Chapter 12.7, Problem 18WE

To determine

To determine the exact value of $\frac{a}{b}$ in triangle ABC without using tables or calculators.

Answer to Problem 18WE

  $\frac{a}{b}=\frac{\sqrt{6}}{2}$

Explanation of Solution

Given information :

  $\cos A=\frac{1}{2},\tan B=1$

According to law of sines,

In any triangle ABC,

  

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

Hence,

  $\frac{a}{b}=\frac{\sin A}{\sin B}$

now we shall use the identity ${\sin }^{2}x+{\cos }^{2}x=1$

hence we get,

  $\begin{array}{l}{\sin }^{2}A+{\cos }^{2}A=1\\ {\sin }^{2}A=1-\frac{1}{4}\\ \sin A=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\end{array}$

Further,

  $\begin{array}{l}\tan B=1\\ B={\tan }^{-1}1\\ B={45}^{\circ }\end{array}$

Hence $\sin B=\sin {45}^{\circ }=\frac{1}{\sqrt{2}}$

Therefore $\frac{a}{b}=\frac{\sin A}{\sin B}=\frac{\left(\frac{\sqrt{3}}{2}\right)}{\left(\frac{1}{\sqrt{2}}\right)}=\frac{\sqrt{6}}{2}$