Question

Solve this ?

Answer 1

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

Explanation:

Solution 1 :

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 :

`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^@`

Answer 2

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^@`