Chapter 9.3, Problem 16WE

(a)

To determine

To Find: the $m\angle B,m\angle BOC,\text{ and }m\stackrel{⌢}{BC}$ in the figure shown such that $m\angle A=35$ and AC is the diameter.

Answer to Problem 16WE

  $35,70\text{ and 70}$

Explanation of Solution

Given:

Figure is shown below.

  

  $m\angle A=35$

Calculation:

In $\Delta AOB$ .

  $OA=OB\therefore \left(\text{Radius}\right)$

If two sides are equal in triangle, then angle opposite to these sides are also equal.

  $\begin{array}{l}m\angle B=m\angle A\\ m\angle B=35\end{array}$

And

  $\begin{array}{l}m\angle B+m\angle A+\angle AOB=180\\ 35+35+\angle AOB=180\\ \angle AOB=110\end{array}$

Since, AOC is a straight line therefore,

  $\begin{array}{l}\angle AOB+\angle BOC=180\\ 110+\angle BOC=180\\ \angle BOC=70\end{array}$

Measure of arc BC is the central angle.

  $\begin{array}{l}m\stackrel{⌢}{BC}=m\angle BOC\\ m\stackrel{⌢}{BC}=70\end{array}$

Therefore, the answer is $35,70\text{ and 70}$ .

(b)

To determine

To Find: the $m\stackrel{⌢}{BC}$ in the figure shown such that $m\angle A=n$ and AC is the diameter.

Answer to Problem 16WE

  $\left(180-2n\right),2n\text{ and }2n$

Explanation of Solution

Given:

Figure is shown below.

  

  $m\angle A=n$

Calculation:

In $\Delta AOB$ .

  $OA=OB\therefore \left(\text{Radius}\right)$

If two sides are equal in triangle, then angle opposite to these sides are also equal.

  $\begin{array}{l}m\angle B=m\angle A\\ m\angle B=n\end{array}$

And

  $\begin{array}{l}m\angle B+m\angle A+\angle AOB=180\\ n+n+\angle AOB=180\\ \angle AOB=180-2n\end{array}$

Since, AOC is a straight line therefore,

  $\begin{array}{l}\angle AOB+\angle BOC=180\\ \left(180-2n\right)+\angle BOC=180\\ \angle BOC=2n\end{array}$

Measure of arc BC is the central angle.

  $\begin{array}{l}m\stackrel{⌢}{BC}=m\angle BOC\\ m\stackrel{⌢}{BC}=2n\end{array}$

Therefore, the answer is $\left(180-2n\right),2n\text{ and }2n$ .

(c)

To determine

To Find: the $m\angle A$ in the figure shown such that $m\stackrel{⌢}{BC}=6k$ and AC is the diameter.

Answer to Problem 16WE

  $3k$

Explanation of Solution

Given:

Figure is shown below.

  

  $m\stackrel{⌢}{BC}=6k$

Calculation:

Measure of arc BC is the central angle.

  $\begin{array}{l}m\stackrel{⌢}{BC}=m\angle BOC\\ m\angle BOC=6k\end{array}$

Since, AOC is a straight line therefore,

  $\begin{array}{l}\angle AOB+\angle BOC=180\\ \angle AOB+6k=180\\ \angle AOB=180-6k\end{array}$

In $\Delta AOB$ .

  $OA=OB\therefore \left(\text{Radius}\right)$

If two sides are equal in triangle, then angle opposite to these sides are also equal.

  $m\angle B=m\angle A$

And

  $\begin{array}{l}m\angle B+m\angle A+\angle AOB=180\\ m\angle A+m\angle A+\left(180-6k\right)=180\\ 2\angle A=6k\\ \angle A=3k\end{array}$

Therefore, the answer is $3k$ .