# Thread: Trigonometric problem

1. ## 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

2. ## Re: Trigonometric problem

Hello, Jaom!

Did you make a sketch?

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?

The answer is 6.2 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
$\displaystyle \text{We are given: }\:AC = 5.$

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

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

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

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

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

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

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

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