Chapter 10, Problem 2CA

To determine

To explain:

The correct order of acute angle from smallest to largest form.

Answer to Problem 2CA

The final order of the acute angle from smallest to largest is:

  $\begin{array}{l}\tan {\theta }_2=\frac{1}{2}\\ \tan {\theta }_3=\frac{\sqrt{3}}{3}\\ \tan {\theta }_1=1\\ \tan {\theta }_6=\sqrt{3}\\ \tan {\theta }_4=\frac{23}{4}\\ \tan {\theta }_5=\frac{38}{5}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}\tan {\theta }_1=1\\ \tan {\theta }_2=\frac{1}{2}\\ \tan {\theta }_3=\frac{\sqrt{3}}{3}\\ \tan {\theta }_4=\frac{23}{4}\\ \tan {\theta }_5=\frac{38}{5}\\ \tan {\theta }_6=\sqrt{3}\end{array}$

Calculation:

The following is the calculation of the acute angles:

  $\begin{array}{l}\tan {\theta }_1=1\\ \tan {\theta }_2=\frac{1}{2}=0.5\\ \tan {\theta }_3=\frac{\sqrt{3}}{3}=0.5774\\ \tan {\theta }_4=\frac{23}{4}=5.75\\ \tan {\theta }_5=\frac{38}{5}=7.6\\ \tan {\theta }_6=\sqrt{3}=1.732\end{array}$

Thus, from above calculation,

  $\tan {\theta }_2<\tan {\theta }_3<\tan {\theta }_1<\tan {\theta }_6<\tan {\theta }_4<\tan {\theta }_5$

Hence, the final order from smallest to largest is:

  $\begin{array}{l}\tan {\theta }_2=\frac{1}{2}\\ \tan {\theta }_3=\frac{\sqrt{3}}{3}\\ \tan {\theta }_1=1\\ \tan {\theta }_6=\sqrt{3}\\ \tan {\theta }_4=\frac{23}{4}\\ \tan {\theta }_5=\frac{38}{5}\end{array}$