Answers

2015-05-07T20:57:01+05:30
Given sinB = sinQ 
we have sin B = AC/AB and sin Q = PR/PQ
Then AC/AB = PR/PQ
Therefore AC/AB = PR/PQ =k
By using Pythagoras theorem 
  (BC^2 =AB^2 - AC^2) / (QR^2 = PQ^2 - PR^2 )    
 (SINCE AC^2=k^2 AB^2,PR^2 = k^2 PQ^2) =AB/PQ 
HENCE, AB/PQ = AC/PR = BC/QR
therefore , B=Q
0