Question

If B and Q are acute angles such that sinB = sinQ, then prove that B = Q.

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

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