Chapter 9.5, Problem 45PS

Solution

To draw: A diagram to represent the given situation.

Given:

One side of the triangular garden is $62\text{ feet}$ long and another side is $54\text{ feet}$ long. The angle opposite the $62\text{ feet}$ side is ${58}^{\circ }.$

Drawing a diagram according to the given information as:

  

Interpretation:

Here, $AC\text{}\text{and}AB$ are the two sides of the triangular garden of $62\text{ feet}$ and $54\text{ feet}$ respectively. $\angle ABC{\text{=58}}^{\circ }$ is the angle opp.to the side of $62\text{ feet}$ .

To solve: The triangle using the law of sines.

The angle $C$ is ${48}^{\circ },$ angle $A$ is ${74}^{\circ }$ and the side $a=70.26\text{ feet}\text{.}$

Given:

One side of the triangular garden is $62\text{ feet}$ long and another side is $54\text{ feet}$ long. The angle opposite the $62\text{ feet}$ side is ${58}^{\circ }.$

Formula:

The law of sines is:

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

Calculation:

Now, calculating the angle $B$ as:

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin C}{c}\\ \frac{\sin {58}^{\circ }}{62}=\frac{\sin C}{54}\\ \sin C=\frac{0.848\times 54}{62}\\ \sin C=\frac{45.792}{62}\\ \sin C=0.739\\ C={\sin }^{-1}\left(0.739\right)\\ C={47.646}^{\circ }\\ C\approx {48}^{\circ }\end{array}$

As the sum of angles of triangle is ${180}^{\circ }$ , we have:

  $\begin{array}{c}\angle A+\angle B+\angle C={180}^{\circ }\\ \angle A={180}^{\circ }-\left(\angle B+\angle C\right)\\ \angle A={180}^{\circ }-\left({58}^{\circ }+{48}^{\circ }\right)\\ \angle A={180}^{\circ }-{106}^{\circ }\\ \angle A={74}^{\circ }\end{array}$

Now, calculating the side $BC=a$ as:

  $\begin{array}{c}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \frac{\sin {74}^{\circ }}{a}=\frac{\sin {58}^{\circ }}{62}\\ a=\frac{\sin {74}^{\circ }\times 62}{\sin {58}^{\circ }}\\ a=\frac{0.961\times 62}{0.848}\\ a=\frac{59.582}{0.848}\\ a=70.26\text{ feet}\end{array}$

Hence, the angle of $A$ is ${74}^{\circ },$ the angle of $C$ is ${48}^{\circ }$ and the side $BC=a=70.26\text{ feet}\text{.}$

To find: the number of bags of fertilizer to cover the entire garden.

  $8$ bags

Given:

One side of the triangular garden is $62\text{ feet}$ long and another side is $54\text{ feet}$ long. The angle opposite the $62\text{ feet}$ side is ${58}^{\circ }.$

Calculation:

The area of a triangular garden is given by:

  $\begin{array}{l}\text{Area}=\frac{1}{2}bc\sin A\\ \text{Area}=\frac{1}{2}\times 62\times 54\times \sin 74\\ \text{Area}=31\times 54\times 0.961\\ \text{Area}=1608.714{\text{ ft}}^2\end{array}$

Hence, the total area of triangular garden is $1608.714{\text{ ft}}^2$ and the area covered by one bag of fertilizer is $200{\text{ ft}}^2.$ So, calculating the number of bags as:

  $\begin{array}{c}\text{Number of bags}=\frac{\text{Total Garden Area}}{\text{Area covered by one bag}}\\ \text{n}=\frac{1608.714}{200}\\ \text{n}=8.04\\ \text{n}\approx \text{8 bags}\end{array}$