# Trig tent question

• January 22nd 2013, 01:17 PM
goldbug78
Trig tent question
Okay, so this problem goes like this: A two-person tent is to be made so that the height at the center is a = 4 feet (see the figure below). If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 5 feet in length, how many square feet of material will be needed to make the tent? (Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.) I am not sure how to set this one up. It looks like an equilateral triangle.

Attachment 26664
• January 22nd 2013, 04:41 PM
Soroban
Re: Trig tent question
Hello, goldbug78!

Quote:

A two-person tent is to be made so that the height at the center is a = 4 feet.
If the sides of the tent are to meet the ground at an angle 60°, and the tent is to be b = 5 feet in length,
how many square feet of material will be needed to make the tent?
(Assume that the tent has a floor and is closed at both ends, and give your answer in exact form.)

Code:

                        *                     *    \                 *        \             *              \           /|\              \           / | \              \       x /  |4 \ x            *         /  |  \        *       /    |    \    * 5       *-----*-----*       : x/2 : x/2 :
The front is an equilateral triangle with altitude 4 feet.

Pythagorus: . $\left(\tfrac{x}{2}\right)^2 + 4^2 \:=\:x^2 \quad\Rightarrow\quad \tfrac{x^2}{4} + 16 \:=\:x^2$

. . . . . . . . . . $\tfrac{3}{4}x^2 \:=\:16 \quad\Rightarrow\quad x^2 \:=\:\tfrac{64}{3} \quad\Rightarrow\quad x \:=\:\tfrac{8\sqrt{3}}{3}$

The front triangle has area: . $\tfrac{1}{2}\left(\tfrac{8\sqrt{3}}{3}\right)(4) \:=\:\tfrac{16\sqrt{3}}{3}$
. . The two triangular panels have area: . $2 \times \tfrac{16\sqrt{3}}{3} \:=\:\tfrac{32\sqrt{3}}{3}$

A rectangular panel has area: . $\left(\tfrac{8\sqrt{3}}{3}\right)(5) \:=\:\tfrac{40\sqrt{3}}{3}$
. . The three rectangular panels have area: . $3 \times\tfrac{40\sqrt{3}}{3} \:=\: 40\sqrt{3}$

The total area is: . $\frac{32\sqrt{3}}{3} + 40\sqrt{3} \;=\;\frac{152\sqrt{3}}{3}\text{ ft}^2$