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Calculus and Beyond Homework Help
Prove the Extended Law of Sines
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[QUOTE="lema21, post: 6504763, member: 691404"] [B]Homework Statement:[/B] Prove the extended Law of Sines. Hint: Let gamma be the circumscribed circle of triangle ABC and let D be the point on gamma such that DB is a diameter of gamma. Prove that Angle BAC is congruent to Angle BDC. Use that result to prove that sin(angle BAC) = BC/2R, where R is the circumradius. [B]Relevant Equations:[/B] Extended Law of sines: [(BC)/sin(angle BAC)] = [AC/sin(angle ABC)] = [AB/sin(angle ACB)] = 2r I just want to know if this proof is okay, and I would like advise on how to improve it. Proof: i. Proving that ∠BAC ⩭∠BDC: Let gamma be the circumscribed circle of ABC. Let D be the point on gamma such that DB is a diameter of gamma. The sum of the angles within a triangle equal to 180°. Adding the angles of BAC together, b+a+c =180° where b=∠B, a=∠A, and c=∠C. Adding the angles of BDC together, b+d+c = 180°, where b=∠B, d=∠D, and c=∠C. Since the sum of both BAC and BDC equals 180°, b+a+c=b+d+c → a=d. Hence, ∠BAC ⩭∠BDC. ii. Proving that sin(∠BAC)=BC/2R: sin(∠BAC) = P/H = BC/BD = BC/sin(∠BDC) = BD = 2R sin(∠ADB) = P/H = AB/BD = AB/sin(∠ADB) = BD = 2R AC/sin(∠ABC)= 2R (BC/sin(∠BAC)) = (AC/sin(∠ABC) = (AB/sin(∠ACB)) = 2R [/QUOTE]
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Prove the Extended Law of Sines
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