Question related to regular pentagon My question is-
ABCDE is e regular pentagon.If AB = 10 then find AC.
Any solution for this question would be greatly appreciated.
Thank you,
Hey all thanks for the solutions using trigonometry....can we also get the solution without using trigonometry? –
 A: Angle ABC = $108^o$, and triangle ABC is isosceles. By the sine rule, $$\frac{\sin(108)}{AC} = \frac{\sin(36)}{10}$$
so $$AC = 10\times\frac{\sin(108)}{\sin(36)} = 5(\sqrt5+1).$$
A: This a classical occurrence of the golden ratio $\phi=\frac{1+\sqrt5}2$: the length of the chord $AC$ is $\phi$ times the length of the side $AB$. The following illustration tries to suggest that if you subtract the length of $AB$ from that of $AC$, the remainder has the same ratio to $BC$ as $AB$ had to $AC$, and that you can therefore continue like this subtracting the shorter from the longer length indefinitely. $A$ is the corner on the left, $B$ at the top, $C$ at the right.

A: Here's a solution without trig:
[Edit: here is a diagram which should make this more intuitive hopefully.]

First, $\angle ABC=108^{\circ}$; I won't prove this here, but you can do it rather trivially by dividing the pentagon into 3 triangles, e.g. $\triangle{ABC}$, $\triangle{ACD}$, and $\triangle{ADE}$, and summing and rearranging their angles to find the interior angle.
Draw segment $\overline{AC}$. Then $\triangle ABC$ is isosceles, and has 180 degrees, therefore $\angle ACB=\angle BAC=(180^{\circ}-108^{\circ})/2=36^{\circ}$.
Draw segment $\overline{BE}$. By symmetry, $\triangle ABC \cong \triangle ABE$, and so therefore $\angle ABE=36^{\circ}$.
Let $F$ be the point where segment $\overline{AC}$ intersects segment $\overline{BE}$. Then $\angle AFB = 180^{\circ}-\angle BAC - \angle ABE = 180^{\circ}-36^{\circ}-36^{\circ} = 108^{\circ}$.
This means that $\triangle ABC \sim \triangle ABF$, as both triangles have angles of $36^{\circ}$, $36^{\circ}$, and $108^{\circ}$.
Next, we can say $\angle CBE=\angle ABC-\angle ABE = 108^{\circ} - 36^{\circ} = 72^{\circ}$, and from there, $\angle BFC = \angle AFC - \angle AFB = 180^{\circ} - 108^{\circ} = 72^{\circ}$. Therefore $\triangle BCF$ is isoceles, as it has angles of $36^{\circ}$, $72^{\circ}$, and $72^{\circ}$. This means that $\overline{CF}=\overline{BC}=\overline{AB}=10$ (adding in the information given to us in the problem).
The ratios of like sides of similar triangles are equal. Therefore,
$$\frac{\overline{AC}}{\overline{AB}}=\frac{\overline{AB}}{\overline{AF}}$$
We know that $\overline{AC}=\overline{AF}+\overline{CF}=\overline{AF}+10$. Let's define $x:=\overline{AF}$. Substituting everything we know into the previous equation,
$$\frac{x+10}{10}=\frac{10}{x}$$
Cross-multiply and solve for $x$ by completing the square. (Or, if you prefer, you can use the quadratic formula instead.)
$$x(x+10)=100$$
$$x^2+10x=100$$
$$x^2+10x+25=125$$
$$(x+5)^2=125$$
$$x+5=\pm 5\sqrt 5$$
$$x=-5\pm 5\sqrt 5$$
Choose the positive square root, as $x:=\overline{AF}$ can't be negative.
$$\overline{AF}=-5 + 5\sqrt 5$$
Finally, recall that earlier we proved $\overline{AC}=\overline{AF}+10$. Plug in to get the final answer:
$$\boxed{ \overline{AC} = 5 + 5\sqrt 5 = 5(1 + \sqrt 5)}$$
Hope this helps! :)
