Chapter 5.5, Problem 122E

(a)

To determine

To calculate: The area of the triangle as a function of $\theta /2$ .

Answer to Problem 122E

The area of the triangle as a function of $\theta /2$ is $\text{100sin}\theta /2\cos \theta /2$ .

Explanation of Solution

Given information:

The following diagram:

  

Formula used:

The following Pythagorean relation:

  $\begin{array}{l}\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}\\ \sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}\end{array}$

The following formula for area,

  $\text{Area=}\frac{1}{2}\times \text{Base×Height}$

Calculation:

Assume base to be $b$ and height to be $h$ . The diagram for $\theta /2$ will be as follows:

  

From the figure,

  $\begin{array}{l}\sin \theta /2=\frac{b/2}{10}\text{ [From Pythagoras relation]}\\ b\text{/2=10sin}\theta /2\text{ [Multiply both sides by 10]}\\ b=20\sin \theta /2\text{ [Multiply both sides by 2]}\\ \cos \theta /2=\frac{h}{10}\text{ [From Pythagoras relation]}\\ h=10\cos \theta /2\text{ [Multiply both sides by 10]}\end{array}$

Therefore,

  $\begin{array}{l}\text{Area=}\frac{1}{2}\left(20\sin \theta /2\right)(10\cos \theta /2)\text{ [Formula for area]}\\ \text{ =100sin}\theta /2\cos \theta /2\text{ [Simplify]}\end{array}$

(b)

To determine

To calculate: The area of the triangle as a function of $\theta$ and value of $\theta$ such that area is maximum.

Answer to Problem 122E

The area of the triangle as a function of $\theta$ is $\text{50sin}\theta$ .

Area is maximum when $\theta \text{ = 90}°$ .

Explanation of Solution

Given information:

The following diagram:

  

Formula used:

The following Double angle formula:

  $\sin 2u=2\sin u\cos u$

Calculation:

From subpart (a), $\text{Area=100sin}\theta /2\cos \theta /2$

Area can be written as a function of $\theta$ as follows:

  $\begin{array}{l}\text{ Area=100sin}\theta /2\cos \theta /2\text{ [Calculated]}\\ =50(2\text{sin}\theta /2\cos \theta /2)\text{ [Factor]}\\ =50\sin \theta \text{ [Double angle formula]}\end{array}$

Area will be maximum when $\sin \theta =1$ . Therefore, $\theta \text{ = 90}°$ (using inverse function)