Chapter 5.5, Problem 37E

Solution

a)

To find: The distance between $C$ and $A$ .

The distance between $C$ and $A$ is $54.6\text{ft}$ .

Given information:

The two angles are $\angle BAC=72°$ and $\angle ABC=53°$ .

Formula used:

The sine law is given by,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Calculation:

In the given figure, there aretwo angles and a non-included side (AAS) is given, so the Law of Sines can be used to solve for the triangle.

Find the angle $\angle ACB$ as follows.

  $\begin{array}{c}\angle ACB+\angle BAC+\angle ABC=180°\\ \angle ACB=180°-72°-53°\\ \angle ACB=55°\end{array}$

Use the Law of sines to find the side $AC$ .

  $\begin{array}{c}\frac{\sin \angle ACB}{AB}=\frac{\sin \angle ABC}{AC}\\ \frac{\sin 55°}{56}=\frac{\sin 53°}{AC}\\ AC=\frac{56\sin 53°}{\sin 55°}\\ \approx 54.6\text{ft}\end{array}$

Therefore, the distance between $C$ and $A$ is $54.6\text{ft}$ .

b)

To find:The distance between the two canyon rims.

The distance between the two canyon rims is $51.9\text{ft}$ .

Calculation:

Draw the altitude from $C$ to $AB$ which will be the distance between the rims $h$ as shown follows.

  

Use the sine function to find $h$ .

  $\begin{array}{c}\sin 72°=\frac{h}{54.6}\\ h=54.6\sin \sin 72°\\ \approx 51.9\text{ft}\end{array}$

Therefore, the distance between the two canyon rims is $51.9\text{ft}$ .