Trying to find an angle in lines layed on top of a circle

[s3a]

Homework Statement

The problem is attached as TheProblem.jpg and the answer is A.

Homework Equations

Geometry rules.

The Attempt at a Solution

The triangle which has angles of 30deg and 10deg also has an angle of 180deg-30deg-10deg = 140deg and the other side of the line intersection “scissor” is also 140deg. The two equivalent angles on the other part of the line intersection “scissor” are (360deg-2*140deg)/2 = 40deg. I don't know if what I did so far is even useful nor do I know how to proceed from here.

Any help would be greatly appreciated!
Thanks in advance!

 

[Villyer]

First try and identify what arcs you need to know the measure of to find angle ACE.
What formula are you going to ultimately use?

 

[s3a]

I'm really unsure but (1) arc AE and (2) comparing an angle ratio with an arc/circumference ratio?

 

[HallsofIvy]

Have you learned "If AB and CD are lines intersecting inside a circle, then the angle between them is the average of the two arcs they make on the circle"? That is the theorem you need.

 

[s3a]

HallsofIvy, did you mean "difference" instead of "average"? Because, if you did, then I get the correct answer by doing:

angle A0E = 2 * angle ABE = 2 * 30deg = 60deg

then

angle ACE = 1/2 * (60-10) deg = 25deg

 

[Curious3141]

Homework Statement

The problem is attached as TheProblem.jpg and the answer is A.

Homework Equations

Geometry rules.

The Attempt at a Solution

The triangle which has angles of 30deg and 10deg also has an angle of 180deg-30deg-10deg = 140deg and the other side of the line intersection “scissor” is also 140deg. The two equivalent angles on the other part of the line intersection “scissor” are (360deg-2*140deg)/2 = 40deg. I don't know if what I did so far is even useful nor do I know how to proceed from here.

Any help would be greatly appreciated!
Thanks in advance!

Consider the chord BD (and the minor arc BD). What can you say about the relationship between the angles BOD and BED?

That should take you most of the way.

 

[HallsofIvy]

HallsofIvy, did you mean "difference" instead of "average"? Because, if you did, then I get the correct answer by doing:

angle A0E = 2 * angle ABE = 2 * 30deg = 60deg

then

angle ACE = 1/2 * (60-10) deg = 25deg

Well, the rule I was thinking of requires that when two lines intersecting a circle come from outside the circle, the arc it cuts is considered negative. The lines BA and BE intersect an arc of measure 0 (at B) and arc AE. The angle is 30 degrees so we have (AE+ 0)/2= 30 so arc AE has measure 2(30)= 60 degrees.

Arc AD is intercepted by lines BO and DO. Of course, the measure of an angle is defined by the central angle it intercepts so it has measure 10 degrees. (Of course, the other arc those two lines intercept would be 10 degrees also so "average" still works.)

That is, lines CA and CE intercept the circle in arcs AE and BD. They come to arc AE from outside the outside so we consider that arc measure negative: the measure of the angle is (-10+ 60)/2= 25 degrees.