<Question>
A triangle $\triangle ABC$ with $\angle A = 14^\circ$ is inscribed in a circle where $AB$ is a diameter. How many degrees are in $\angle B$?
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this geometry problem. So, we have a triangle ABC inscribed in a circle, with AB as the diameter. Angle A is 14 degrees, and we need to find angle B. Hmm, let's start by recalling some geometry concepts.

First, if a triangle is inscribed in a circle and one of its sides is the diameter, then that triangle is a right-angled triangle. I remember that from Thales' theorem. So, since AB is the diameter, the angle opposite to AB, which would be angle C, should be 90 degrees.

<prompt_1.1> Apply Thales' theorem to determine the type of triangle formed when a side is the diameter of the circumscribed circle. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Right? Because Thales' theorem states that if you have a triangle inscribed in a circle where one side is the diameter, then the angle opposite that side is a right angle. So angle C is 90 degrees.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Wait, let me confirm that.
Yes, Thales' theorem says exactly that. So angle C is 90 degrees. That simplifies things a bit.

Now, in any triangle, the sum of the interior angles is 180 degrees. So if we know two angles, we can find the third one. We already know angle A is 14 degrees and angle C is 90 degrees. Let's check that.
14 + 90 + angle B = 180. So angle B would be 180 - 14 - 90 = 76 degrees.

But wait, is that all?

<prompt_2.1> Use the triangle angle sum property to calculate angle B. </prompt_2.1>

But maybe I need to verify this using another method to make sure I didn't skip any steps or make an assumption.

Let me think. Since AB is the diameter, the circle has its center at the midpoint of AB.

<prompt_2.2> Analyze the position of point C on the circle and its implications for triangle ABC. </prompt_2.2>

But how does that help with the angles? Well, maybe we can use some trigonometry here. Let's consider the circle with AB as diameter.

<prompt_2.3> Use trigonometric relationships and circle properties to verify the calculation of angle B. </prompt_2.3>

Alternatively, maybe we can use the property that in a circle, the angle subtended by a diameter is a right angle, so angle C is 90 degrees. Then, in triangle ABC, angles at A and B must add up to 90 degrees.

<prompt_2.4> Explore alternative methods to verify the calculation of angle B using properties of circles and triangles. </prompt_2.4>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me verify the steps again. AB is the diameter, so triangle ABC is right-angled at C. Therefore, angle C is 90 degrees. Given angle A is 14 degrees, then angle B must be 180 - 90 - 14 = 76 degrees.
</execution_2.1>

<execution_2.2>
The radius is half of AB. Point C is somewhere on the circumference of the circle, forming triangle ABC. Since angle C is subtended by the diameter AB, it's indeed a right angle. 

But how does angle A being 14 degrees relate to the rest of the triangle? Let's denote the center of the circle as O, which is the midpoint of AB. If I draw radii from O to C, then OC is also a radius, so OC = OA = OB. Maybe triangle AOC or BOC can help here?
</execution_2.2>

<execution_2.3>
Let's denote the center as O, which is the midpoint of AB. The radius of the circle would then be AB/2. Points A, B, and C are all on the circumference. Since AB is the diameter, the length of AB is 2r, where r is the radius. 

But maybe using coordinates would help. Let's place the circle on a coordinate system with center at the origin (0,0). Then, point A would be at (-r, 0) and point B at (r, 0). Point C is somewhere on the circumference. Since angle C is 90 degrees, triangle ABC is right-angled at C. 

The coordinates of point C must satisfy the equation x² + y² = r². Also, since angle at C is 90 degrees, the vectors AC and BC should be perpendicular. The coordinates of AC would be (x + r, y - 0) = (x + r, y), and BC would be (x - r, y). The dot product of AC and BC should be zero. So:

(x + r)(x - r) + y * y = 0
x² - r² + y² = 0
But since x² + y² = r² (because C is on the circle), substituting that in:

r² - r² = 0, which checks out.
</execution_2.3>

<execution_2.4>
Since angle A is 14, angle B must be 76. That seems straightforward. But let me make sure there are no other factors here. For example, is there a possibility that the triangle is not acute or something? Wait, angle C is 90, so it's a right-angled triangle. So angles at A and B must add up to 90. 14 and 76 do add up to 90.
</execution_2.4>
</Execution_2>

<Plan_3>
Based on execution_2.1:
But maybe I need to verify this using another method to make sure I didn't skip any steps or make an assumption.
Let me try using coordinates to model this.

