Circle Thereoms, how do I find x?

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http://tinypic.com/view.php?pic=j6raqf&s=6

and http://tinypic.com/view.php?pic=xm48w3&s=6

The first one is 94 degrees.

ruveyn

Here's what I used for the first one:

A tangent to a circle forms a right angle with the circle's radius, at the point of contact of the tangent.

Also, if two tangents are drawn on a circle and they cross, the lengths of the two tangents (from the point where they touch the circle to the point where they cross) will be the same.

Then find x using by forming two triangles with the angle and adding their measure.

I found the theorems here:
Circle Theorems

I've taken the liberty to mangle your figure:



As the poster above me mentioned, the lines AC and BC make a 90 degree angle with the tangential lines. Angle ACB (this notation means: the angle at point C between lines AC and CB) is double that of the indicated 43 degree angle (I don't recall the name of this theorem, but it's a basic one. It's called "Angle at the Centre" in the link posted by starkid), so 86 degrees. So now we have the quadrilateral ACBX, with the sum of angles of 360 degrees. Since XAC and XBC are right angles (90 degrees) and ACB is 86 degrees, we find that the answer, angle AXB is 94 degrees.

The second problem uses the "alternate segment theorem" from the Circle Theorems link that starkid posted. The corner of the triangle that is unmarked in the figure has an angle of 76 degrees according to this theorem. Now it's simply a matter of computing X as the sum of angles in a triangle is 180 degrees: X = 180 - 60 - 76 = 44 degrees.

Actually, there's no need to construct anything that isn't already in the figure. There is already circle theorem that says that the angle between a chord and the tangent to the circle at the point where they meet is equal to the opposite angle in the circumscribed triangle. This theorem is already enough to solve both problems without drawing any extra lines on the figure.

For the first figure:



Simply label vertices of the circumscribed triangle A, B and C, with C at the 43 degree angle. Now, according to the theorem stated above, both the angle xAB and the angle xBA must equal to 43 degrees because the 43 degree angle at ACB is opposite to those angles in the circumscribed triangle ABC. Now, we know that all the angles in the triangle AxB must add up to 180 degrees, therefore x + 43 + 43 = 180. Thus x = 94 degrees.

For the second figure:



Let's label the vertices of the circumscribed triangle A, B and C with A at the angle x and B at the 60 degree angle. Also, let's further label the ends of the tangent at point B, points D and E with point D on the left hand side and point E on the right hand side. Now, using the same theorem we used before, we see that the angle CBE must be equal to x. Now notice that the angles CBE, ABC and ABD fall on a straight line. Because they fall on a straight line, these angles must add up to 180 degrees. Now, since the angles ABC and ABD are given in the figure as 60 degrees and 76 degrees respectfully, we have x + 60 + 76 = 180. Thus x = 44 degrees.