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?

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

$\text{We are given: }\:AC = 5.$

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

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

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

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

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

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

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

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