Law of Reflection

Since the ray reflects $6$ times, it follows a straight line through a series of $7$ reflected congruent triangles, as shown below. Let $A$ be the base angle that the ray starts from, $B$ be the base angle that the ray ends on, $O$ be the vertex angle, $C$ be the the angle the ray ends on in the last reflected triangle, and $\angle AOB = x$ .

Since there are $7$ congruent angles at $O$ , $\angle AOC = 7x$ . Since $\triangle OAC$ is an isosceles triangle with a vertex angle of $7x$ , $\angle OAC = \frac{180° - 7x}{2}$ , and since we are given that $\angle BAC = 60°$ , $\angle BAO = \frac{180° - 7x}{2} + 60°$ .

However, we also know that since $\triangle AOB$ is an isosceles triangle with a vertex angle of $x$ , $\angle BAO = \frac{180° - x}{2}$ .

Therefore, we have $\angle BAO = \frac{180° - 7x}{2} + 60° = \frac{180° - x}{2}$ , and this solves to $x = 20°$ , which means $\angle BAO = 80°$ and $\angle OAC = 20°$ .

Using the law of sines on $\triangle AOB$ gives $\frac{AO}{\sin 80°} = \frac{1}{\sin 20°}$ , which means $AO = \frac{\sin 80°}{\sin 20°}$ , and using the law of sines on $\triangle AOC$ gives $\frac{AC}{\sin 140°} = \frac{\frac{\sin 80°}{\sin 20°}}{\sin 20°}$ , which means $AC = \frac{\sin 140° \sin 80°}{\sin 20° \sin 20°}$ .

Using the sine double angle formula, $AC = \frac{2 \sin 70° \cos 70° \sin 80°}{2 \sin 10° \cos 10° \sin 20°}$ , and using the product-to-sum formulas and the identity $\sin x = \sin (180° - x)$ , $AC = \frac{2 \sin 70° (\sin 30° + \sin 10°)}{2 \sin 10° (\sin 30° + \sin 10°)}$ , which simplifies to $AC = \frac{\sin 70°}{\sin 10°} \approx \boxed{5.411}$ .

Hint:

$\text{The exact answer is } \frac{ \sin 70^{\circ} }{ \sin 10^{\circ} } .$

Another way of angle calculation as 20°

If the acute angle of the triangle is theta, then, after each reflection from the slanted side of the triangle, the ray which began with 60° angle with base of the triangle, loses this angle by (theta).

Therefore, if after 3 reflections from the slanted sides of the triangle, the ray has become parallel to the base of the triangle, i.e. 0°, we have the relation,

3*(theta)=60°,

giving theta=20°.

Length calculations are given by the following sum 'S=5.411',

S= 1+2{cos (60°)+cos(40°)+cos(20°)}

In an octagon, the angles add up to $1080^\circ$ . Since we know that $\angle BAH = \angle AHG = 60^\circ$ , the other angles in the octagon $ABCDEFGH$ must equal $160^\circ$ . Thus the base angle of the original triangle is half of this, $80^\circ$ , and the vertical angle is $20^\circ$ . This further tells us that in triangle $AOH$ , the vertical and base angles are $140^\circ$ and $20^\circ$ respectively.

Using the sine rule on triangle $ABO$ , since $AB = 1$ , we deduce that $AO = \frac{\sin 80^\circ}{\sin 20^\circ}$ . We can use the sine rule again on $AOH$ , giving us that $AH = \frac{AO \sin 140^\circ}{\sin 20^\circ} \\ = \frac{\sin 80^\circ \sin 140^\circ}{\sin^2 20^\circ} \\ \approx \boxed{5.411}.$

I will do a step by step breakdown: First notice the angle of light relative to horizontal decrease by same amount after every reflection. 6 reflection = decrease 6 times. Notice that the light from its initial position to its final position, the angle changed by 120 degrees (vector diagram or just visualize) Thus each reflection the angle changes by 120/6= 20 degrees.

Next, know that when the mirror angle changes by x degrees, the reflection changes by 2x degrees. This can be proven or just google. Thus we know the bottom most line must have moved up 20/2=10 degrees relative to the horizontal. Then look at the bottom picture and you can figure out everything using trigonometry =)