# Trigonometric problem

• Dec 4th 2012, 06:44 AM
Jaom
Trigonometric problem
Hey I need help with a problem regarding a septangular.

In a regular septangular there are two diagonals with 2 different lengths. How long is the longer one if the shorter one is 5.0 cm?

The answer is 6.2 cm. To think about is that it can be divided into 5 triangles and the sum of the angles is 900o

I think I have to use the law of sines and cosines. Im really stuck, I appreciate the help :)
• Dec 4th 2012, 09:44 AM
Soroban
Re: Trigonometric problem
Hello, Jaom!

Did you make a sketch?

Quote:

In a regular heptagon, there are two diagonals with different lengths.
How long is the longer one if the shorter one is 5.0 cm?

Consider the upper "half" of the heptagon.
Code:

          B            C             o  *  *  *  o           * 5θ    θ * 4θ*           *      *      *         *    *          *         * θ *  5          *       * * θ              2θ *       o  *  *  *  *  *  *  *  o     A            x            D
$\text{We are given: }\:AC = 5.$

$\text{Let }\,x \:=\:AD.$

$\text{We have: }\:\angle ABC = \tfrac{5\pi}{7}$

$\text{In the diagram, let }\theta \,=\, \tfrac{\pi}{7}$

$\text{Since }\Delta ABC\text{ is isosceles, }\angle BAC = \angle BCA = \tfrac{\pi}{7}$

$\text{Hence: }\:\angle ACD =\tfrac{4\pi}{7}$
. $\text{And: }\:\angle BAD = \angle CDA = \tfrac{2\pi}{7}$

$\text{In }\Delta ACD,\text{ Law of Sines: }\:\frac{x}{\sin\frac{4\pi}{7}} \:=\:\frac{5}{\sin\frac{2\pi}{7}}$

. . . . . . . . $x \;=\;\frac{5\sin\frac{4\pi}{7}}{\sin\frac{2\pi}{7} } \;=\; 6.234898019$

$\text{Therefore: }\:x \;\approx\;6.2\text{ cm}$