Chapter 11.6, Problem 21WE

a.

To determine

To determine the measure of angle AOX and AOB.

Answer to Problem 21WE

  $\begin{array}{l}\angle AOX={48.59}^{\circ }\\ \angle AOB={97.18}^{\circ }\end{array}$

Explanation of Solution

Given information :

  

Here we know that the triangle AXO is a right angle triangle.

Hence,

  $\begin{array}{l}\sin \angle AOX=\frac{AX}{OA}\\ \sin \angle AOX=\frac{9}{12}\\ \angle AOX={\sin }^{-1}\frac{9}{12}\\ \angle AOX={48.59}^{\circ }\end{array}$

Now we know that,

  $\begin{array}{l}\angle AOB=2\angle AOX\\ \angle AOB=2\left({48.59}^{\circ }\right)={97.18}^{\circ }\end{array}$

b.

To determine

To determine the area of shaded region.

Answer to Problem 21WE

Area of shaded region is 50.56 cm².

Explanation of Solution

Given information :

  

Area of shaded region is equal to area of arc AOB minus area of triangle AOB.

Hence we get,

  $\begin{array}{l}A=\frac{{97.18}^{\circ }}{{360}^{\circ }}\times \left(3.14\right){\left(12\right)}^2-\frac{1}{2}\left(18\right)\left(3\sqrt{7}\right)\\ A=122.06-71.43\\ A=50.56\end{array}$

Hence area is 50.56 cm².