Proof: Sine 3

Let's prove the following theorem:

if ∠ABC is a right angle, then sine of (m∠CAB) = (distance BC) / (distance AC)

Proof:

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Given
1 ∠ABC is a right angle
Proof Table
# Claim Reason
1 sine of (m∠CAB) = (distance CB) / (distance CA) if ∠ABC is a right angle, then sine of (m∠CAB) = (distance CB) / (distance CA)
2 (distance CB) / (distance CA) = (distance BC) / (distance AC) (distance CB) / (distance CA) = (distance BC) / (distance AC)
3 sine of (m∠CAB) = (distance BC) / (distance AC) if sine of (m∠CAB) = (distance CB) / (distance CA) and (distance CB) / (distance CA) = (distance BC) / (distance AC), then sine of (m∠CAB) = (distance BC) / (distance AC)

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