Question

Ninth Grade Geometry - Circles and in desperate need of help.

This is what I did -

146 = 1/2(6x+6-x+10

146 = 1/2(5x+16)

146 = 5x+16

-16 -16

130/5 = 5x/5

26 =x


mED = 6x+6 x+10

6(26)+6 26+10

=162 - 36

162 - 36 = 126°

Answer 1

Explanation:

Your setup was not quite right. The formula is

: (the outer part of the circle minus innner)1/2 = angle

(146 - (6x+6))1/2 = x+10 Be careful to distrubut the negative

(146-6x-6)1/2 = x+10

(140-6x)1/2=x+10

70-3x=x+10

60=4x

x=15

<ECB = x+10 = 15+10= 25

Answer 2

25°

Explanation:

You want the measure of exterior angle ECB marked as (x+10°) given that it intercepts arcs marked as 146° and (6x+6°).

External angle

The measure of the exterior angle is half the difference of the intercepted arcs.

x +10° = 1/2(146° -(6x +6°))

x +10° = 70° -3x . . . . . . . . . . . simplify

4x = 60° . . . . . . . . . . . add 3x-10°

x = 15° . . . . . . . . . divide by 4

The measure of the external angle is ...

∠ECB = x +10° = 15° +10°

∠ECB = 25°

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Additional comment

Often, you don't need to do any math. You only need to do a "reasonableness check" on the offered answer choices.

You know that any inscribed angle in the circle that intercepts arc BD (146°) will have a measure of 146°/2 = 73°. The vertex of an external angle is necessarily farther away from arc BD than any inscribed angle. This means the external angle will have a smaller measure than 73°. That matches only one answer choice.

The inscribed angles we're concerned with here would have their vertex on long arc BED.