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