Since the lines appear to be tangent to the circle at point 0, we know that the radius (line segment connecting the center of the circle to the point of tangency) is perpendicular to the tangent line. Therefore, the angles formed by the radius and tangent line are right angles.
We can draw a diagram with the given information:
[asy]
pair O,A,B;
O=(0,0);
A=(1,0);
B=(0.342,-0.94);
draw(circle(O,1));
draw(O--A--B--cycle);
label("$O$",O,N);
label("$A$",A,E);
label("$B$",B,S);
label("$x$",(O+A)/2,N);
label("$140^\circ$",(0.15,0.05));
[/asy]
Let's label the point of intersection between the tangent lines as point $A$. We can draw in the radius from the center of the circle to point $A$ as shown. Let's label the length of that radius as $x$.
Since angle $OAB$ is perpendicular to $AB$, we know that it measures $90^\circ$. We also know that angle $OAB$ is part of an isosceles triangle (since the radius is the same length on both sides), so angle $OBA$ also measures $90^\circ$.
Therefore, we can see that angle $OAB$ + angle $OBA$ + angle $O$ = $360^\circ$, or:
$90^\circ + 90^\circ + 140^\circ + x = 360^\circ$
Simplifying:
$320^\circ + x = 360^\circ$
$x = 40$
Therefore, the value of $x$ is $\boxed{\textbf{(A) }40}$
The measure of angle o equals 140 degrees assume that the lines that appear to be tangent are tangent 0 is the center of the circle find the value of x
A.40
B.70
C.220
D.240
3 answers
what?!?
O is the center of the circle.
No tangent line can pass through the center!
O is the center of the circle.
No tangent line can pass through the center!
I apologize for my mistake. If the tangent lines do not pass through the center of the circle, we cannot find the value of $x$ with the given information and diagram.