Chapter 10.7, Problem 15PSB

To determine

Find angle $m\angle Q$ .

Answer to Problem 15PSB

  $m\angle Q=116,80$

Explanation of Solution

Given information:

  

Calculation:

Consider the given figure, if a quadrilateral is inscribed in a circle then its opposite angles are supplementary,

  $\begin{array}{l}m\angle Q+m\angle A=180°\\ 100-2x+x^2=180\\ x^2-2x-80=0\left[quadraticequaton\right]\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{2\pm \sqrt{{\left(-2\right)}^2-4\times 1\times \left(-80\right)}}{2\left(1\right)}\\ x=\frac{2+18}{2}=10,x=\frac{2-18}{2}=-8\\ putx=10andsolveforangle,\\ m\angle Q=100-2\left(10\right)\\ m\angle Q=80\\ putx=-8andsolveforangle,\\ m\angle Q=100-2\left(-8\right)\\ m\angle Q=116\end{array}$

Hence, $m\angle Q=116,80$ .