Chapter 12.6, Problem 6WE

To determine

To find the indicated part of triangle ABC.

Answer to Problem 6WE

  $b=290.93$

Explanation of Solution

Given information :

  

Here we shall use the cosine formula.

Cosine formula states that,

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

Hence we put the values given,

  $\begin{array}{l}\cos {150}^{\circ }=\frac{{100}^2+{200}^2-b^2}{2\left(100\right)\left(200\right)}\\ -0.866=\frac{{100}^2+{200}^2-b^2}{2\left(100\right)\left(200\right)}\\ -34641=50000-b^2\\ b^2=84641\\ b=290.93\end{array}$