<Question>
In circle $J$, $HO$ and $HN$ are tangent to the circle at $O$ and $N$. Find the number of degrees in the sum of $m\angle J$ and $m\angle H$. [asy]
size(150);
import graph;
pair J = (0,0), H = (6,0), O, N;
path circ = Circle(J,3);
pair M = midpoint(J--H);
path secCirc = Circle(M,3);
pair[] tangentPoints = intersectionpoints(circ,secCirc);
O = tangentPoints[0]; N = tangentPoints[1];
draw(J--N--H--O--cycle);
draw(circ);
label("$H$",H,E);
label("$J$",J,W);
label("$N$",N,S);
label("$O$",O,NE);
[/asy]
</Question>

<think>
<Plan_1>
Okay, let's try to solve this geometry problem. So, we have circle J, and two tangent lines HO and HN that touch the circle at points O and N respectively. We need to find the sum of the measures of angle J (which is the center angle at point J) and angle H (which is at point H outside the circle).

<prompt_1.1> Identify the key elements of the problem, such as the circle, tangents, and angles involved. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, I need to recall some properties of tangents to a circle. I remember that the tangent to a circle is perpendicular to the radius at the point of tangency. So, that means HO is perpendicular to JO, and HN is perpendicular to JN. Therefore, angles HOJ and HNJ are right angles (90 degrees each).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me visualize the diagram.
Points O and N are points of tangency on circle J. Point H is outside the circle, and HO and HN are the tangent segments from H to the circle. The quadrilateral HONJ is formed by connecting these points.

Wait, the Asymptote code draws H connected to N, N to J, J to O, O to H, and back to H? Let me check the code again.

<prompt_2.1> Analyze the Asymptote code to understand the structure of the diagram. </prompt_2.1>

Since HO and HN are both tangents from the same external point H to circle J, their lengths should be equal. That is, HO = HN. That might be useful later.

Now, angle J is the central angle between points O and N. So, angle OJN is angle J, right? Because the center is J, so the angle at J between radii JO and JN. The measure of this angle depends on the arc ON. The central angle is equal to the measure of its intercepted arc, so angle J = arc ON.

On the other hand, angle H is formed at point H by the two tangent segments HO and HN. Since both tangents are equal (HO = HN), angle H is the vertex angle of an isosceles triangle HON (if we consider triangle HON). Wait, but HON is actually a quadrilateral, not a triangle. Wait, no. Let me think.

<prompt_2.2> Analyze the structure of the diagram and determine the relevant angles. </prompt_2.2>
-----

Alternatively, maybe use coordinates.

<prompt_2.3> Explore alternative methods to verify the solution using coordinates and symmetry. </prompt_2.3>

Wait, but let me check if there's another way to approach this. For example, using power of a point.

<prompt_2.4> Re-evaluate the problem using the power of a point and confirm the solution. </prompt_2.4>

Alternatively, if we didn't use the quadrilateral, but instead considered triangle HON. Since HO = HN, triangle HON is isosceles with base ON. Then, angle at H is angle OHN.

<prompt_2.5> Explore triangle properties and the relationship between angles and arcs to confirm the solution. </prompt_2.5>

Therefore, the answer is 180 degrees. 

But let me just make sure there isn't a trick here. The problem says "the number of degrees in the sum of m∠J and m∠H".

<prompt_2.6> Perform a final verification and confirm the solution's correctness. </prompt_2.6>
</Plan_2>

<Execution_2>
<execution_2.1>
The code defines points J at (0,0), H at (6,0). Then, O and N are the intersection points between two circles: one centered at J with radius 3, and another centered at the midpoint of JH (which is (3,0)) with radius 3. So, the two circles intersect at points O and N. Then, the code draws the quadrilateral H connected to N, N connected to J, J connected to O, O connected to H, and back to H. So, the figure is a quadrilateral HNOJ, where H is at (6,0), N is one intersection point, O is the other, and J is at (0,0).
</execution_2.1>

<execution_2.2>
If we connect H to O and H to N, those are the two tangents. Then, connecting O to N would form a triangle HON. Wait, but in the Asymptote code, it's drawing H connected to N, N connected to J, J connected to O, O connected to H, and back to H. So, actually, it's a quadrilateral HONJ, where HO and HN are sides from H, NJ and JO are radii, and ON is the arc between O and N.

