The tangent at a point C of a circle and a diameter AB when extended intersect at P. If `anglePCA=110^(@) "find" angleCBA`.

Options

A. `20^@`
B. `70^@`
C. `60^@`
D. `80^@`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Draw the Circle and Label the Points
- Draw a circle with center O.
- Draw the diameter AB of the circle.
- Extend the diameter AB to point P.
- Mark point C on the circumference of the circle such that line CP is tangent to the circle at point C.

### Step 2: Identify the Given Angle
- We are given that \( \angle PCA = 110^\circ \).

### Step 3: Use Properties of the Circle
- Since AB is a diameter, by the property of a circle, we know that \( \angle BCA = 90^\circ \) (the angle inscribed in a semicircle is a right angle).
- Also, the tangent at point C is perpendicular to the radius OC at point C, so \( \angle PCO = 90^\circ \).

### Step 4: Relate the Angles
- We can express \( \angle PCA \) in terms of \( \angle PCO \) and \( \angle OCA \):
\[ \angle PCA = \angle PCO + \angle OCA \]
- Substituting the known values:
\[ 110^\circ = 90^\circ + \angle OCA \]
- Solve for \( \angle OCA \):
\[ \angle OCA = 110^\circ - 90^\circ = 20^\circ \]

### Step 5: Use Isosceles Triangle Property
- In triangle OCA, since OC = OA (both are radii of the circle), triangle OCA is isosceles.
- Therefore, the angles opposite the equal sides are equal:
\[ \angle OCA = \angle OAC = 20^\circ \]

### Step 6: Find the Remaining Angle in Triangle ABC
- Now, consider triangle ABC. The sum of the angles in a triangle is \( 180^\circ \):
\[ \angle CAB + \angle BCA + \angle CBA = 180^\circ \]
- We know:
- \( \angle CAB = \angle OCA = 20^\circ \)
- \( \angle BCA = 90^\circ \)
- Substitute these values into the equation:
\[ 20^\circ + 90^\circ + \angle CBA = 180^\circ \]
- Simplifying gives:
\[ 110^\circ + \angle CBA = 180^\circ \]
- Solve for \( \angle CBA \):
\[ \angle CBA = 180^\circ - 110^\circ = 70^\circ \]

### Final Answer
Thus, the value of \( \angle CBA \) is \( 70^\circ \).

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