1451}{A}$

$A=\dfrac{1451}{\tan 14.2^{\circ}}$

$A=5734.2961674...$

$A=5734.3\; \sf ft\;(nearest\;tenth)$

Therefore, the distance between the bases of the two skyscrapers is 5,734.3 feet (rounded to the nearest tenth).

Answer 2

Answer:

5734.3 feet

Step-by-step explanation:

The distance between the two skyscrapers can be found using trigonometry, specifically the tangent function.

Let's denote:

• $d$ as the distance between the skyscrapers,
• $h$ as the height of skyscraper B,
• $\theta$ as the angle of inclination.

The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. In this case:

$\tan(\theta) = \dfrac{h}{d}$

We have the angle of inclination ($\theta = 14.2^\circ$) and the height of skyscraper B ($h = 1451$ feet). We want to find $d$.

$\tan(14.2^\circ) = \dfrac{1451}{d}$

Now, solve for $d$:

$d = \dfrac{1451}{\tan(14.2^\circ)}$

Use a calculator to find the value:

$d \approx \dfrac{1451}{0.253038901}$

$d \approx 5734.296167$

$d \approx 5734.3 \textsf{(in nearest tenth)}$

Therefore, the distance between the two skyscrapers is approximately 5734.3 feet.