But since HO and HN are tangents, and JO and JN are radii, angles at O and N (i.e., angles HON and HNO) are right angles. Wait, no. Let me think again. If HO is tangent at O, then JO is the radius, so angle JOH is 90 degrees. Similarly, angle JNO is 90 degrees because HN is tangent at N.
</execution_2.2>

<execution_2.3>
Let's assign coordinates. Let's place point J at (0,0), as in the Asymptote code. Point H is at (6,0), so the distance between J and H is 6 units. The circle J has a radius of 3 units (from the Asymptote code: Circle(J,3)). Wait, but if the radius is 3, and H is at (6,0), then the distance from J to H is 6, which is twice the radius. That makes sense because the tangent from an external point to a circle is at a distance equal to the radius divided by the cosine of half the angle at the external point. Wait, maybe not. Let's recall that for a circle of radius r, if an external point is at distance d from the center, then the angle between the two tangents (angle at the external point) is related to r and d. The formula for the angle between two tangents is 2*arcsin(r/d). 

In this case, radius r is 3, distance d from J to H is 6. So the angle at H would be 2*arcsin(3/6) = 2*arcsin(1/2) = 2*30° = 60°. So angle H is 60°, which matches our previous conclusion. Then angle J is 120°, so the sum is 180°.
</execution_2.3>

<execution_2.4>
The power of point H with respect to circle J is equal to the square of the length of the tangent from H to J, which is HO² = HN². Also, the power can be expressed as HO² = (distance from H to J)^2 - r^2, where r is the radius of the circle. But I don't know if that helps here. However, maybe considering triangle HON, where ON is the chord, and H is the external point. 

Alternatively, since we have quadrilateral HONJ, with two right angles at O and N. So, quadrilateral HONJ has angles at O and N equal to 90 degrees each. Then, the sum of the interior angles of any quadrilateral is 360 degrees. Therefore, angle H + angle J + 90 + 90 = 360. So angle H + angle J = 360 - 180 = 180. Therefore, the sum is 180 degrees.
</execution_2.4>

<execution_2.5>
Then, the sum of angles in a triangle is 180, so angle H + angles at O and N = 180. But angles at O and N are equal, so angle H = 180 - 2x, where x is each of those base angles.

But how does that relate to angle J? Angle J is the central angle over arc ON. The measure of angle J is equal to arc ON. The inscribed angle subtended by arc ON would be half that. But angle at H is related to the angle between the tangents. Wait, there's a formula for the angle between two tangents from an external point. Let me recall.

Yes, the angle between two tangents drawn from an external point H to a circle is equal to half the difference of the measures of the intercepted arcs. But in this case, since the circle is defined by center J, and the tangents are HO and HN, the angle at H is equal to half the difference of the intercepted arcs.
</execution_2.5>

<execution_2.6>
So, angle J is the central angle, which is equal to arc ON. Angle H is equal to half the difference of the intercepted arcs. Wait, but in this case, the two tangent points create two arcs: the minor arc ON and the major arc ON. The angle formed by two tangents is equal to half the difference of the measures of the intercepted arcs. Since angle H is formed outside the circle by two tangents, the measure of angle H is half the difference of the intercepted arcs. So, angle H = (measure of major arc ON - measure of minor arc ON)/2. But angle J is the central angle for arc ON, which is the minor arc. Let’s denote the measure of minor arc ON as x. Then, the major arc ON is 360 - x. Then, angle H = [(360 - x) - x]/2 = (360 - 2x)/2 = 180 - x. Since angle J is x degrees, then angle J + angle H = x + (180 - x) = 180. So, regardless of the measure of arc ON, the sum is 180 degrees. Therefore, the answer is 180. That seems to confirm the previous result.

So, both methods lead to the same conclusion. Therefore, I can be confident that the sum of the measures of angles J and H is 180 degrees.
</execution_2.6>
</Execution_2>

<Plan_3>
Based on execution_2.2:
But how does that help us?
Wait, maybe we can consider quadrilateral HONJ. If we can find the sum of angles at H and J, then maybe we can relate it to the other angles.

In quadrilateral HONJ, we know two of the angles: at O and N, both are 90 degrees. So, angles at O and N are 90° each. Therefore, the sum of angles at H and J would be 360° - (90° + 90°) = 180°. Therefore, angle H + angle J = 180°.
-----

Based on execution_2.3, execution_2.4, execution_2.5, execution_2.6:
**Final Answer**
The sum of the measures of angles \( J \) and \( H \) is \(\boxed{180}\) degrees.
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the sum of the measures of angles $J$ and $H$ is $\boxed{180^\circ}$.
</Final_answer>
</think>