Chapter 12.8, Problem 14WE

To determine

Find the length of the median to $\overline{AC}$ .

Answer to Problem 14WE

7.79

Explanation of Solution

Given:

In $\Delta ABC$ , c = 10 , $\angle A=\angle B=40°$ .

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 :

  

Since $\angle A=\angle B=40°$ , so , the triangle is an isosceles triangle .

Hence , a = b.

  $\angle C=180°-40°-40°=100°$

Use Law of Sine:

  $\begin{array}{l}\frac{a}{\sin 40°}=\frac{10}{\sin 100°}\\ \Rightarrow a=\sin 40°\cdot \frac{10}{\sin 100°}\mathrm{.................}\text{[multiply each side by sin40}°]\\ \Rightarrow a=b\approx 6.527\end{array}$

The median will be from B to the side AC and it bisects the side AC and the angle B.

  

Now, Use Law of Cosine to find BD:

  $\begin{array}{l}B{D}^2=c^2+A{D}^2-2c(AD)\cos A\\ \Rightarrow B{D}^2={10}^2+{3.26}^2-2(10)(3.26)\cos 40°\\ \Rightarrow BD\approx 7.79\end{array}$

So, the length of the median to $\overline{AC}$ is about 7.79