Question

Find the measure of ∠ECB.

Circle A is intersected by line CD at points D and E and line CB at point B, forming angle ECB outside of the circle, the measure of angle ECB is x plus 10 degrees, arc EB is 6x plus 6 degrees, and arc DB 146 degrees.

Answer 1

This might be somewhat difficult to visualize without a drawing, but here's what I figure:
You can make an isocoles triangle with points D, B, and the center of the circle. It must be an isoceles triangle because D-center and B-center are both the same length, which is the length of the radius. And we are given one of the angles in this problem as 146, in the problem (I'll attempt to illustrate below)

Answer 2

Now, remember that there are always 180 degrees in a triangle. And we have an isosceles triangle, so the remaining two angles must be equal. One of the angles is already 146, which leaves 34 degrees remaining. Split these up evenly between the two remaining angles and we get 17 degrees per angle

Answer 3

Now, since the question says that line CB intersects the circle at exactly 1 point, B, we know that CB is a tangent line. And one property of tangent lines is that if you draw an angle between the tangent line connecting the center of the circle to the point of intersection, you get a right angle.

Answer 4

We can then find the measure of angle CBD by subtracting 17 from 90, which gives us 77. The drawing is getting pretty cluttered now, so I'll try to redraw to summarize what we have so far.

Answer 5

I also added the angles of arc EB and angle ECB (apologies for the bad handwriting),
6x+6 and x+10, respectively.

Answer 6

Truth be told, I'm a bit stumped here. Maybe it's because my trigonometry is a bit rusty.
I'll try to come back to this one, but for now, see if you can reach out to others for help

Answer 7

Ok, so from what I have so far, I've drawn a quadrilateral with E, C, B, and the Center (I'll call that point X), and we have the angles of 3 of these points as (x+10), (90), and (6x+6). These three, plus angle XEC must add up to 360.
Meanwhile, we have a triangle marked by points B, X, and the intersection of EX and DB (I'll call this point Y). We know angle YBX is 17, angle BXY is 6x+6, and these must add together with XYB to total 180 (the degrees of a triangle).
In addition, we can look at the quadrilateral made by points B, Y, E, and C, all of which must add up to 360. CBY is 73, YEC=XEC (the unknown from above), ECX is still x+10, and BYE adds up with XYB to total 180 (the degrees of a straight line).
I'll represent the measure of angles YEC and XEC as a and the measure of angle XYB as b.

Answer 8

To summarize:
x+10+6x+6+90+a=360
17+6x+6+b=180
73+a+x+10+(180-b)=360

Answer 9

Simplifying this:
7x+106+a=360;
7x+a=254
23+6x+b=180;
6x+b=157
a+x-b+263=360
a+x-b=97

Answer 10

Unfortunately, I'm still not sure that gets us anywhere closer to the answer. It was worth a try, but I must still be missing something here...