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:
Given
1 | ∠ABC is a right angle |
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# | 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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