Chapter 9.6, Problem 23WE

To determine

To calculate: Measure of the four non − overlapping arcs.

Answer to Problem 23WE

The four arcs are 90, 86, 84 and 100.

Explanation of Solution

Given information: 

Angles Quadrilateral: $80°,90°,94°\text{, and }96°$ .

Formula used: 

The measure of an angle formed by two tangents is equal to half the difference of the two arcs.

Calculation:

Let the angles be  $80°,90°,94°\text{, and }96°$ , and let the points of tangency be $x$ (between A and B), y, z, w in order.

The measure of an angle formed by two tangents is equal to half the difference of the two arcs, we write:

  $\begin{array}{l}A=\frac{1}{2}\left(\stackrel{⌢}{XYZ}-\stackrel{⌢}{XW}\right),\\ B=\frac{1}{2}\left(\stackrel{⌢}{YZX}-\stackrel{⌢}{YX}\right),\\ C=\frac{1}{2}\left(\stackrel{⌢}{ZWY}-\stackrel{⌢}{ZY}\right),\text{ and}\\ D=\frac{1}{2}\left(\stackrel{⌢}{WXZ}-\stackrel{⌢}{WZ}\right).\end{array}$

Let $a=\left(\stackrel{⌢}{XY}\right)$ , $b=\left(\stackrel{⌢}{YZ}\right)$ , $c=\left(\stackrel{⌢}{ZW}\right)$ , and $d=\left(\stackrel{⌢}{WX}\right)$ .

So,

  $\begin{array}{l}80=\frac{1}{2}\left(a+b+c-d\right),\\ 90=\frac{1}{2}\left(a+b+c-a\right),\\ 94=\frac{1}{2}\left(a+b+c-b\right),\text{ and}\\ 96=\frac{1}{2}\left(a+b+c-c\right).\end{array}$

Since $a=90$ , we get:

  $\begin{array}{l}70=b+c-d\mathrm{....}(1),\\ 270=b+c+d\text{.}\mathrm{...}(2),\\ 98=-b+c+d\mathrm{....}(3),\text{ and}\\ 102=b-c+d\text{.}\mathrm{...}(4)\\ \text{Adding eq - (1) and (2), yields:}\\ 340=2b+2c\text{.}\mathrm{...}(5)\\ \text{Adding eq - (3) and (4), yields:}\\ 200=2d,\\ \Rightarrow d=100.\\ \text{Again, adding eq - (2) and (3), yields:}\\ 368=2c+2d,\\ \Rightarrow 2c=168,\\ \Rightarrow c=84.\\ \text{Thus, substituting above values in eq - (5), we obtain:}\\ \text{b}=\text{86}\text{.}\end{array}$