Chapter 9.3, Problem 12PSB

(a)

To determine

To Find: The coordinates of the point $C$ .

Answer to Problem 12PSB

The coordinates of point $C$ are (0, − 12).

Explanation of Solution

Given information :

  $A\left(12,0\right),B\left(0,12\right)\text{ and }C (x,y)$

Formula used :

Point lies on negative axis will have coordinates with − ive sign

Calculation :

Since, point $B$ is lying on positive axis of $Y$ at a distance of 12. Similarly, point $C$ is lying on negative axis of $Y$ at a distance of -12.

Thus, the coordinates of point $C$ are (0, − 12)

(b)

To determine

To find: The measure of arc $\stackrel{⌢}{ABC}\text{, i}\text{.e}\text{., }m(\stackrel{⌢}{ABC)}$ .

Answer to Problem 12PSB

The required measure of the arc $m(\stackrel{⌢}{ABC)}$ is ${270}^{\text{o}}$

Explanation of Solution

Given information :

The points $A\left(12,0\right),B\left(0,12\right)\text{ and }C$ are lying on the circumference.

Formula used :

  $\text{The measure of the arc = Angle formed by the arc at the center of the circle}$

Calculation :

From the question figure, the arc is making an angle of 270 $°$ at the center of the circle.

Therefore, the measure of the arc $\stackrel{⌢}{ABC}$ is 270 $°$

(c)

To determine

To find: The length of the arc $\stackrel{⌢}{ABC}$

Answer to Problem 12PSB

The length of arc $\stackrel{⌢}{ABC}$ is $18\pi$ .

Explanation of Solution

Given information :

The points $A\left(12,0\right),B\left(0,12\right)\text{ and }C$ are lying on the circle.

  $\text{Μeasure of the arc (}\theta )={270}^{\text{o}}$ (From section b)

  $\text{Radius (}r\text{) = 12}$

  $\text{Circumference}=2\pi r$

Formula used :

  $\text{The length of the arc}=\frac{\text{Measure of the Arc (}\theta \text{) }}{{360}^{\text{o}}}\times \text{Circumference}$

Calculation :

  $\begin{array}{l}\text{The length of the arc}=\frac{\theta }{{360}^{\text{o}}}\times \text{Circumference}\\ \text{=}\frac{{270}^{\text{o}}}{{360}^{\text{o}}}\times 2\text{π}r\\ =\frac{{270}^{\text{o}}}{{360}^{\text{o}}}\times 2\text{π}\times 12=18\text{π}\end{array}$