If $Δ A_{1} A_{2} A_{3}$ is equilateral and $A_{n + 3}$ is the mid - point of line segment $A_{n} A_{n + 1}$ for all positive integers $n$

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The angle $∠ A_{44} A_{45} A_{46}$ is equal to $120^{\circ}$

The angle $∠ A_{4} A_{6} A_{5}$ is equal to $90^{\circ}$

The angle $∠ A_{57} A_{58} A_{59}$ is equal to $60^{\circ}$

The angle $∠ A_{98} A_{99} A_{100}$ is equal to $30^{\circ}$

answer is A, B, C, D.

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Detailed Solution

${ΔA}_{2} A_{3} A_{4}$ has vertex angles $60^{\circ}, 30^{\circ}, 90^{\circ}$ respectively

now $A_{2} A_{4} = A_{2} A_{5}$

$\Rightarrow {ΔA}_{2} A_{4} A_{5}$ is equilateral

now ${ΔA}_{3} A_{4} A_{5}$ has vertex angle $30^{\circ}, 30^{\circ}, 120^{\circ}$

$\Rightarrow {ΔA}_{4} A_{5} A_{6}$ has vertex angle $30^{\circ}, 60^{\circ}, 90^{\circ}$

Now $A_{5} A_{6} = A_{5} A_{7} \Rightarrow Δ A_{5} A_{6} A_{7}$ is again equilateral is

$Δ A_{1} A_{2} A_{3} ~ Δ A_{5} A_{6} A_{7}$

Similarly we have ingeneral

$\begin{array}{l} △ A_{n} A_{n + 1} A_{n + 2} ~ A_{n + 4} A_{n + 5} A_{n + 6} \\ \Rightarrow ∠ A_{44} A_{45} A_{46} = ∠ A_{4} A_{5} A_{6} = 60^{\circ} \\ ∠ A_{98} A_{99} A_{100} = ∠ A_{2} A_{3} A_{4} = 30^{\circ} \\ ∠ A_{57} A_{58} A_{59} = ∠ A_{5} A_{6} A_{7} = 60^{\circ} \end{array}$