Chapter 10, Problem 15RP

a)

To determine

To find : Probability than a point lies on $\stackrel{⌢}{BA}$ .

Answer to Problem 15RP

  $\frac{2}{9}$

Explanation of Solution

Given Information :

A point is chosen at random on $\odot P$ .

  

Construction :

Join TA, TP and PB.

Formula Used:

Angle subtended by an arc at center is twice the angle subtended at any point on the circle.

Calculation :

Since, $\angle ATB=40°,\angle APB=80°=\stackrel{⌢}{BA}$

Again, angle subtended by AT in the center = $70°$

  $\begin{array}{l}\therefore \angle TAB=\stackrel{⌢}{TUB}=360°-(70+80°)=210\\ \therefore P(\stackrel{⌢}{BA})=\frac{80}{360}=\frac{2}{9}\end{array}$

b)

To determine

To find : Probability that a point lies on $\stackrel{⌢}{TUB}$ .

Answer to Problem 15RP

  $\frac{7}{12}$

Explanation of Solution

Given Information :

A point is chosen at random on $\odot P.$

Formula Used:

Angle subtended by ar arc at center is twice the angle subtended at any point on the circle.

Calculation :

  $\begin{array}{l}\angle TPB=\stackrel{⌢}{TUB}=360°-(70°+80°)=210°\\ \therefore P(\stackrel{⌢}{TUB})=\frac{210°}{360°}=\frac{7}{12}\end{array}$