<prompt_3.1> Use coordinate geometry to model the problem and verify the calculation of angle B. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Let's assume the circle has a radius of 1 for simplicity. Then, the diameter AB would be 2 units long. Let's place point A at (-1, 0) and point B at (1, 0). The center O is at (0,0). Point C is somewhere on the circumference, so its coordinates satisfy x² + y² = 1. 

Since angle C is 90 degrees, triangle ABC is right-angled at C. So, the coordinates of C must satisfy the condition that vectors AC and BC are perpendicular. As I thought earlier, the dot product of vectors AC and BC is zero. 

Let me compute vectors AC and BC. If point C is (x, y), then vector AC is (x - (-1), y - 0) = (x + 1, y). Vector BC is (x - 1, y - 0) = (x - 1, y). The dot product is (x + 1)(x - 1) + y * y = x² - 1 + y². Since x² + y² = 1 (because C is on the unit circle), substituting gives 1 - 1 = 0. So that's consistent. 

Now, angle A is 14 degrees. Let's see if that holds in this coordinate system. Angle at A is 14 degrees. In triangle ABC, angle at A is 14 degrees. Let's compute angle at A in terms of coordinates. 

In triangle ABC, angle at A is between sides AB and AC. The vector AB is from A to B: (2, 0). The vector AC is from A to C: (x + 1, y). The angle between AB and AC is 14 degrees. 

The formula for the angle between two vectors u and v is cos(theta) = (u . v) / (|u| |v|). 

So, the angle at A is 14 degrees, so:

cos(14°) = [AB . AC] / (|AB| |AC|)

Compute AB . AC: AB is (2, 0), AC is (x + 1, y). The dot product is 2*(x + 1) + 0*y = 2(x + 1).

|AB| is the magnitude of AB, which is 2. 

|AC| is the magnitude of AC, which is sqrt((x + 1)^2 + y^2).

Therefore:

cos(14°) = [2(x + 1)] / [2 * sqrt((x + 1)^2 + y^2)]

Simplify:

cos(14°) = (x + 1) / sqrt((x + 1)^2 + y^2)

But since point C is on the unit circle, x² + y² = 1. Let's express y^2 as 1 - x^2.

So, substitute y^2 = 1 - x^2 into the denominator:

sqrt((x + 1)^2 + 1 - x^2) = sqrt(x² + 2x + 1 + 1 - x²) = sqrt(2x + 2) = sqrt(2(x + 1))

Therefore, the equation becomes:

cos(14°) = (x + 1) / sqrt(2(x + 1)) ) = sqrt(x + 1) / sqrt(2)

So:

cos(14°) = sqrt( (x + 1)/2 )

Square both sides:

cos²(14°) = (x + 1)/2

Multiply both sides by 2:

2 cos²(14°) = x + 1

Therefore:

x = 2 cos²(14°) - 1

Recall the double-angle identity: 2 cos²(theta) - 1 = cos(2 theta). Therefore:

x = cos(28°)

So, x coordinate of point C is cos(28°). Then, since x² + y² = 1, y² = 1 - cos²(28°) = sin²(28°), so y = sin(28°) or y = -sin(28°). Since the triangle is inscribed in the circle, and angle C is 90°, the point C can be in either the upper or lower half of the circle. But angle A is 14°, which is acute, so the triangle is above the diameter AB. Let's take y positive.

So, coordinates of C are (cos(28°), sin(28°)).
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Now, we need to find angle B, which is at point B (1,0). Let's compute angle B.

<prompt_4.1> Calculate angle B using trigonometric relationships and verify the result. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
In triangle ABC, angle at B is between sides BA and BC. Vector BA is from B to A: (-2, 0). Vector BC is from B to C: (cos(28°) - 1, sin(28°)). 

