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 . Please show all work in order to earn full credit.

Answer 1

To find $CB$ in the given scenario while utilizing the provided information, we will follow these steps:

Step 1: Understand the setup

Assuming you have a circle with center $O$ and a tangent line at point $A$ intersecting the line segment $CB$. The angle $\mathrm{\angle }ABD={38}^{\circ }$.

Step 2: Use the properties of a tangent line

When a tangent touches a circle, the radius to the point of tangency is perpendicular to the tangent line. Thus, we have:

$$\mathrm{\angle }OAB={90}^{\circ }$$

Step 3: Use the concept of angles

Since we know:

$$\mathrm{\angle }ABD={38}^{\circ }$$

we can calculate $\mathrm{\angle }OBD$:

$$\mathrm{\angle }OBD=\mathrm{\angle }OAB-\mathrm{\angle }ABD={90}^{\circ }-{38}^{\circ }={52}^{\circ }$$

Step 4: Utilize the law of sines

In triangle $OBD$,

• We have $OB$ as the radius, which is $12.6$ m.
• $OD$ is also a radius of the circle, which is $12.6$ m.
• $\mathrm{\angle }OBD={52}^{\circ }$.
• $\mathrm{\angle }ODB$ can be calculated knowing that the angles in a triangle add up to ${180}^{\circ }$:

We know that: $$\mathrm{\angle }ODB={180}^{\circ }-\mathrm{\angle }OAB-\mathrm{\angle }OBD={180}^{\circ }-{90}^{\circ }-{52}^{\circ }={38}^{\circ }$$

Step 5: Calculating $CB$

Using the law of sines in triangle $OBD$:

$$\frac{CB}{\sin (\mathrm{\angle }OBD)}=\frac{OB}{\sin (\mathrm{\angle }ODB)}$$

Substituting the known values:

$$\frac{CB}{\sin ({52}^{\circ })}=\frac{12.6}{\sin ({38}^{\circ })}$$

Now, solving for $CB$:

$$CB=\frac{12.6\cdot \sin ({52}^{\circ })}{\sin ({38}^{\circ })}$$

Step 6: Plug in the sine values

Using a calculator gives:

$$\sin ({52}^{\circ })\approx 0.788$$ $$\sin ({38}^{\circ })\approx 0.615$$

Now substituting these values to find $CB$:

$$CB=\frac{12.6\cdot 0.788}{0.615}\approx \frac{9.91}{0.615}\approx 16.12\text{ meters}$$

Final Answer

CB is approximately $16.12$ meters.