Solve this ?

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2 Answers
Oct 17, 2017

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^@

Oct 19, 2017

drawndrawn

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^@