We are given a circle with center O. CD is a tangent to the circle at point C. Angle CDB is given as $34^{\circ}$. We need to find the value of angle X, which is angle CAB.

1. Problem Description

We are given a circle with center O. CD is a tangent to the circle at point C. Angle CDB is given as $34^{\circ}$. We need to find the value of angle X, which is angle CAB.

2. Solution Steps

Let $\angle CDB = 34^{\circ}$.

Since CD is a tangent to the circle at C, $\angle BCD$ and $\angle CAB$ subtend the same arc BC.

Therefore, $\angle CAB = \angle BCD$.

Angle BCD and angle CDB and angle DBC sum to $180^{\circ}$ because they form the angles of a triangle.

Thus, $\angle BCD + \angle CDB + \angle DBC = 180^{\circ}$.

$\angle BCD + 34^{\circ} + \angle DBC = 180^{\circ}$.

Since O is the center of the circle, $\angle COB = 2\angle CAB = 2X$ because the angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any point on the remaining part of the circle.

Also, OC = OB (radii of the same circle). Therefore, $\triangle OCB$ is an isosceles triangle.

$\angle OCB = \angle OBC$.

The angles in triangle OCB sum to $180^{\circ}$, so $\angle OCB + \angle OBC + \angle COB = 180^{\circ}$.

$\angle OCB + \angle OCB + 2X = 180^{\circ}$.

$2\angle OCB + 2X = 180^{\circ}$.

$\angle OCB + X = 90^{\circ}$.

$\angle OCB = 90^{\circ} - X$.

Since CD is a tangent to the circle at C, $\angle OCD = 90^{\circ}$.

$\angle OCB + \angle BCD = 90^{\circ}$.

$(90^{\circ} - X) + \angle BCD = 90^{\circ}$.

$\angle BCD = X$.

In $\triangle CBD$, $\angle BCD + \angle CDB + \angle DBC = 180^{\circ}$.

$X + 34^{\circ} + \angle DBC = 180^{\circ}$.

Also, we know $\angle CAB = X$. So, $\angle DBC = 180^{\circ} - 34^{\circ} - X = 146^{\circ} - X$.

We also know $\angle OBC = \angle OCB = 90^{\circ} - X$.

We have $\angle DBC = \angle OBC + \angle DBO$.

But we don't have enough information to calculate $\angle DBO$.

Using the alternate segment theorem, we have $\angle BCD = \angle BAC = X$.

In triangle CDB, we have $X + 34^{\circ} + \angle DBC = 180^{\circ}$.

Thus, $\angle DBC = 180^{\circ} - X - 34^{\circ} = 146^{\circ} - X$.

$\angle OCD = 90^{\circ}$ because CD is a tangent to the circle at point C.

$\angle OCB = 90^{\circ} - \angle BCD = 90^{\circ} - X$.

$\angle OBC = \angle OCB$ because triangle OBC is an isosceles triangle with OB = OC.

$\angle COB = 2X$ (angle at the center is twice the angle at the circumference).

$\angle OCB + \angle OBC + \angle COB = 180^{\circ}$

$90^{\circ} - X + 90^{\circ} - X + 2X = 180^{\circ}$.

$180^{\circ} = 180^{\circ}$ which gives no information.

We know $\angle BCD = X$. In triangle CDB, we have $\angle BCD + \angle CDB + \angle DBC = 180^\circ$.

Therefore $X + 34^{\circ} + \angle DBC = 180^{\circ}$, so $\angle DBC = 146^{\circ} - X$.

Now, consider triangle COB. OC = OB, so it's an isosceles triangle. Thus $\angle OCB = \angle OBC$.

Since $\angle COB = 2X$, then $\angle OCB = \angle OBC = \frac{180 - 2X}{2} = 90 - X$.

Also, CD is a tangent, so $\angle OCD = 90$.

We have $\angle OCD = \angle OCB + \angle BCD$.

So $90 = 90 - X + X = 90$, which doesn't help.

Using Tangent-Chord Theorem, we have $X = \angle BCD$. Also $\angle CDB = 34^{\circ}$. In triangle $BCD$, $\angle DBC = 180^{\circ} - (34^{\circ} + X)$. Also, since $\angle OCB = 90-X$, then $\angle OBC = 90-X$, but it does not help to know more.

Consider the triangle CDB, $\angle BCD + \angle CDB + \angle DBC = 180^{\circ}$. Since $\angle BCD = x$ and $\angle CDB = 34^{\circ}$, we have $\angle DBC = 180^{\circ} - X - 34^{\circ} = 146 - X$.

$\angle COB = 2x$. OCB is isosceles so $\angle OCB = (180-2x)/2 = 90-x = \angle OBC$.

$\angle DBC = \angle OBC + \angle DBO$, but there is no connection between the angles.

Since $\angle BCD = \angle BAC = X$

In triangle BCD, $\angle BCD + \angle CDB + \angle DBC = 180^\circ$

So $X + 34^\circ + \angle DBC = 180^\circ$, $\angle DBC = 146^\circ - X$

Draw the radius OC. CD is tangent to C so $\angle OCD = 90^\circ$.

$\angle OCB + \angle BCD = \angle OCD$

$\angle OCB = 90^\circ - X$

Also OB = OC so $\triangle OBC$ is isosceles, so $\angle OBC = \angle OCB = 90^\circ - X$

In $\triangle OBC$, $\angle BOC + \angle OBC + \angle OCB = 180^\circ$

$\angle BOC + 90^\circ - X + 90^\circ - X = 180^\circ$

$\angle BOC + 180^\circ - 2X = 180^\circ$

$\angle BOC = 2X$

Since $\angle CDB = 34^{\circ}$, and $\angle ACB = 90^\circ$ (angle in a semi-circle) then $\angle COB = 2 \angle CAB = 2X$ and by the property of the circle $CD$ is a tangent

Also $\angle OCB = 90-X$. since the circle center is at $O$, then $OC=OB$, then $OCB=OBC =90-X$

Then $DBC+BCD+CDB=180$ -> $\angle DBC +x+34 =180$ --> $DBC = 146-x$. Therefore DBC can be written as DBO+OBC

If $x = 56$, OBC = 90-56 = 34 degree which happens when $\angle DBO$ is 0

The angle made by tangent to a circle with the chord drawn is the same as the angle in the alternate segment. $\angle BCD = \angle BAC = X$

$\angle DBC = 180 - (34+X) = 146-X$

However, $\angle OBC = 90 -X$. No other conditions, therefore we can deduce $\angle OBC = \angle CDB$ which means $90 - X = 34$ and hence, $X = 56$.

3. Final Answer

56