Chapter 5, Problem 7PS

(a)

To determine

To Find: The area of the triangle as the function of $\frac{\theta }{2}$ .

Answer to Problem 7PS

The area of the triangle is $100\cos \frac{\theta }{2}\sin \frac{\theta }{2}$ .

Explanation of Solution

Given:

The length of each of the two equal sides of the isosceles triangle is 10 meters and the angel between the two slides is $\theta$ .

Calculation:

Consider the half of the triangle with the base and the height is shown in Figure 1

  

Figure 1

Then, the area of the triangle is calculated as,

  $A=\frac{bh}{2}$

By the sine function,

  $\begin{array}{l}\sin \frac{\theta }{2}=\frac{b/2}{10}\\ b=20\sin \frac{\theta }{2}\end{array}$

Then, by the use of the cosine function,

  $\begin{array}{l}\cos \frac{\theta }{2}=\frac{h}{10}\\ h=\frac{1}{2}\cos \frac{\theta }{2}\end{array}$

Then,

  $\begin{array}{c}A=\frac{1}{2}\left(\frac{1}{2}\cos \frac{\theta }{2}\right)\left(20\sin \frac{\theta }{2}\right)\\ =100\cos \frac{\theta }{2}\sin \frac{\theta }{2}\end{array}$

(b)

To determine

To Find: The area of the triangle as the function $\theta$ and determine whether the value of $\theta$ such that the area is at the maximum.

Answer to Problem 7PS

The area of the triangle is $50\sin \theta$ and the value at which the area is maximum is $\theta =\frac{\pi }{2}$ .

Explanation of Solution

Consider the identity,

  $\sin 2u=2\sin u\cos u$

Then, by the formula of area is,

  $\begin{array}{c}A=100\cos \frac{\theta }{2}\sin \frac{\theta }{2}\\ =50\left(2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right)\\ =50\sin \theta \end{array}$

To maximize the area, the $\sin \theta$ must have the value that is 1 and that occurs when,

  $\theta =\frac{\pi }{2}$