Chapter 10.5, Problem 28PSB

To determine

To Find : $m\stackrel{⌢}{BC}$

Answer to Problem 28PSB

  $120°$

Explanation of Solution

Given information :

1) $\text{AB = 6, AC = 6, BC = 6}\sqrt{\text{3}}$

2) $\Delta \text{ABC}$ is inscribed in a circle.

Formula Used:

1) Law of cosine : $a^2=b^2+c^2-2bc\cos (A)$

2) Inscribed angle is half of the enclosed arc.

  $\begin{array}{l}B{C}^2=C{A}^2+A{B}^2-2(CA)(AB)\cos (A)\\ $ {(6\sqrt{3})}^2=6^2+{12}^2-2(6)(12)\cos (A)$108=36+144-144\cos (A)\\ 144\cos (A)=72\\ \cos (A)=\frac{1}{2}\\ A=60°\\ \text{Now, }\frac{1}{2}BC=\angle A\\ \frac{1}{2}\stackrel{⌢}{BC}=60°\\ \stackrel{⌢}{BC}=120°\end{array}$