Chapter 10.5, Problem 7PSA

To determine

To find : $\angle AXB\text{ and }\angle y$

Answer to Problem 7PSA

  $\angle AXB=85°,\angle y=31°$

Explanation of Solution

Given information : $\stackrel{⌢}{AB}=108°\text{and}\stackrel{⌢}{CD}=62°$

  

  $\text{From }\Delta BXC$

  $\angle AXB=\angle XBC+\angle XCB\text{[Exterior angle of a triangle is sum}$

  $\text{of the interior angles]}$

  $\angle ACB=\frac{1}{2}\angle AOB\text{[Central angle is twiceof inscribed angle]}$

  $\therefore \angle ACB=\frac{1}{2}\times 108°=54°$

  $\angle DBC=\frac{1}{2}\angle DOC=\frac{1}{2}\times 62°=31°$

  $\therefore \angle AXB=\angle XBC+\angle XCB=(54°+31°)=85°$

  $\angle DBC=\angle DAC\text{[They are on same arc DC]}$

  $\text{From }\Delta ACY,\angle ACB=\angle CAY+\angle AYC$

  $\begin{array}{l}\Rightarrow \angle AYC=\angle ACB-\angle CAY\\ \Rightarrow \angle y=54°-31°\\ \text{Hence, }\angle AXB=85°,\angle y=31°\end{array}$