Okay, let me try to figure out this geometry problem. So, we have circle J with two tangent lines HO and HN touching the circle at points O and N respectively. We need to find the sum of the measures of angle J and angle H. Hmm, angle J is the central angle at the center of the circle, right? And angle H is the angle at point H outside the circle. 

First, let me recall some properties of tangent lines to a circle. If a line is tangent to a circle, then it is perpendicular to the radius at the point of tangency. So, that means HO is perpendicular to JO, and HN is perpendicular to JN. So, angles HOJ and HNJ are right angles (90 degrees each). 

Looking at the diagram, points O and N are points where the tangents from H touch the circle. So, H is an external point from which two tangents are drawn to the circle. There's a theorem that says the lengths of two tangent segments from the same external point to a circle are equal. So, HO should be equal to HN. That might come in handy later.

Now, angle J is the central angle between points O and N. The measure of a central angle is equal to the measure of its intercepted arc. So, angle J is equal to the measure of arc ON. Let me denote that as m∠J = measure of arc ON.

Angle H is formed by the two tangents HO and HN. There's a theorem that relates the measure of an angle formed outside the circle by two tangents to the difference of the intercepted arcs. Specifically, the measure of angle H is half the difference of the measures of the intercepted arcs. The formula is: m∠H = (1/2)(measure of major arc ON - measure of minor arc ON). 

But wait, the problem is asking for the sum of angles J and H. Let me write that down:

m∠J + m∠H = measure of arc ON + (1/2)(major arc ON - minor arc ON)

But the major arc ON is equal to 360 degrees minus the minor arc ON. Let me denote the measure of minor arc ON as x. Then major arc ON is 360 - x. Plugging that into the equation:

m∠J + m∠H = x + (1/2)((360 - x) - x) = x + (1/2)(360 - 2x) = x + 180 - x = 180 degrees.

Wait, that simplifies to 180 degrees? So, regardless of the measure of arc ON, the sum of angles J and H is always 180 degrees? That seems too straightforward. Let me check my steps again.

First, angle J is the central angle, so it's equal to the measure of arc ON. Then angle H is formed by two tangents from point H. The formula for the angle outside the circle is indeed half the difference of the intercepted arcs. The intercepted arcs here are the major arc ON and the minor arc ON. The formula is:

m∠H = (1/2)(measure of major arc ON - measure of minor arc ON)

Since major arc ON = 360 - minor arc ON, substituting:

m∠H = (1/2)(360 - minor arc ON - minor arc ON) = (1/2)(360 - 2 * minor arc ON) = 180 - minor arc ON

But angle J is equal to minor arc ON. Therefore, m∠J + m∠H = minor arc ON + (180 - minor arc ON) = 180 degrees.

Yes, that seems correct. The sum is 180 degrees. Let me think if there's another way to approach this to verify.

Alternatively, consider triangle HJO and HJN. Since HO and HN are tangents, they are equal in length (from the same external point H). JO and JN are radii of the circle, so they are equal. Therefore, triangles HJO and HJN are congruent right triangles (since the radii are perpendicular to the tangents). 

In triangle HJO, angle HOJ is angle J, and angle HJO is 90 degrees. Similarly, in triangle HJN, angle HJN is angle J, and angle HJN is 90 degrees. Then, angle at H in triangle HJO is angle JHO, which is equal to angle JHN in triangle HJN. Since these two angles are adjacent and form angle H, their sum is angle H. 

But since triangles HJO and HJN are congruent, the angles at J (angle HOJ and angle HJN) are equal. Wait, but angle HOJ is angle J, and angle HJN is also angle J. So, angle J is the same in both triangles. Hmm, maybe this approach is complicating things. Let me think again.

Alternatively, in quadrilateral JONH, we have two right angles at O and N. The sum of the interior angles of a quadrilateral is 360 degrees. So, angle J + angle H + 90 + 90 = 360. Therefore, angle J + angle H = 360 - 180 = 180 degrees. That's another way to see it. 

Yes, that's a much simpler approach. Since quadrilateral JONH has four angles: angle J, angle H, and two right angles (each 90 degrees). The total sum is 360 degrees, so angle J + angle H = 360 - 90 - 90 = 180 degrees. That confirms the previous result.

So, regardless of the specific positions or lengths, the sum of angles J and H is always 180 degrees. Therefore, the answer should be 180 degrees.

**Final Answer**
\boxed{180}
Therefore, the sum of $m\angle J$ and $m\angle H$ is $\boxed{180}$ degrees.
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