Question

PQRS is square and angle abc =90° as shown in fig. If AP = BQ = CR, then prove that angleBAC = 45°

Answer 1

given square PQRS , prove that angle BAC= $45$°

Explanation:

1. given square PQRS hence we have, $PQ=QR=RS=SP=$$2x$.
2. in rectangle APQC , $AC=PQ=2x$ and $AP=QC$
3. Since it is given that $AP=BQ=CR$ we can say $QC=CR$ also $QR=QC+CR=2x$ and $PQ=PB+BQ=2x$        
4. hence from above points we get , $AP=PB=BQ=QC=x$  
5. in triangles APB and BQC we have , $AB=BC=\sqrt{x^2+x^2} =x\sqrt{2}$  
6. now in triangle ABC let angle BAC be θ then,                                                         $tan$θ$=\frac{BC}{AB} = \frac{x\sqrt{2} }{x\sqrt{2} }=1$                                                                                                     θ$=45$°          
7. hence proved that angle BAC is $45$°

                                                         

Answer 2

Answer:

in the figure pqrs is a square