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Oct 17, 2017

Answer:

Option #(B) " " : " "45^@#

Explanation:

Solution 1 :
enter image source here
Let #AP=AQ=a, => PB=QD=1-a, and PQ=asqrt2#,
given #AP+AQ+PQ=2#,
#=> a+a+asqrt2=2#,
#=> a=2-sqrt2#
#=> PB=QD=1-a=sqrt2-1#
#=> y=tan^-1(sqrt2-1)=22.5^@#
#=> anglePCQ=x=90-2y=90-2*22.5=45^@#

Solution 2 :
enter image source here
#AP+AQ+PQ=2#,
As #AQ+QD=1, and AP+PB=1#,
#=> AP+AQ+QD+PB=2#,
#=> color(red)(PQ=PB+DQ)#
Now, extend #AB# to #E# such that #BE=DQ#,
#=> DeltaCBE and DeltaCDQ# are congruent,
#=> CE=CQ#
As #angleDCB=90^@, => color(red)(angleQCE=90^@#
#color(red)(PE)=PB+BE=color(red)(PB+DQ=PQ)#,
Now, as #CQ=CE, PQ=PE, and CP# is the common side,
#=> DeltaCPQ and DeltaCPE# are congruent,
#=> anglePCQ=anglePCE=(angleQCE)/2=90/2=45^@#

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Oct 19, 2017

drawn

Let #/_QCD=alpha#

#/_PCB=beta#

So #/_PCQ=90^@-(alpha+beta)#

Now for #Delta QCD,/_QDC=90^@ and DC = 1unit#
So #DQ=DCtanalpha=tanalpha#

And for #Delta PCB,/_PBC=90^@ and BC = 1unit#
So #PB=BCtanbeta=tanbeta#

Hence for #DeltaAPQ#

#AQ=1-tanalpha#

#AP=1-tanbeta#

and

#PQ=sqrt((1-tanalpha)^2+(1-tanbeta)^2)#

Given

#AQ+AP+PQ=2#

#=>1-tanalpha+1-tanbeta+sqrt((1-tanalpha)^2+(1-tanbeta)^2)=2#

#=>sqrt((1-tanalpha)^2+(1-tanbeta)^2)=tanalpha +tanbeta#

#=>1-2tanalpha+tan^2alpha+1-2tanbeta+tan^2beta=tan^2alpha+tan^2beta+2tanalphatanbeta#

#=>1-tanalpha-tanbeta=tanalphatanbeta#

#=>tanalpha+tanbeta=1-tanalphatanbeta#

#=>(tanalpha+tanbeta)/(1-tanalphatanbeta)=1=tan45^@#

#=>tan(alpha+beta)=tan45^@#

#=>alpha +beta=45^@#

Hence #/_PCQ=90^@-(alpha+beta)=90^@-45^@=45^@#

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