Proof: Sine 2
Let's prove the following theorem:
if ∠ABC is a right angle, then sine of (m∠CAB) = (distance CB) / (distance CA)
    
    
    
    Proof:
  
      
      Given
      
    
    
      
  
  
| 1 | ∠ABC is a right angle | 
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| # | Claim | Reason | 
|---|---|---|
| 1 | sine of (m∠BAC) = (distance CB) / (distance CA) | if ∠ABC is a right angle, then sine of (m∠BAC) = (distance CB) / (distance CA) | 
| 2 | m∠BAC = m∠CAB | m∠BAC = m∠CAB | 
| 3 | sine of (m∠BAC) = sine of (m∠CAB) | if m∠BAC = m∠CAB, then sine of (m∠BAC) = sine of (m∠CAB) | 
| 4 | sine of (m∠CAB) = (distance CB) / (distance CA) | if sine of (m∠BAC) = sine of (m∠CAB) and sine of (m∠BAC) = (distance CB) / (distance CA), then sine of (m∠CAB) = (distance CB) / (distance CA) | 
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