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Answer 1 · 2017-11-19T11:45:23

#sin(A/2)=sqrt((65-sqrt(65))/130)#

Here, #secA=sqrt(65)#

or, #cosA=1/sqrt(65)#

We have, #sin(A/2)=+-sqrt((1-cosA)/2#

Taking +sign only as #A# is acute angle.

Then #sin(A/2)=sqrt((1-1/sqrt(65))/2#

#sin(A/2)=sqrt(((sqrt(65)-1)/sqrt(65))/2)#

#=sqrt((sqrt(65)-1)/(2*sqrt(65))#

#=sqrt((sqrt(65)-1)/(2*sqrt(65))*(sqrt65)/sqrt(65))#

#=sqrt((65-sqrt(65))/130)#