To find angle B, we can use the Law of Sines or Law of Cosines. Alternatively, compute the angle using vectors. Let's try the vector approach again.

The angle at B is between vectors BA and BC. 

Vector BA is (-2, 0), and vector BC is (cos(28°) - 1, sin(28°)). 

The angle between BA and BC is angle B. Let's compute the cosine of angle B using the dot product formula.

cos(theta) = (BA . BC) / (|BA| |BC|)

Compute BA . BC:

(-2)(cos(28°) - 1) + 0 * sin(28°) = -2(cos(28°) - 1) = -2 cos(28°) + 2

|BA| is 2.

|BC| is sqrt( (cos(28°) - 1)^2 + sin²(28°) )

Let's compute |BC|:

= sqrt( cos²(28°) - 2 cos(28°) + 1 + sin²(28°) )

= sqrt( (cos² + sin²)(28°) - 2 cos(28°) + 1 )

= sqrt( 1 - 2 cos(28°) + 1 )

= sqrt( 2 - 2 cos(28°) )

= sqrt(2(1 - cos(28°)) )

Using the identity 1 - cos(theta) = 2 sin²(theta/2):

= sqrt(2 * 2 sin²(14°) )

= sqrt(4 sin²(14°) )

= 2 sin(14°)

Therefore, |BC| = 2 sin(14°)

Now, back to the dot product:

BA . BC = -2 cos(28°) + 2 = 2(1 - cos(28°))

Again, using the identity 1 - cos(theta) = 2 sin²(theta/2):

= 2 * 2 sin²(14°) = 4 sin²(14°)

Therefore, cos(theta) = (4 sin²(14°)) / (2 * 2 sin(14°)) ) = (4 sin²(14°)) / (4 sin(14°)) ) = sin(14°)

Therefore, cos(theta) = sin(14°)

But sin(14°) = cos(90° - 14°) = cos(76°)

Therefore, cos(theta) = cos(76°), so theta = 76°
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, let me check that again. If cos(theta) = sin(14°), then theta is arccos(sin(14°)). But sin(14°) is equal to cos(90° - 14°) = cos(76°), so arccos(cos(76°)) = 76°, since 76° is between 0 and 180°, which is the range of arccos. Therefore, angle B is 76 degrees.
-----

Based on execution_2.4, execution_3.1, execution_4.1:
Alternatively, since we found that the x-coordinate of point C is cos(28°), and y-coordinate is sin(28°), we can compute the angles at A and B using slopes or trigonometric ratios. 

From point A (-1,0) to point C (cos28, sin28), the slope of AC is [sin28 - 0]/[cos28 - (-1)] = sin28 / (cos28 + 1). The angle at A is the angle between AB (which is along the x-axis) and AC. The angle can be found using arctangent of the slope. 

tan(theta) = slope = sin28 / (cos28 + 1)

Let me compute this:

sin28° / (cos28° + 1)

Using the identity sin(theta) / (1 + cos(theta)) = tan(theta/2)

So, sin28° / (1 + cos28°) = tan(14°)

Which is exactly the slope. Therefore, the angle at A is 14°, which matches the given. So that's consistent.

Similarly, from point B (1,0) to point C (cos28, sin28), the slope of BC is [sin28 - 0]/[cos28 - 1] = sin28 / (cos28 - 1). This slope is equal to -sin28 / (1 - cos28). Using the identity sin(theta)/(1 - cos(theta)) = cot(theta/2)

So, -sin28 / (1 - cos28) = -cot(14°). Therefore, the angle at B is arctangent of |slope|, but since it's negative, the angle is measured from the negative x-axis. The angle between BC and the x-axis is 180° - arctan(cot14°). Wait, maybe this approach is complicating things. 

Alternatively, since we already used vectors and found angle B to be 76°, which matches the earlier calculation, and the coordinate system confirms angle A is 14°, everything checks out.

Therefore, angle B is 76 degrees.
</Plan_5>

<Execution_5>
No prompt to execute.
</Execution_5>

<Final_answer>
Therefore, the measure of angle $\angle B$ is $\boxed{76^\circ}$.
</Final_answer>
</think>