Chapter 16.3, Problem 5PS

To determine

To find the measure of a side of the triangle if the other two sides and the bisector of angle they include have measures of $3,5$ and $2$ respectively.

Answer to Problem 5PS

  $BC=\frac{8\sqrt{165}}{15}$

Explanation of Solution

Given:

The following is the drawn figure-

  

Calculation:- The angle bisector divides the opposite side in segments proportional to the angle's sides.

  $\begin{array}{l}\frac{AB}{AC}=\frac{BD}{DC}\\ \frac{3}{5}=\frac{BD}{DC}\\ 3DC=5BD\\ DC=\frac{5}{3}BD\end{array}$

Using Stewart’s theorem-

  $A{B}^2\cdot DC+A{C}^2\cdot BD=BC\cdot A{D}^2+BC\cdot BD\cdot DC$

Now, substituting the values-

  $\begin{array}{l}{3}^2\cdot DC+5^2\cdot BD=BC\cdot 2^2+BC\cdot BD\cdot DC\\ 9\cdot \frac{5}{3}BD+25BD=(BD+\frac{5}{3}BD)\cdot 4+(BD+\frac{5}{3}BD)\cdot BD\cdot \frac{5}{3}BD\\ 15BD+25BD=\frac{32}{3}BD+\frac{40}{9}B{D}^3\\ 360BD=96BD+40B{D}^3\\ 40B{D}^3-264BD=0\\ BD=0\\ BD=\sqrt{\frac{33}{5}}=\frac{\sqrt{165}}{5}\end{array}$

The only positive solution is $BD=\frac{\sqrt{165}}{5}$ ,

Therefore, the length of the side BC is $BC=DC+BD=\frac{5}{3}\cdot \frac{\sqrt{165}}{5}+\frac{\sqrt{165}}{5}=\frac{8\sqrt{165}}{15}$