Chapter 8.1, Problem 66AYU

Willis Tower Willis Tower in Chicago is the second tallest building in the United States and is topped by a high antenna. A surveyor on the ground makes the following measurements $:$

The angle of elevation from his position to the top of the building is ${34}^{\circ }.$

The distance from his position to the top of building is $2595$ feet.

The distance from his position to the top of antenna is $2760$ feet.

How far away from the (base of the) building is the surveyor located $?$

How tall is the building $?$

What is the angle of elevation from the surveyor to the top of the antenna $?$

How tall is the antenna $?$

To determine

The distance between the surveyor located and the base of the Willis Tower in

Chicago the tallest building in the United States and is topped by a high antenna.

Answer to Problem 66AYU

Solution:

The distance between the surveyor and the base of the building is approximately $2151$ feet.

Explanation of Solution

Given information:

A high antenna is mounted on top of the Willis Tower in Chicago, which is the second tallest building in the United States.

The measurements by surveyor on the ground are:

The angle of elevation from the surveyor’s position to the top of the building is ${34}^{\circ }$.

The distance from the surveyor’s position to the top of the building is $2595$ feet.

The distance from the surveyor’sposition to the top of the antenna is $2760$ feet.

Explanation:

From the given information, the diagram of the building is as shown below:

In the right angled triangle $BCD$, an angle $B$ is ${34}^{\circ }$ and the hypotenuse is $2595\text{ft}$ and $BC$ is the

adjacent side of the triangle which represents the distance between the surveyor and the base of

the building.

$BC$ is found by using cos ratio in the right triangle $BCD$,

$\cos (B)=\frac{BC}{BD}$

By substituting the values of angle $B={34}^{\circ }$ and $BD=2595$, it gives,

$\cos ({34}^{\circ })=\frac{BC}{2595}$

$\Rightarrow BC=2595\cdot \cos ({34}^{\circ })$

$\Rightarrow BC=2151.35250\approx 2151\text{ft}$

Therefore, the surveyor is located approximately $2151$ feet from the base of the building.

To determine

The height of the Willis Tower.

Answer to Problem 66AYU

Solution:

The height of the Willis Tower is $1451$ feet.

Explanation of Solution

Given information:

A high antenna is mounted on top of the Willis Tower in Chicago, which is the second tallest building in the United States.

The measurements by surveyor on the ground are:

The angle of elevation from the surveyor’s position to the top of the building is ${34}^{\circ }$.

The distance from the surveyor’s position to the top of the building is $2595$ feet.

The distance from the surveyor’s position to the top of the antenna is $2760$ feet.

Explanation:

From the given information, the diagram of the building is as shown below:

In the right angled triangle $BCD$, angle $B$ is ${34}^{\circ }$ and the hypotenuse is $2595\text{ft}$ and $DC$ is the

opposite side of the triangle which represents the height of the building from the ground.

By using the sin ratio in the right triangle $BCD$,

$\sin (B)=\frac{CD}{BD}$

By substituting the values of angle $B={34}^{\circ }$ and $BD=2595$, it gives,

$\Rightarrow \sin \left({34}^{\circ }\right)=\frac{\text{CD}}{2595}$

$\Rightarrow \text{CD=2595}\cdot \text{sin}\left({34}^{\circ }\right)$

$\Rightarrow \text{CD=1451}\text{.1056}\approx \text{1451}\text{ft}$

Therefore, the height of the building from the ground is about $1451$ feet.

To determine

The angle of elevation from the surveyor to the top of the antenna.

Answer to Problem 66AYU

Solution:

The angle of elevation from the surveyor to the top of the antenna is ${38.8}^{\circ }$.

Explanation of Solution

Given information:

A high antenna is mounted on top of the Willis Tower in Chicago, which is the second tallest building in the United States.

The measurements by surveyor on the ground are:

The angle of elevation from the surveyor’s position to the top of the building is ${34}^{\circ }$.

The distance from the surveyor’s position to the top of the building is $2595$ feet.

The distance from the surveyor’s position to the top of the antenna is $2760$ feet.

Explanation:

From the given information, the diagram of the building is as shown below:

Consider the right angled triangle $BCA$, the angle $B$ is the angle of elevation from the surveyor

to the top of the building.

Here the hypotenuse is $2760\text{ft}$ and by the subpart a the adjacent side BC is $2151\text{ft}$.

By using the cos ratio in the right triangle $BCA$,

$\cos (B)=\frac{BC}{AB}$

By substituting the values of $BC=2151$ feet, $AB=2760$ feet,

$\cos (B)=\frac{2151}{2760}=0.77935$

$\Rightarrow B={\cos }^{-1}(0.77935)={38.8}^{\circ }$

Therefore, the angle of elevation from the surveyor to the base of the triangle is ${38.8}^{\circ }$.

To determine

The height of the antenna

Answer to Problem 66AYU

Solution:

The height of the antenna is $\text{278}$ feet.

Explanation of Solution

Given information:

A high antenna is mounted on top of the Willis Tower in Chicago, which is the second tallest building in the United States.

The measurements by surveyor on the ground are:

The angle of elevation from the surveyor’s position to the top of the building is ${34}^{\circ }$.

The distance from the surveyor’s position to the top of the building is $2595$ feet.

The distance from the surveyor’s position to the top of the antenna is $2760$ feet.

Explanation:

From the given information, the diagram of the building is as shown below:

Consider the right triangle $BCA$, $AD$ represents the antenna.

From part (c), the angle $B={38.8}^{\circ }$.

By using the sin ratio in the right triangle $BCA$,

$\sin (B)=\frac{\text{AC}}{\text{AB}}$

By substituting the values of $B={38.8}^{\circ }$ and $AB=2760$, it gives,

$\sin (B)=\frac{\text{AC}}{2760}$.

$\Rightarrow \sin ({38.8}^{\circ })=\frac{\text{AC}}{2760}$

$\Rightarrow \text{CA}=\sin ({38.8}^{\circ })\cdot 2760$

$\Rightarrow \text{AC}=1729.4265\approx 1729.43$ feet.

From the diagram $\text{AC}\text{=}\text{}\text{AD}\text{+}\text{}\text{DC}$

$\Rightarrow \text{AD}\text{=}\text{AC}\text{-}\text{DC}$

$\text{AC}\text{=}1729.43$ feet, and from part b, $\text{DC}\text{=}\text{1451}$ feet.

$\text{AD}\text{=}1729.43\text{-}\text{1451}\text{=}\text{278}\text{.43}\approx \text{278}$

Therefore, the height of the antenna is approximately $\text{278}$ feet.