Understanding the Geometry of the Circle, Tangent, and Diameter

The problem describes a circle with diameter AB. A tangent at a point C on the circle is drawn. This tangent line intersects the line formed by extending the diameter AB at point D. We are given that the angle DCA is $110^\circ$, and we need to find the measure of the angle CBA. This setup involves key geometric properties of tangents, chords, and angles in a circle.

Relating Angles in Triangle ABC

Since AB is the diameter AB of the circle, the angle subtended by the diameter at any point on the circumference is $90^\circ$. Therefore, $\angle \text{ACB} = 90^\circ$.

Let the angle we want to find, $\angle \text{CBA}$, be denoted by $\theta$.

In $\triangle \text{ABC}$, the sum of angles is $180^\circ$. So, $\angle \text{BAC} + \angle \text{ABC} + \angle \text{ACB} = 180^\circ$.

Substituting the known values, we get $\angle \text{BAC} + \theta + 90^\circ = 180^\circ$.

This gives us $\angle \text{BAC} = 180^\circ - 90^\circ - \theta = 90^\circ - \theta$.

Using Properties involving the Center and Tangent

Let O be the center of the circle. O is the midpoint of the diameter AB. OC is the radius.

A property of tangents is that the radius drawn to the point of tangency is perpendicular to the tangent. Thus, $\angle \text{OCD} = 90^\circ$.

Since O is the center and OB and OC are radii, $\triangle \text{BOC}$ is an isosceles triangle with OB = OC. The angles opposite these equal sides are equal, so $\angle \text{OBC} = \angle \text{OCB}$.

We defined $\angle \text{CBA} = \theta$, which is the same as $\angle \text{OBC}$. So, $\angle \text{OBC} = \theta$.

In $\triangle \text{BOC}$, $\angle \text{OCB} = \angle \text{OBC} = \theta$.

The sum of angles in $\triangle \text{BOC}$ is $180^\circ$: $\angle \text{BOC} + \angle \text{OBC} + \angle \text{OCB} = 180^\circ$.

$\angle \text{BOC} + \theta + \theta = 180^\circ$, which means $\angle \text{BOC} = 180^\circ - 2\theta$.

Analyzing Triangle OCD and Angle ADC

The line AB is extended to intersect at D. So, the points A, O, B, D are collinear in that order. This means $\angle \text{COD}$ is the same angle as $\angle \text{BOC}$ since O, B, D are on a straight line.

So, $\angle \text{COD} = \angle \text{BOC} = 180^\circ - 2\theta$.

Now consider $\triangle \text{OCD}$. The angles are $\angle \text{ODC}$, $\angle \text{OCD}$, and $\angle \text{COD}$.

$\angle \text{ODC}$ is the same as $\angle \text{ADC}$, as D, B, A are collinear.

We know $\angle \text{OCD} = 90^\circ$ and $\angle \text{COD} = 180^\circ - 2\theta$.

The sum of angles in $\triangle \text{OCD}$ is $180^\circ$: $\angle \text{ODC} + \angle \text{OCD} + \angle \text{COD} = 180^\circ$.

$\angle \text{ADC} + 90^\circ + (180^\circ - 2\theta) = 180^\circ$.

$\angle \text{ADC} + 270^\circ - 2\theta = 180^\circ$.

$\angle \text{ADC} = 180^\circ - 270^\circ + 2\theta = 2\theta - 90^\circ$.

Combining Angles in Triangle ADC to Solve for Angle CBA

Now consider the $\triangle \text{ADC}$. The angles are $\angle \text{DAC}$, $\angle \text{DCA}$, and $\angle \text{ADC}$.

We know $\angle \text{DAC} = \angle \text{BAC} = 90^\circ - \theta$.

We are given $\angle \text{DCA} = 110^\circ$. This is the angle DCA.

We found $\angle \text{ADC} = 2\theta - 90^\circ$.

The sum of angles in $\triangle \text{ADC}$ is $180^\circ$: $\angle \text{DAC} + \angle \text{DCA} + \angle \text{ADC} = 180^\circ$.

Substituting the expressions for the angles:

$(90^\circ - \theta) + 110^\circ + (2\theta - 90^\circ) = 180^\circ$.

Final Calculation of Angle CBA

Let's simplify the equation:

$90^\circ - \theta + 110^\circ + 2\theta - 90^\circ = 180^\circ$

Combine the constant terms: $(90^\circ + 110^\circ - 90^\circ) = 110^\circ$.

Combine the $\theta$ terms: $-\theta + 2\theta = \theta$.

The equation becomes: $110^\circ + \theta = 180^\circ$.

Solving for $\theta$: $\theta = 180^\circ - 110^\circ = 70^\circ$.

Therefore, the measure of $\angle \text{CBA}$ is $70^\circ$. This confirms that the given angle DCA of $110^\circ$ is consistent with $\angle \text{CBA} = 70^\circ$ in this geometric configuration where the tangent at a point C and diameter AB extended intersect at D.