$解：(2)因为 A_{1}A_{2}=A_{2}A_{3}，所以∠A_{2}A_{1}A_{3}=∠A_{2}A_{3}A_{1}.$
$又因为A_{1}A_{2}⊥A_{2}A_{3}，所以∠A_{1}A_{2}A_{3}=90°.$
$所以∠A_{2}A_{1}A_{3} +∠A_{2}A_{3}A_{1}= 90°，即∠A_{2}A_{1}A_{3}=45°.$
$同理得∠A=∠AA_{2}A_{1}.$
$又∠A_{2}A A_{3}=∠A+∠AA_{2}A_{1},$
$所以∠A =\frac{1}{2}∠A_{2}A_{1}A_{3}=22.5°.$
$又∠A=θ，所以θ=22.5°.$
$(4)由题意得\begin{cases}{50＜90°, }\\{6θ≥90°，}\end{cases}解得15°≤0＜18°\ $
$则θ 的取值范围为15°≤0＜18°.$
