Chapter 9.4, Problem 18WE

Solution

To find: $HG$

  $HG=32$

Given:

The figure:

  $OE=8\sqrt{3}$

Concept Used:

Theorem 9-5: A diameter that is perpendicular to a chord bisects the chord and its arc.

  $\cos A=\frac{\text{leg adjacent to}\angle A}{\text{hypotenuse}}$

Explanation:

In the given figure join O and F .

It is given that the measure of the intercepted arc $HF$ is $150°$ .

So, the measure of $\angle HOF=150°$

Now, $\overline{HG}$ is the diameter of the circle, so

  $m\angle HOG=180$

Thus,

  $\begin{array}{c}m\angle HOG=180\\ m\angle HOF+m\angle FOG=180\\ 150+m\angle FOG=180\end{array}$

  $\begin{array}{c}m\angle FOG=180-150\\ m\angle FOG=30\end{array}$

Thus, in the triangle $△FOE$

  $m\angle FOE=30$

Consider the right triangle $△FOE$ .

  $\begin{array}{c}\cos \left(FOE\right)=\frac{\text{leg adjacent to}\angle FOE}{\text{hypotenuse}}\\ \cos 30°=\frac{OE}{OF}\\ \frac{\sqrt{3}}{2}=\frac{8\sqrt{3}}{OF}\end{array}$

Now perform cross multiplication,

  $\begin{array}{c}OF=\frac{2\times 8\sqrt{3}}{\sqrt{3}}\\ OF=16\end{array}$

So, the radius of the given circle is 16.

Then, the diameter of this circle $\overline{HG}$ is given by:

  $\begin{array}{c}HG=2\times \text{radius}\\ =2\times 16\\ =32\end{array}$