Question

determine $\angle dba$.

Explanation:

Step1: Identify the type of angle

$\angle DAE$ is a straight angle, so it measures $180^\circ$. Given $\angle CAB = 66^\circ$, the central angle $\angle DAB$ is $180^\circ - 66^\circ = 114^\circ$? Wait, no, maybe $AD$ and $AE$ are diameters? Wait, $A$ is the center, so $AD$ and $AB$ are radii? Wait, no, $AD$ is a diameter? Wait, let's correct. Since $A$ is the center, $AD$ and $AB$ are radii, and $DE$ is a diameter. Wait, the angle at $A$: $\angle CAB = 66^\circ$, so the vertical angle or the adjacent angle? Wait, maybe $\triangle DBA$ is isosceles with $AD = AB$ (radii of the circle). Wait, no, maybe the inscribed angle? Wait, no, $\angle DBA$: let's see, $AD$ is a diameter? Wait, the key is that $A$ is the center, so $AD$ and $AB$ are radii, so $AD = AB$. Wait, but the central angle: wait, maybe the angle at $A$: $\angle DAB$: if $\angle CAB = 66^\circ$, then $\angle DAB = 180^\circ - 66^\circ = 114^\circ$? No, that can't be. Wait, maybe $DE$ is a diameter, so $\angle DAE = 180^\circ$, and $\angle CAB = 66^\circ$, so $\angle DAB = 180^\circ - 66^\circ = 114^\circ$? But then in $\triangle DAB$, $AD = AB$ (radii), so it's isosceles. Wait, no, maybe I made a mistake. Wait, the problem is to find $\angle DBA$. Let's recall that an angle inscribed in a semicircle is a right angle? No, wait, maybe $\angle DBE$? No, the question is $\angle DBA$. Wait, maybe $AD$ and $AB$ are radii, so $AD = AB$, so $\triangle DAB$ is isosceles with $AD = AB$. Wait, but the central angle: wait, no, maybe the angle $\angle DBA$ is an inscribed angle? Wait, no, $A$ is the center. Wait, maybe the correct approach: since $A$ is the center, $AD$ and $AB$ are radii, so $AD = AB$. The angle at $A$: $\angle DAB$: if $\angle CAB = 66^\circ$, then $\angle DAB = 180^\circ - 66^\circ = 114^\circ$? No, that's obtuse. Then in $\triangle DAB$, angles sum to $180^\circ$, so $\angle DBA = \angle ADB = \frac{180^\circ - 114^\circ}{2} = 33^\circ$. Ah, that makes sense. Because $AD = AB$ (radii), so $\triangle DAB$ is isosceles with base angles equal. So:

Step1: Find $\angle DAB$

Since $DE$ is a straight line (diameter), $\angle DAE = 180^\circ$. $\angle CAB = 66^\circ$, so $\angle DAB = 180^\circ - 66^\circ = 114^\circ$ (adjacent angles on a straight line).

Step2: Calculate $\angle DBA$

In $\triangle DAB$, $AD = AB$ (radii of the circle), so $\triangle DAB$ is isosceles with $\angle DBA = \angle ADB$. The sum of angles in a triangle is $180^\circ$, so:

$\angle DBA + \angle ADB + \angle DAB = 180^\circ$

Since $\angle DBA = \angle ADB$, let's call each $x$:

$2x + 114^\circ = 180^\circ$

$2x = 180^\circ - 114^\circ = 66^\circ$

$x = \frac{66^\circ}{2} = 33^\circ$

Answer:

$\boxed{33^\circ}$