Chapter 12.4, Problem 41PPS

a.

To determine

To find: Sketch a drawing of the situation and Label the missing sides $a$ and $b$ .

Explanation of Solution

Given information:

One of a triangular cycling path if $4$ miles long.

The angle opposite this side is ${64}^0$ .

Another angle formed by the triangular path measures ${66}^0$ .

Given

One of a triangular cycling path if $4$ miles long.

The angle opposite this side is ${64}^0$ .

Another angle formed by the triangular path measures ${66}^0$ .

Sketch of given data is

  

b.

To determine

To find: Equations that could be used to find the lengths of the missing sides.

Answer to Problem 41PPS

  $\frac{\sin {50}^0}{b}=\frac{\sin {64}^0}{4}$

  $\frac{\sin {66}^0}{a}=\frac{\sin {64}^0}{4}$

Explanation of Solution

Given information:

One of a triangular cycling path if $4$ miles long.

The angle opposite this side is ${64}^0$ .

Another angle formed by the triangular path measures ${66}^0$ .

It means

  $\begin{array}{l}A={66}^0\\ C={64}^0\\ c=4\end{array}$

Given

  $\begin{array}{l}A={66}^0\\ C={64}^0\\ c=4\end{array}$

We know

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {66}^0}{a}=\frac{\sin {64}^0}{4}\end{array}$

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {66}^0+B+{64}^0={180}^0\\ \Rightarrow {130}^0+B={180}^0\\ \Rightarrow B={180}^0-{130}^0\\ \Rightarrow B={50}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {50}^0}{b}=\frac{\sin {64}^0}{4}\end{array}$

Therefore,

Equations that could be used to find the lengths of missing sides are

  $\frac{\sin {50}^0}{b}=\frac{\sin {64}^0}{4}$

  $\frac{\sin {66}^0}{a}=\frac{\sin {64}^0}{4}$

c.

To determine

To find: Perimeter of the triangle.

Answer to Problem 41PPS

Perimeter of triangle is $11.5mi$

Explanation of Solution

Given information:

One of a triangular cycling path if $4$ miles long.

The angle opposite this side is ${64}^0$ .

Another angle formed by the triangular path measures ${66}^0$ .

It means

  $\begin{array}{l}A={66}^0\\ C={64}^0\\ c=4\end{array}$

Given

  $\begin{array}{l}A={66}^0\\ C={64}^0\\ c=4\end{array}$

We know

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {66}^0}{a}=\frac{\sin {64}^0}{4}\\ \Rightarrow a=\frac{4\sin {66}^0}{\sin {64}^0}\\ \Rightarrow a=\frac{4\left(0.9135\right)}{0.8988}\\ \Rightarrow a=\frac{3.654}{0.8988}\\ \Rightarrow a=4.1\end{array}$

Sum of angles of a triangle is ${180}^0$

  $\begin{array}{l}\Rightarrow A+B+C={180}^0\\ \Rightarrow {66}^0+B+{64}^0={180}^0\\ \Rightarrow {130}^0+B={180}^0\\ \Rightarrow B={180}^0-{130}^0\\ \Rightarrow B={50}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin B}{b}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {50}^0}{b}=\frac{\sin {64}^0}{4}\\ \Rightarrow b=\frac{4\sin {50}^0}{\sin {64}^0}\\ \Rightarrow b=\frac{4\left(0.766\right)}{0.8988}\\ \Rightarrow b=\frac{3.064}{0.8988}\\ \Rightarrow b=3.4\end{array}$

Therefore,

Perimeter of triangle is

  $\begin{array}{l}P=a+b+c\\ \Rightarrow P=4.1+3.4+4\\ \Rightarrow P=11.5mi\end{array}$

Therefore,

Perimeter of triangle is $11.5mi$