Chapter 8.4, Problem 44AYU

(a)

To determine

To find: The area of the triangle OBC in terms of $\sin \theta$ and $\cos \theta$ .

Answer to Problem 44AYU

The area of the triangle is $\frac{\sin \theta }{2}$ .

Explanation of Solution

Given:

The given diagram is shown in Figure

  

Figure

The radius of the circle is 1 unit, the angle $\angle COB=\theta$ and the line segment DB is tangent to the circle at B $\angle OBD=90°$ .

Calculation:

Consider the triangle OBC with sides OB and OC with the angle COB. Then the area of the triangle is,

  $\begin{array}{c}A=\frac{1}{2}\left(OB\right)\left(OC\right)\sin \angle COB\\ =\frac{\sin \theta }{2}{\text{unit}}^{\text{2}}\end{array}$

(a)

To determine

To find: The area of the triangle OBD in terms of $\sin \theta$ and $\cos \theta$ .

Answer to Problem 44AYU

The area of the triangle is $\frac{\sin \theta }{2}$ .

Explanation of Solution

Given:

The radius of the circle is 1 unit, the angle $\angle COB=\theta$ and the line segment DB is tangent to the circle at B $\angle OBD=90°$ .

Calculation:

Consider the side DB in triangle OBD is,

  $\tan \theta =\frac{DB}{OB}$

Then,

  $\begin{array}{l}\tan \theta =\frac{DB}{1}\\ DB=\tan \theta \end{array}$

Then the area of the triangle is,

  $\begin{array}{c}A=\frac{1}{2}\left(DB\right)\left(OB\right)\\ =\frac{1}{2}\left(\tan \theta \right)\\ =\frac{\sin \theta }{2\cos \theta }\end{array}$

(c)

To determine

To Prove: The expression $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$ .

Answer to Problem 44AYU

The

Explanation of Solution

Given:

The given fact is $\text{area}\Delta OBC<\text{area}\Delta OBC<\text{area}\Delta OBD$ .

Calculation:

Consider the formula for the area of the sector is,

  $OBC=\frac{\theta }{2}$

Then,

  $\frac{\sin \theta }{2}<\frac{\theta }{2}<\frac{\sin \theta }{2\cos \theta }$

Divide both the side by $\sin \theta$ .

  $1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta }$

Hence, proved