Question #c2a16

1 Answer
P dilip_k
May 5, 2017

drawn

#tana=tan/_CPA=(AC)/(AP)=(1/2AB)/(nAB)=1/(2n)#

#tan(a+b)=tan/_BPA#

#=>(tana+tanb)/(1-tanatanb)=(AB)/(AP)=1/n#

#=>(ntana+ntanb=(1-tanatanb)#

#=>(nxx1/(2n)+ntanb)=(1-1/(2n)tanb)#

#=>1/2+ntanb=1-1/(2n)tanb#

#=>ntanb+1/(2n)tanb=1-1/2#

#=>(n+1/(2n))tanb=1/2#

#=>((2n^2+1)/(2n))tanb=1/2#

#=>tanb=n/(2n^2+1)#

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