Chapter 5.6, Problem 39E

Solution

(a)

To find: The measure of angle $\angle CAE$ .

The measure of angle $\angle CAE$ is $18.4°$ .

Given information:

Length of the opposite side is $6\text{ft}$ and adjacent side is $18\text{ft}$ .

Formula used:

The tangent function is given as, $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}$ .

Calculation:

Substitute $6\text{ft}$ for the opposite, $18\text{ft}$ for adjacent, and $\angle CAE$ for $\theta$ in tangent fuction to calculate the measure of $\angle CAE$ .

  $\begin{array}{c}\tan \angle CAE=\frac{6}{18}\\ \angle CAE={\tan }^{-1}\left(\frac{6}{18}\right)\\ =18.4°\end{array}$

Therefore, measure of angle $\angle CAE$ is $18.4°$ .

(b)

To find: The length of $DF$ .

The length of $DF$ is $4.48\text{ft}$ .

Given information:

The length of $AF$ is $12\text{ft}$ .

The length of $AD$ is $9\text{ft}$ .

From part (a), $\angle A=18.4°$

Formula used:

Law of Cosine

  $D{F}^2=\left(A{F}^2\right)+{\left(AD\right)}^2-2\times AF\times AD\cos \left(A\right)$

Calculation:

Substitute $12\text{ft}$ for $AF$, $9\text{ft}$ for $AD$, and $18.4°$ for $\angle A$ in $D{F}^2=\left(A{F}^2\right)+{\left(AD\right)}^2-2\times AF\times AD\cos \left(A\right)$ to calculate the length of $DF$ .

  $\begin{array}{c}D{F}^2=\left(A{F}^2\right)+{\left(AD\right)}^2-2\times AF\times AD\cos \left(A\right)\\ D{F}^2=\left({12}^2\right)+{\left(9\right)}^2-2\times 12\times 9\cos \left(18.4\right)\\ DF=4.48\text{ft}\end{array}$

Therefore, length of $DF$ is $4.48\text{ft}$ .

(c)

To find: The length of $EF$ .

The length of $EF$ is $7.6\text{ft}$ .

Given information:

The length of $AE$ is $18\text{ft}$ .

From part (a), $\angle A=18.4°$

From part (b), $DF=4.48\text{ft}$

Formula used:

Law of Cosine

  $E{F}^2=\left(D{F}^2\right)+{\left(AE\right)}^2-2\times DF\times AE\cos \left(A\right)$

Calculation:

Substitute $18\text{ft}$ for $AE$, $4.48\text{ft}$ for $DF$, and $18.4°$ for $\angle A$ in $E{F}^2=\left(D{F}^2\right)+{\left(AE\right)}^2-2\times DF\times AE\cos \left(A\right)$ to calculate the length of $EF$ .

  $\begin{array}{c}E{F}^2=\left(D{F}^2\right)+{\left(AE\right)}^2-2\times DF\times AE\cos \left(A\right)\\ $ E{F}^2=\left({12}^2\right)+{(18)}^2-2\times 12\times 18\cos \left(18.4\right)$EF=7.6\text{ft}\end{array}$

Therefore, length of $EF$ is $7.6\text{ft}$ .