1 Answer

Yallie · Answer 1 · 2021-02-06T11:24:50+0000

Answer:

Part 12)
$Center\ (2,-3),r=2\ units, (x-2)^2+(y+3)^2=4$

Part 13)
$m\angle ABC=47^o$

Explanation:

Part 12) we know that

The equation of a circle in center-radius form is equal to

(x-h)^2+(y-k)^2=r^2

where

(h,k) is the center of the circle

r is the radius of the circle

In this problem

Looking at the graph

The center is the
$point\ (2,-3)$

The radius is
$r=2\ units$

substitute in the expression above

(x-2)^2+(y+3)^2=2^2

(x-2)^2+(y+3)^2=4

Part 13) we know that

The measure of the external angle is the semi-difference of the arcs it covers.

so

$m\angle ABC=(1)/(2)[arc\ DE-arc\ AC]$

we have

$arc\ DE=142^o$

$arc\ AC=48^o$

$m\angle ABC=(1)/(2)[142^o-48^o]$

$m\angle ABC=47^o$

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