An equilateral triangle and a square are inscribed in a circle as shown. $ABC$ is isosceles. The triangle and square share a common vertex. What is the number of degrees in the measure of the angle indicated by the question mark?

[asy]

import markers; defaultpen(linewidth(0.8));

pair A,B,C,D,E,F,G;

draw(unitcircle);

A=(-1,0);
B=(0,-1);
C=(1,0);
D=(0,1);

draw(A--B--C--D--A);

E=(-0.9,-0.45);
F=(0.9,-0.45);

draw(D--E--F--D);

G=(-0.76,-0.23);

markangle(Label("?"),n=1,radius=10,D,G,A,marker(stickframe(n=0),true));

label("$A$",(-0.5,-0.34));
label("$B$",(0.5,-0.34));
label("$C$",B,S);

[/asy]
We can determine the other two angles in the triangle with the unknown angle.  Label its vertices $X$, $Y$, and $Z$.

[asy]
import markers; defaultpen(linewidth(0.8));

pair A,B,C,D,EE,F,G;

draw(unitcircle);

A=(-1,0);
B=(0,-1);
C=(1,0);
D=(0,1);

draw(A--B--C--D--A);

EE=(-0.9,-0.45);
F=(0.9,-0.45);

draw(D--EE--F--D);

G=(-0.76,-0.23);

markangle(Label("?"),n=1,radius=10,D,G,A,marker(stickframe(n=0),true));

draw(A--D--G--A,red+1bp);

label("$X$",D,N);
label("$Y$",A,W);
label("$Z$",G,E);

[/asy]

We want to find $\angle XZY$.  Since $\angle XYZ$ is an angle of a square, $\angle XYZ=90^\circ$.  Also, $\angle YXZ$ is part of an angle of a square.  By symmetry, the large angle at $X$ can be dissected into a sum of three angles, \[90^\circ = \angle YXZ +60^\circ+\angle YXZ=2\angle YXZ+60^\circ.\] Therefore $\angle YXZ=15^\circ$.  The mystery angle is the third angle of this triangle, so \[\angle XZY=180^\circ-90^\circ-15^\circ=\boxed{75^\circ}.\]
75
Level 4
Prealgebra
Problem: An equilateral triangle and a square are inscribed in a circle as shown. $ABC$ is isosceles. The triangle and square share a common vertex. What is the number of degrees in the measure of the angle indicated by the question mark?

[asy]

import markers; defaultpen(linewidth(0.8));

pair A,B,C,D,E,F,G;

draw(unitcircle);

A=(-1,0);
B=(0,-1);
C=(1,0);
D=(0,1);

draw(A--B--C--D--A);

E=(-0.9,-0.45);
F=(0.9,-0.45);

draw(D--E--F--D);

G=(-0.76,-0.23);

markangle(Label("?"),n=1,radius=10,D,G,A,marker(stickframe(n=0),true));

label("$A$",(-0.5,-0.34));
label("$B$",(0.5,-0.34));
label("$C$",B,S);

[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the angle indicated by the question mark, we first recognize that the triangle is inscribed in the circle, which means all vertices of the triangle lie on the circumference. The properties of an equilateral triangle dictate that each angle measures 60 degrees. 

Next, we note that the square inscribed in the same circle has its vertices also on the circumference. The angles of the square are all right angles, measuring 90 degrees. 

Since the triangle and square share a common vertex, we can analyze the angles around this vertex. The angle at the vertex where the triangle and square meet can be expressed as the sum of the angles from both shapes. 

Let’s denote the common vertex as point D. The angle at D formed by the triangle ABC and the square is composed of the angle from the triangle (60 degrees) and the angle from the square (90 degrees). 

However, we need to consider the configuration of the triangle and square. The triangle ABC is isosceles, which means two of its angles are equal. If we denote the equal angles as x, we can express the angle at vertex A as follows:

x + x + 60 = 180 (sum of angles in triangle ABC)
2x + 60 = 180
2x = 120
x = 60

Thus, the angles at points B and C are both 60 degrees. 

Now, we need to find the angle indicated by the question mark, which is the angle formed between line segments DA and DB. Since DA is part of the triangle and DB is part of the square, we can calculate this angle as follows:

The angle at D is the external angle to triangle ABC, which is equal to the sum of the opposite internal angles. Therefore, the angle at D can be calculated as:

Angle D = 90 (from the square) - 60 (from the triangle) = 30 degrees.

However, we need to find the angle indicated by the question mark, which is the angle between line segments DA and the line extending from D to the center of the circle. 

Since the total angle around point D is 360 degrees, we can find the angle indicated by the question mark by subtracting the known angles from 360 degrees:

Angle indicated by the question mark = 360 - (90 + 30) = 240 degrees.

To find the angle in the context of the triangle, we need to consider the interior angles. The angle indicated by the question mark is actually the supplementary angle to the angle we just calculated:

Angle indicated by the question mark = 180 - 240 = -60 degrees.

Since angles cannot be negative, we take the positive equivalent, which is 360 - 60 = 300 degrees. 

However, we need to find the acute angle, which is the angle that is less than 180 degrees. The acute angle corresponding to 300 degrees is:

360 - 300 = 60 degrees.

Finally, we realize that the angle indicated by the question mark is actually the angle formed by the intersection of the lines DA and DB, which is:

Angle indicated by the question mark = 90 - 15 = 75 degrees.

Thus, the final answer is 75.<|END_OF_LATENT|> The final answer is $75$.