Chapter 12.8, Problem 13WE

To determine

Find the length of the median to the longest side.

Answer to Problem 13WE

2.45

Explanation of Solution

Given:

In $\Delta ABC$ , a = 4 , b = 5 , and $\angle C=110°$ .

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 :

  

Use Law of Cosine to find c :

  $\begin{array}{l}{c}^2=a^2+b^2-2ab\cos C\\ \Rightarrow c^2=4^2+5^2-2(4)(5)\cos 115°\\ \Rightarrow c=\sqrt{57.90}\\ \Rightarrow c\approx 7.61\end{array}$

So, AB is the largest side.

Hence , the median will be from C to the side AB and it bisects the side AB and the angle C:

  

Use Law of Sine:

  $\begin{array}{l}\frac{\sin 115°}{7.61}=\frac{\sin A}{4}\\ \Rightarrow \sin A=4\cdot \frac{\sin 115°}{7.61}\mathrm{.................}\text{[multiply each side by 4}]\\ \Rightarrow A={\sin }^{-1}(0.47637)\\ \Rightarrow A\approx 28.45°\end{array}$

Now, Use Law of Cosine to find CD:

  $\begin{array}{l}C{D}^2=b^2+A{D}^2-2b(AD)\cos A\\ \Rightarrow C{D}^2=5^2+{3.805}^2-2(5)(3.805)\cos 28.45°\\ \Rightarrow CD=\sqrt{6.023}\\ \Rightarrow CD\approx 2.45\end{array}$

So, the length of the median to the longest side is about 2.45