Chapter 10.5, Problem 36PSC

To determine

To Find : $\begin{array}{l}\text{The difference between measured of}\stackrel{⌢}{AB}\\ \text{of the smaller circle and}\stackrel{⌢}{AB}\text{of the larger circle}\text{.}\end{array}$

Answer to Problem 36PSC

$100°$

Explanation of Solution

Given information : $\angle AXB=70°,\stackrel{⌢}{CD}=20°,\stackrel{⌢}{EF}=160°$

Formula Used:

1) The angle between two secants is equal to half the difference of the enclosed arcs.

2) The angle between two chords is equal to half the sum of the intercepted arcs.

  $\begin{array}{l}\text{Let us denote }\stackrel{⌢}{AB}\text{ of smaller circle as }\stackrel{⌢}{AB}\text{(s)}\\ \text{and that of larger circle as }\stackrel{⌢}{AB}(\text{l}).\end{array}$

  $\begin{array}{l}\frac{1}{2}(\stackrel{⌢}{CD}+\stackrel{⌢}{AB}(s))=\angle AXB\\ \frac{1}{2}(20°+\stackrel{⌢}{AB}(s))=70°\\ 20°+\stackrel{⌢}{AB}(s)=140°\\ \stackrel{⌢}{AB}(s)=120°\end{array}$

  $\begin{array}{l}\text{And }\frac{1}{2}(\stackrel{⌢}{EF}-\stackrel{⌢}{AB}(\text{l}))=\angle AXB\\ \frac{1}{2}(160°-\stackrel{⌢}{AB}(\text{l}))=70\\ 160°-\stackrel{⌢}{AB}(\text{l})=140°\\ \stackrel{⌢}{AB}(\text{l})=20°\\ \text{Now}\stackrel{⌢}{AB}(s)-\stackrel{⌢}{AB}(\text{l})=120°\\ =100°\end{array}$