Chapter 12.8, Problem 19WE

To determine

Find the lengths of the three sides.

Answer to Problem 19WE

The lengths of the sides of the triangle are 78.43 units, 81.05 units , and 83.43 units.

Explanation of Solution

Given:

The angles of a triangle measure $57°,60°,\text{ and 63}°$ . The longest side is 5 units longer than the

shortest side.

Formula Used:

Law of Cosine :

  $a^2=b^2+c^2-2bc\cos A$

Law of Sine:

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

  

Calculation:

Represent the situation and solve :

  

Since the side opposite to the largest angle will be longest.

So, shortest side is opposite to vertex A and the longest side is opposite to vertex C.

Now, use Law of Sine :

  $\begin{array}{l}\frac{a}{\sin 57°}\text{=}\frac{a+5}{\sin 63°}=\frac{b}{\sin 60°}\\ \Rightarrow \frac{a}{a+5}\text{=}\frac{\sin 57°}{\sin 63°}\\ \Rightarrow \frac{a}{a+5}=0.94\\ \Rightarrow a=(a+5)\mathrm{0.94..................}[\text{multiply each side by a+5]}\\ \Rightarrow a=0.94a+4.706\\ \Rightarrow a-0.94a=\mathrm{4.706............}[\text{subtract 0}\text{.94a from each side]}\\ \Rightarrow 0.06a=4.706\\ \Rightarrow a=\frac{4.706}{0.06}=78.43\\ \Rightarrow a+5=83.43\end{array}$

Use Law of Cosine:

  $\begin{array}{l}{b}^2=a^2+{(}^a-2a(a+5)\cos 60°\\ \Rightarrow b^2={78.43}^2+{83.43}^2-(78.43)(83.43)\frac{1}{2}\\ \Rightarrow b\approx 81.05\end{array}$

So, the lengths of the sides of the triangle are 78.43 units, 81.05 units , and 83.43 units.