Chapter 12.4, Problem 46HP

To determine

To find: The side lengths of two different triangles ABC that can be formed.

Answer to Problem 46HP

  $a=20,b=23.6,c=8.4$

  $a=10,b=11.8,c=4.2$

Explanation of Solution

Given information:

  $\begin{array}{l}A={55}^0\\ C={20}^0\end{array}$

Given

  $\begin{array}{l}A={55}^0\\ C={20}^0\end{array}$

We know

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

Using that

We get

  $\begin{array}{l}A+B+C={180}^0\\ \Rightarrow {55}^0+B+{20}^0={180}^0\\ \Rightarrow {75}^0+B={180}^0\\ \Rightarrow B={180}^0-{75}^0\\ \Rightarrow B={105}^0\end{array}$

Using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin B}{b}\\ \Rightarrow \frac{\sin {55}^0}{a}=\frac{\sin {105}^0}{b}\\ \Rightarrow b=\frac{a\sin {105}^0}{\sin {55}^0}\\ \Rightarrow b=\frac{a\left(0.966\right)}{0.82}\\ \Rightarrow b=1.18a\end{array}$

Again, using law of sines

We get

  $\begin{array}{l}\frac{\sin A}{a}=\frac{\sin C}{c}\\ \Rightarrow \frac{\sin {55}^0}{a}=\frac{\sin {20}^0}{c}\\ \Rightarrow c=\frac{a\sin {20}^0}{\sin {55}^0}\\ \Rightarrow c=\frac{a\left(0.342\right)}{0.82}\\ \Rightarrow c=0.42a\end{array}$

Therefore,

The side lengths will be in the ratio

  $\begin{array}{l}a:b:c=a:1.18a:0.42a\\ \Rightarrow a:b:c=1:1.18:0.42\end{array}$

Therefore,

If $a=10$

Then

  $\begin{array}{l}b=1.18a\\ \Rightarrow b=1.18a\times 10\\ \Rightarrow b=11.8\end{array}$

  $\begin{array}{l}c=0.42a\\ \Rightarrow c=0.42a\times 10\\ \Rightarrow c=4.2\end{array}$

One possible set is

  $a=10,b=11.8,c=4.2$

If $a=20$

Then

  $\begin{array}{l}b=1.18a\\ \Rightarrow b=1.18a\times 20\\ \Rightarrow b=23.6\end{array}$

  $\begin{array}{l}c=0.42a\\ \Rightarrow c=0.42a\times 20\\ \Rightarrow c=8.4\end{array}$

Another set is

  $a=20,b=23.6,c=8.4$