In the figure above, assume that any lines that appear to be tangent, are tangent to the circle. If m < ABD = 38°, and the radius of the circle is 12.6 meters, find CB.

Answer 1

To solve for \( CB \) using the given information:

1. Identify Known Angles: The problem states that \( m \angle ABD = 38^\circ \). Since \( AB \) is tangent to the circle at point \( A \), by the properties of tangents, we know:

   \[ \angle ABD = \angle ACB \]

   Therefore, \( m \angle ACB = 38^\circ \).

2. Use the Inscribed Angle Theorem: The angle \( ACB \), being an inscribed angle that intercepts arc \( AB \) of the circle, is half the measure of the central angle \( AOB \) that subtends the same arc. Thus:

   \[ m \angle AOB = 2 \cdot m \angle ACB = 2 \cdot 38^\circ = 76^\circ \]

3. Apply Triangle Properties: In triangle \( AOB \), we know two sides: \( AO \) (which is the radius) and \( OB \) (which is also the radius), both equal to 12.6 meters. To find side \( AB \), we will use the Law of Cosines:

   \[ AB^2 = AO^2 + OB^2 - 2 \cdot AO \cdot OB \cdot \cos(\angle AOB) \] \[ AB^2 = 12.6^2 + 12.6^2 - 2 \cdot 12.6 \cdot 12.6 \cdot \cos(76^\circ) \]

4. Calculate \( AB \): \[ AB^2 = 2 \cdot 12.6^2 (1 - \cos(76^\circ)) \] Now substitute \( 12.6^2 = 158.76 \): \[ AB^2 = 2 \cdot 158.76 \cdot (1 - \cos(76^\circ)) \] Using a calculator to find \( \cos(76^\circ) \): \[ \cos(76^\circ) \approx 0.2419 \ AB^2 = 2 \cdot 158.76 \cdot (1 - 0.2419) \ AB^2 = 2 \cdot 158.76 \cdot 0.7581\ AB^2 = 2 \cdot 120.18 \ AB^2 \approx 240.36 \ AB \approx \sqrt{240.36} \approx 15.5 \text{ meters (approx)} \]

5. Determine Length \( CB \): In right triangle \( ACB \): Since \( AC \) is the radius and equals 12.6 meters, apply Pythagorean theorem: \[ AC^2 + CB^2 = AB^2 \] \[ 12.6^2 + CB^2 = 15.5^2 \] This simplifies to: \[ 158.76 + CB^2 = 240.36 \] \[ CB^2 = 240.36 - 158.76 \] \[ CB^2 = 81.6 \] \[ CB \approx \sqrt{81.6} \approx 9.03 \text{ meters} \]

Thus, the length \( CB \) is approximately \( 9.03 \) meters.