✨ Best Answer ✨
t^2a/2=[(1+c)/2]/[(1-c)/2]
=(1/10)/(9/10)
=1/9
π<a<3/2πより、0<ta/2
よって、ta/2=1/3
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✨ Best Answer ✨
t^2a/2=[(1+c)/2]/[(1-c)/2]
=(1/10)/(9/10)
=1/9
π<a<3/2πより、0<ta/2
よって、ta/2=1/3
Users viewing this question
are also looking at these questions 😉
