Question

Solve this ?

Answer 1

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^@`